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1 E X C E L L E N C I A J U N I O R C O L L E G E SSHAMIRPET | MADHAPUR | SUCHITRA | EC IL

KINEMATICS

1. Mechanics : The branch of physics which deals with the motion of objects in everydaylife is called mechanics.

2. It is divided into , Kinematics, Dynamics and Statics.

i.Kinematics :

Kinematics which is derived from a Greek word kinema meaning motion, is a branchof physics, which deals with the motion of a body without taking into account thecause of motion.

ii.Dynamics :

Dynamics , which is derived from the Greek word dyna meaning power, is a branch ofphysics which deals with the motion of bodies by taking into account the cause ofmotion (force).

iii. Statics :

Statics deals with bodies at rest under the effect of different forces.

3. Point object :

An object is said to be a point object if its size is very small as compared to the distancetravelled by it in the given time interval.

Ex :

i) A bus travelling a distance of 100 km can be considered as a point object. This isbecause the size of the bus is very small as compared to distance travelled by it.

ii) A car travelling a distance of 50 km can be considered as a point object. This isbecause the size of car is very small as compared to the distance travelled by it.

4. Reference point :

A fixed point or a fixed object with respect to which the given body changes its positionis known as reference point.

5. Rest :

A body is said to be at rest if it does not change its position with respect to thereference point. The objects which remain stationary at a place and do not changetheir position are said to be at rest.

Ex : A chair lying in a room is in the state of rest, because it doesnot change itsposition with respect to the surroundings of the room.

KINEMATICSSYNOPSIS-1


KINEMATICS

7. Motion :

A body is said to be in motion if it changes its position with respect to the surroundingswith the passage of time. All moving things are said to be in motion.

Ex : A car is changing its position w.r.t trees, houses etc. is in the state of motion.

8. Rest and motion are relative terms :

Rest and motion are relative terms. A body can be at rest as well as in motion at thesame time.

When we say that a body or an object is in motion, then it is essential to see whetherthe body or object changes its position with respect to other bodies or objects around itor with respect to any fixed point known as reference point. For example, when a busmoves on a road, then the bus as well as the passengers sitting in it change theirposition with respect to a person standing on the road side. So, the bus and thepassengers sitting in it are in motion with respect to the person standing on the roadside. However, the passengers sitting in the bus do not change their positions withrespect to each other. It means, the passengers sitting in a moving bus are not inmotion with respect to each other.

Ex : A person sitting in the compartment of a moving train is in the state of rest, withrespect to the surroundings of compartment. Yet he is in the state of motion, if hecompares himself with surroundings outside the compartment.


KINEMATICSNote : In order to describe the motion of an object we need to keep in mind threethings.

1. The distance of the body from a reference point. This reference point is called theorigin of the motion of the body.

2. The direction of motion of the body. 3. The time of motion.

Terms related to Kinematics :

Distance : The distance travelled by a body is actual length of the path covered by amoving body irrespective of the direction in which the body travels.

4 km

3 km

BA

C

N

S

EW

Fig. (ii)

(i) Suppose a man lives at place A as in figure (ii), and he has to reach another place

C, but first he has to meet his friend living at place B.

(ii) Now, the man starts from point A and travels a distance of 4 km to reach B

towards East, and then travels another 3 km from B to reach C toward North.

(iii) Thus, the total length of the path i.e. AB + BC = 4 km + 3 km = 7 km gives the

distance travelled by the man and it has no specific direction.

Displacement : When a body moves from one position to another, the shortest distance

between the initial position and final position.

5 km

4 km

3 km

BA

C

N

S

EW

Fig. (ii)

(i) In the above example, the shortest distance between the initial position A and

final position C of the man is 5 km, [according to pythagorous principle, i.e.,

(hypotenous)2 = (adjacent side)2 + (opposite side)2 in the North – East direction.

Note : 1. The displacement of an object in a given interval of time can be positive,zero or negative whereas distance travelled is always positive.

2. Distance is a scalar quantity whereas displacement is a vector quantity.


KINEMATICS

KINEMATICSWORKSHEET 1

1) Displacement is a ________________ [ ]

1) Vector quantity 2) Scalar quantity

3) Either vector quantity or scalar quantity

4) Neither vector quantity nor scalar quantity

*2) A player hits a baseball at some angle. The ball goes high up in space. The player runsand catches the ball before it hits the ground. Which of the two has greaterdisplacement ?

1) the player 2) the ball

3) data in sufficient 4) both have same displacement

3) A ball is thrown up with a certain velocity. It attains a height of 40m and comes backto the thrower. Then the [ ]

1) total distance covered is 40m 2) magnitude of displacement is 80m

3) displacement is zero 4) total distance covered is zero

4) Assertion : The numerical ratio of displacement to distance is equal to one or less than one.

Reason : Displacement is a vector quantity and distance is a scalar quantity.[ ]

1) assertion and reason are true and the reason is correct explanation of the assertion.

2) assertion and reason are true, but reason is not correct explanation of the assertion.

3) assertion is true, but the reason is false.

4) assertion is false, but the reason is true.

*5) Statement - I : Displacement of a body may be zero even when distance travelled byit is not zero.

Statement - II : Displacement of a body cannot be zero even when distance travelledby it is not zero. [ ]

1) Statement I and II are true 2) Statement I is true and II are false

3) Statement I is false and II are true 4) Statement I and II are false

6) Pick out the false statement from the following

1) displacement is a vector quantity and hence direction is important

2) displacement can be both positive and negative

3) distance is always positive. It never decreases with time

4) distance can be negative


KINEMATICS7) When a moving particle returns to its initial point

1) displacement is not zero 2) distance is zero

3) displacement is zero 4) both distance and displacement will be zero

8) Which of the following is not a characteristic of displacement ?

1) it is always positive

2) the magnitude is equal to the shortest distance between the initial and final positionsof the particle

3) it can be represented geometrically

4) it has both magnitude and direction

KINEMATICSSYNOPSIS 2

1. Recap of Concept of displacementConsider the one dimensional motion of a particle along OX axis as shown in figure.

P1 P2

XO

x1

x2

Let the time be measured from the instant the particle is at the origin O. Supposethe particle, the particle is at point P1 with coordinate x1 at time t1 and at time t2 it isat P2 where its coordinate is x2. The change in the position of the particle (i.e., x2 –x1) is the magnitude of displacement of the particle.

Magnitude of displacement, 2 1x x x

The direction of displacement is from initial point from final point.

2. Displacement (Vector notation) : We know distance is the actual path length covered

by a moving particle or body in a given time interval, while displacement is the change

in position vector, i.e., a vector joining initial to final position. If a particle moves from

A to C through a path ABC.

Then distance s travelled is the actual path length ABC, while the displacement is


KINEMATICS

C Ar r r

If a particle moves in a straight line without change in direction, the magnitude ofdisplacement is equal to the distance travelled otherwise it is always less than it.Thus

displacement dis tance


*1. A cyclist moves from a certain point X and completes one revolution around a circularpath of radius ‘r’. The distance travelled and magnitude of displacement of the cyclistare respectively [ ]

1) 2r , r 2) 2

, 2 3) 2 r, zero 4) r, 2r

2. In the above problem of the cyclist reaches exactly the other side of the point X (i.e.diametrically opposite point to X). Then the distance travelled and magnitude ofdisplacement of cyclist are respectively [ ]

1) r , 2r 2) r

2

, 2 3) 2 r, zero 4) r, 2r

*3. Considering the above problem if the cyclist reaches a point ‘Z’ as shown in thefigure,

O rX

Z

then the distance travelled and magnitude of displacement of the cyclist arerespectively

1) r

2

, 2r 2) r

4

, 4r 3) r

2

, 2 r 4) 2 r, zero

4) A player completes a circular path of radius ‘r’ in 40s. At the end of 2 minutes 20seconds, displacement will be [ ]1) 2r 2) 2 r 3) 7 r 4) Zero

5) A man walks 8m towards East and then 6m towards north. His displacement is_________________ [ ]1) 10 m N – E 2) 14 m S – E 3) 2 m S – W 4) Zero

6) A man has to go 50m due north, 40m due east and 20m due south to reach a field.

a) What distance he has to walk to reach the field ?b) What is his displacement from his house to the field ?


KINEMATICS1) 110m, 20 m NE 2) 100m, 50 m NE 3) 110m, 50 m NE 4) 100m, 100 m NE

7) A particle starts from the origin, goes along the X-axis to the point (20m, 0) and thenreturns along the same line to the point (–20m, 0). Find the distance anddisplacement of the particle during the trip.

1) 80m, 20m in the negative direction 2) –20m, 50m in the negative direction

3) 60m, 20m in the negative direction 4) –20m, 80m in the negative direction

8) A spy report about a suspected car reads as follows. “The car moved 2.00 kmtowards east, made a perpendicular left turn, ran for 500m, made a perpendicularright turn, ran for 4.00 km and stopped”. Find the displacement of the car.

1) 5 km, tan–1 (1) 2) 6.02 km, tan–1(1/12) 3) 5 km, tan–1(1/6) 4) 10 km, tan–1(1/4)

*9) A carrom board (4ft × 4ft square) has the queen at the centre. The queen, hit by thestriker moves to the front edge, rebounds and goes in the hole being the striking line.The magnitude of displacement of the queen from the centre to the front edge is

1) 2 10 ft3

2) 4 10 ft3

3) 2 2 ft 4) 5 10 ft3

10) In the above problem (i) find the magnitude of the displacement of the queen fromthe front edge to the hole and (ii) from the centre to the hole

1) (i) 2 10 ft3

(ii) 2 ft 2) (i)4 10 ft3

(ii) 2 2 ft

3) (i) 2 2 ft (ii) 2 10 ft3

4) (i) 5 10 ft3

(ii) 2 2 ft

*11) A mosquito net over a 7 ft × 4 ft bed is 3 ft high. The net has a hole at one corner of thebed through which a mosquito enters the net. It flies and sits at the diagonally oppositeupper corner of the net. The magnitude of the displacement of the mosquito is

1) 7 ft 2) 7 ft 3) 3 ft 4) 74 ft

12) In the above problem find the magnitude of the displacement of the mosquito. Takingthe hole as the origin, the length of the bed as the X–axis, its width as the Y–axis,and vertically up as the Z–axis, write the components of the displacement vector.1) 4 ft, 3 ft, 2 ft 2) 5 ft, 3 ft, 2 ft 3) 7 ft, 4 ft, 3 ft 4) 7 ft, 5 ft, 3 ft

*13) A body covers an arc of a circle of radius ‘r’, subtending an angle of 120° at the centreof the circle. The magnitude of the displacement of the body is

1) r 3 2) r 2 3) 2r 4) 3r

[Hint : Any line passing through the centre of the circle bisects the chord of the circle]

*14) A man travels 30m due north, 20m due east and 30 2 m due south west. Thedisplacement of man is

1) 10m west 2) 10m west 3) 20m west 4) 20 2 m west


KINEMATICS

trip. KINEMATICSSYNOPSIS 3

1. Introduction to Speed :Imagine two boys Ram and Shyam running a 200m race.Let us imagine Ram finishesthe race in 20 seconds and Shyam finishes it in 25 seconds.

Who ran faster?Obviously Ram.

It is because the rate at which Ram was running is 200m 20s = 10m/s, whereas therate at which Shyam was running is 200m25s = 8m/s.

It must be clear that both the boys have ran the same distance, but one of them isfaster than the other. Simply because he covers more distance in unit time.

The distance covered by a body in unit time is called Speed.

2. Mathematical Expression of Speed : The speed can be found by dividing the distance

covered by the time in which the distance is covered. i.e., distance Speed =

time

3. Units of Speed: We know the C.G.S. unit of distance is centimetre(cm) and time issecond(s). Hence the C.G.S unit of speed is centimetre per second (cm/s). Simi-larly the S.I. unit of distance is metre(m) and time is second(s). Hence the S.I. unit ofspeed is metre per second (m/s).

Bigger unit of speed is km/hr. The relation between km/hr and m/s is

1km 1000m 51kmph= = = m/s1hr 3600s 18

51km/hr = m/s

184 Types of Speedi. Uniform Speed : Observe a car moving on a straight road:

We observe that the distance covered by the car during A to B is 30km and from B toC is also 30km. The time taken by the car from A to B is 1hour (5pm– 4pm=1hour)and the time taken by the car from B to C is also 1hour (6pm – 5pm).So we can say that the car covers equal distances in equal intervals of time. Herethe car is said to have uniform speed.


KINEMATICSIf a body covers equal distances in equal intervals of time (however small thetime intervals may be), then the body is said to have uniform speed or constantspeed.Note : If the ratio of distance travelled and time taken by a moving body is constant,

i.e., dis tance cons tan t

time then we say that the body is moving with uniform speed.

ii. Non-Uniform Speed : Observe a car moving on a straight road:

We observe that the distance covered by the car during A to B is 40km, B to C is 20kmand from C to D is 30 km. The time taken by the car from A to B is 1 hour (5.00pm–4.00pm=1 hour), the time taken by the car from B to C is 1 hour (6.00pm – 5.00pm) andthe time taken by the car from C to D is 1 hour (7.00pm – 6.00pm).

or we observe that the car covers unequal distances in equal intervals of time (orequal distances in unequal intervals of time). Here we say that the car is moving withnon uniform speed or variable speed.

Note : If the ratio of distance travelled and time taken by a moving body is not con-

stant, i.e., distance constant

time then we say that the body is moving with variable speed.

iii. Average speed: When the body is moving with nonuniform speed, then we calculateits average speed by dividing the total distance covered by the total time taken tocover the distance.

In the above figure the total distance covered is (40 + 20 + 30)km = 90km. The totaltime taken is three hours.

Average speed = Total distance covered

Total time taken to cover the distance =

90km3hr

= 30km/hr


1) A horse runs a distance of 1200m in 3 min and 20 s. The speed of the horseis_________

1) 60 ms–1 2) 65 ms–1 3) 40 ms–1 4) 6ms–1

2) A car is moving at a speed of 15 ms–1. In how much time will it cover a distance of1.2 km ?

1) 70 s 2) 80 s 3) 18 s 4) 45 s

3) A bus is moving at 20ms–1. How much distance in kilometres will the bus cover in 25minutes ?


KINEMATICS1) 30 km 2) 20 km 3) 40 km 4) 50 km

4) A scooterist covers a distance of 3 kilometres in 5 minutes. Calculate his speed inkilometres per hour (km/h)

1) 28 2) 36 3) 42 4) 78

5) The train ‘A’ travelled a distance of 120 km in 3hr whereas another train ‘B’travelled a distance of 180 km in 4 hours. Which train travelled faster ?

1) train A 2) train B 3) Both are equally faster 4) none

*6) A particle travels different distances s1 & s2 with different speeds v1 & v2 in thesame direction. Find the expression for the average speed.

1) 1 2

1 1 2 2

s ss v s v

2) 1 2

21

1 2

s sss

v v

3) 1 1 2 2

1 2

s v s vt t 4) none of these

7) A runner runs 100 m in 10 s, then turns around and jogs 50m back toward thestarting point in 30 s. The average speed of the runner is

1) 1.25ms–1 2) 2.50ms–1 3) 3.75ms–1 4) 4ms–1

*8) If a particle travels with speeds v1, v2 and v3 during the time intervals t1,t2 and t3

respectively, then average speed is equal to ____________ [ ]

1) 1 1 2 2 3 3

1 2 3

v t v t v tt t t 2)

1 2 3

1 2 3

v v vt t t 3) 1 4)

1 2 3

1 1 2 2 3 3

t t tv t v t v t


Recap of Average Speed : The average speed of a particle in a time interval is definedas the distance travelled by the particle divided by the time interval. If the particle

travels a distance s in time t1 to t2, the average speed is defined as av2 1

svt t

The average speed gives the overall “rapidity” with which the particle moves in thisinterval.

In a one-day cricket match, the average run rate is quoted as the total runs divided bythe total number of overs used to make these runs. Some of the overs may be expensiveand some may be economical.

Similarly, the average speed gives the total effect in the given interval.

INSTANTANEOUS SPEED : The rapidity or slowness may vary from instant to instant.When an athlete starts running, he or she runs slowly and gradually increases the rate.We define the instantaneous speed at a time t as follows.

Let s be the distance travelled in the time interval t to t + t. The average speed in

this time interval is avsvt

.

Now make t vanishingly small and look for the value of st

. Remember s is the

distance travelled in the chosen time interval t. As t approches zero, the distance


KINEMATICS

s also approches zero but the ratio st

has a finite limit.

The instantaneous speed at a time t is defined as v = lim s dst dt

t 0

where s is the distance travelled in time t.

The average speed is defined for a time interval and the instantaneous speed is definedat a particular instant.


*1) When a body covers first one third distance with speed 1m/s, the second one thirddistance with speed 2m/s and the last one third distance with speed 3m/s thenaverage speed is ____________ [ ]

1) 2.24 m/s 2) 1.79 m/s 3) 2.66 m/s 4) 1.64 m/s

2) A bus travelling the first one-third distance at a speed of 10 km/h, the next onefourth at 20km/h and the remaining at 40 km/h. The average speed of the bus isnearly. [ ]

1) 9 km/h 2) 16 km/h 3) 18 km/h 4) 48 km/h

*3) A particle moving in a straight line covers half the distance with speed of 3m/s. Theother half of the distance is covered in two equal time intervals with speed of4.5 m/s and 7.5 m/s respectively. The average speed of the particle during thismotion is

1) 4.0 m/s 2) 5.0 m/s 3) 5.5 m/s 4) 4.8 m/s [ ]

4) An aeroplane flies round a square field PQRS of each side 100 km. Its speed alongPQ is 400 km/h, along QR is 500 km/h, along RS is 600 km/h and along SP is 700km/h. The average speed of the aeroplane over the entire trip is nearly [ ]

1) 550 km/h 2) 600 km/h 3) 527 km/h 4) 500 km/h

5) A car travels the first half of a distance between two places at a speed of 30 km/hand the second half of the distance at 50 km/h. The average speed of the car for thewhole journey is _____________ [ ]

1) 42.5 km/h 2) 40.0 km/h 3) 37.5 km/h 4) 35.0 km/h

6) A car moves for half of its time at 80 km/h and for next half of time at 40 k/h. Totaldistance covered is 60 km. The average speed of the car is____________ [ ]

1) 60 km/h 2) 80 km/h 3) 120 km/h 4) 180 km/h

*7) The distance travelled by a particle in time t is given by s = (2.5 m/s2) t2. Find (a) theaverage speed of the particle during the time 0 to 5.0s, and (b) the instantaneousspeed at t = 5.0 s.

1) 1.25m/s, 20m/s 2) 20m/s, 25m/s 3) 12.5m/s, 20m/s 4) 12.5m/s, 25m/s


KINEMATICS*8) A car travels 30 km at a uniform speed of 40 km/h and the next 30 km at a

uniform speed of 20 km/h. Find its average speed.

1) 35km/h 2) 40km/h 3) 22km/h 4) 26.6km/h

9) A train travels at a speed of 60 km/h for 0.5 h, at 30 km/h for the next 0.2 h andthen at 70 km/h for the next 0.7 h. What is the average speed of the train ?

1) 60 km/h 2) 38 km/h 3) 54 km/h 4) 70 km/h


Introduction to Velocity : If someone states that speed of the car is 10 m/s, it is veryclear that the car will cover 10m in one second.However, there is one question which comes up in mind. As you have not seen the car,you can always question, in which direction the car moves, i.e., whether it movestowards east or west etc., or whether it goes up a hill or down the hill, etc.

In other words, the statement “speed of car is 10 m/s” is not complete, as the directionis not specified. To have a clear idea another term is used, which is called velocity.Thus, in a way, velocity is speed along a specified direction.

For example, when we say that a car is moving at 10 m/s towards East, then we arereferring to velocity, rather than the speed.Mathematical Expression of Velocity : Velocity of a body can be found by dividing the

displacement with time. i.e., Velocity = displacement time taken

Units of Velocity : We know that the C.G.S. unit of displacement is centimetre(cm)and time is second(s).

Hence the C.G.S unit of velocity is centimetre per second(cm/s).

Similarly the S.I. unit of displacement is metre(m) and time is second(s). Hence theS.I. unit of velocity is metre per second (m/s).

Note: The units of velocity is the same as the units of speed. However in case ofvelocity the direction is specified.

Types of velocity

Uniform velocity : Observe a car moving along a straight road towards east.

We observe that the car covers equal distances in equal intervals of time in a specifieddirection.


KINEMATICSHere we say that the car is said to be moving with uniform velocity.

Note: A body has uniform velocity only if :

i) If it covers equal displacements in equal intervals of time

ii) Its direction of motion remains the same.Non Uniform velocity : Observe a car moving on a straight road:

We observe that the car covers unequal displacements in equal intervals of time [orequal displacements in unequal intervals of time (however small these intervals maybe]].

Here we say that the car is said to be moving with non uniform velocity or variablevelocity.

Average Velocity : When the body moves with non uniform velocity, then we calculateits average velocity by dividing the total displacement by the total time taken.

In the above figure the total displacement is (40 + 20 + 30)km = 90km. The total timetaken is three hours.

Average velocity = Total displacementTotal time taken =

90km3hr =30km/hr towards east.


1) A person travelling on a straight line moves with a uniform velocity v1 for sometime and with uniform velocity v2 for the next equal time. The average velocity

1) 1 2v vv2

2) 1 2v v v 3) 1 2

2 1 1v v v 4)

1 2

1 1 1v v v

2) A person travelling on a straight line moves with a uniform velocity v1 for a distancex and with a uniform velocity v2 for the next equal distance. The average velocity vis given by

1) 1 2v vv2

2) 1 2v v v 3) 1 2

2 1 1v v v 4)

1 2

1 1 1v v v

*3) Consider the motion of the tip of the minute hand of a clock. In one hour (morethan one answer is possible).

1) the displacement is zero 2) the distance covered is zero

3) the average speed is zero 4) the average velocity is zero


KINEMATICS4) Mark the correct statements :

1) The magnitude of the velocity of a particle is equal to its speed.

2) The magnitude of average velocity in an interval is equal to its average speed inthat interval

3) It is possible to have a situation in which the speed of a particle is always zero butthe average speed is not zero

4) It is possible to have a situation in which the speed of the particle is never zero butthe average speed in an interval is zero

*5) There is a square field of side ‘a’ units. An insect starts from one corner and reachesthe diagonally opposite corner in a time t. The magnitude of its average velocity is

1) 2at

2) 2at

3) 4 2a

t4) 2at [ ]

6) If a body starts from a point and returns back to the same point, then its [ ]1) average velocity is zero but not average speed2) average speed is zero, but not average velocity3) both average speed and velocity are zero4) average speed and velocity depends upon the path.

7) A man leaves his house for a cycle ride. He comes back to his house after half-an-hour after covering a distance of one km. What is his magnitude of average velocityfor the ride? [ ]

1) 2kmh–1 2) 0 3) 2 kms–1 4) 11 kms2

8) The numerical ratio of velocity to speed of a particle is always [ ]1) zero 2) equal to one

3) equal to or less than one 4) less than one

*9) A stopwatch is used to time the motion of a car. At time t1 = 12s, the car is atx1= 50 m. At t2 = 15 s, the car is at x2 = 5m. Find the average velocity of the car.

1) 5m/s 2) –5m/s 3) 15m/s 4) –15m/s

10) A particle is moving along X–axis. At time t1= 3s, its position is x1= 6m and att2 = 8s, its position is x2 = 21m. What is its magnitude of average velocity ?

1) 1 ms–1 2) 2 ms–1 3) 3 ms–1 4) 4 ms–1

11) A car travels a distance of 200 km from Delhi to Ambala towards North in 5 hours.The velocity of the car for this journey is

1) 40m/h towards east 2) 40km/h towards south

3) 40km/h towards north 4) 40m/h towards west


KINEMATICS


1. Recap of Average Velocity :

The average velocity of a particle in a time interval t1 to t2 is defined as its displacementdivided by the time interval. If the particle is at a point A at time t = t1 and at B at time

t = t2, the displacement in this time interval is the vector AB

. The average velocity inthis time interval (i.e. from t1 to t2) is then,

ave

2 1

ABVt t

= 2 1

2 1

r rt t

Y

Z

X

B

A

O

1r

2r

Like displacement, it is a vector quantity.

[Note that only the positions of the particle at time t = t1 and t = t2 are used incalculating the average velocity. The positions in between t1 and t2 are not needed,hence the actual path taken in going from A to B is not important in calculating theaverage velocity].

ii. Instantaneous velocity : The Instantaneous velocity of a particle at a time t is definedas follows. Let the average velocity of the particle in a short time interval t to t +

avt be v

. This average velocity can be written as avrvt

Where r is the displacement in the time interval t . We now make t very very...

small and find the limiting value of rt

. This value is instantaneous velocity v

of the

particle at time t. v r dr

t dt

limt 0

=

For very small intervals the displacement r is along the line of motion of the particle.

Thus, the length r equals the distance s travelled in that interval. So the magnitude

of the velocity isdrdr dsv

dt dt dt

Which is the instantaneous speed at time t.


KINEMATICS


1) It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30minutes to go from Patna to Ranchi where as a deluxe bus takes 8 hours.

Find the average velocity of the plane is1) 52 km/hr, Patna to Ranchi 2) 520 km/hr Patna to Ranchi

3) 400 km/hr Patna to Ranchi 4) 500 km/hr Patna to Ranchi

2) From the above problem the average velocity of the bus is

1) 12.5 km/hr, Patna to Ranchi 2) 10.5 km/hr Patna to Ranchi

3) 48.5 km/hr Patna to Ranchi 4) 32.5 km/hr Patna to Ranchi

*3) When a person leaves his home for sight-seeing by his car, the meter reads 12352km. When he returns home after two hours the reading is 12416 km.

What is the average speed and average velocity of the car during this period ?1) 12 km/hr, zero 2) 40 km/hr, zero 3) 32 km/hr, zero 4) 48 km/hr, zero

4) A car travels equal distances in the same direction with velocities 60 km h–1,20 kmh-1 and 10 km h–1 respectively. The magnitude of average velocity of the carover the whole journey of motion is

1) 8 m s–1 2) 7 m s–1 3) 6 m s–1 4) 5 m s–1

5) A bus covers a distance of 250 km from Delhi to Jaipur towards west in 5 hours inthe morning and returns to Delhi in the evening covering the same distance of250 km in the same time of 5 hours. Find the magnitude of average velocity of thebus for the whole journey .

1) zero 2) 100km/h 3) 50m/h 4) none of these6) The Narayana IIT Olympiad conference theatre of a school is 15 m wide and has a

door at a corner. A teacher enters at 8.00 am through the door and makes 10 roundsalong the 15m wall back and forth during the period and finally leaves the class-roomat 9.30a.m. through the same door. Compute his average speed and average velocity.1) (5/3)m/min, zero2) 6m/min, –2m/min 3) 8m/min, –2m/min 4) 8m/m, 4m/min

*7) The position of a particle moving on x-axis is given by x = At3 + Bt2 + Ct + D.The average velocity during the interval t = 0 to t = 4s1) 4A + 4B + C 2) 16A + 4B + C 3) 16A – 4B + C 4) 16A – 16B – C

*8) The position of a stone dropped from rest from a cliff is given by x = 5 t2. What is themagnitude of the velocity of the stone at t = 2s ?1) 10m/s 2) 20m/s 3) 40m/s 4) 25m/s

*9) The position of a ball rolling in a straight line is given by ;

x = 2 + 6.6 t – 1.1 t2

What is its velocity at (i) t = 2s and (ii) t = 3s ?1) 1m/s, zero 2) zero, m/s 3) 2m/s, zero 4) 4m/s, 2m/s


KINEMATICS10) Pick the wrong statements (More than one answer is possible):

1) Average speed of a particle in a given time is never less than the magnitude of theaverage velocity.

2) It is possible to have a situation in which d v 0dt

but d v 0dt

3) The average velocity of a particle is zero in a time interval. It is possible that theinstantaneous velocity is never zero in the interval.

4) The average velocity of a particle moving on a straight line is zero in a time interval.It is possible that the instantaneous velocity is never zero in the interval. (Infiniteaccelerations are not allowed.)


Introduction to Acceleration : Consider a car at point A moving with a velocity of20km/h, towards east.If the driver of the car presses the accelerator, the car will start moving fast. Let thevelocity of the car be 40km/h at B, 60km/h at C and 80km/h at D.

We observe that the velocity of the car increases continuously with respect to time. Herewe say that the car accelerates.

The increase in velocity per unit time is called acceleration.Similarly now consider a car at point A moving with a velocity of 80km/h, towards east.If the driver of the car presses the brakes, the velocity of the car decreases. Let the ve-locity of the car be 60km/h at B, 40km/h at C and 20km/h at D.

We observe that the velocity of the car decreases continuously with respect to time.Here we say that the car decelerates or retards.

The decrease in velocity per unit time is called deceleration or retardation.

Negative acceleration is called Retardation or Deceleration.


KINEMATICS(a) Mathematical Expression of Acceleration : In the figure of introduction to accelera-

tion the change in velocity from A to B in 1hour is 40 – 20 = 20 km/h. Similarly thechange in velocity from B to C in 1hour is 60 – 40 = 20 km/h.

Total change in velocity from A to D = (80 –20) = 60 km/h Total time required for above change in velocity = 3 hours.

Rate of change of velocityTotal change in velocity=

Total time required for change

22

60km 1 20km= = = 20km/hrhr 3hr hr

But the rate of change of velocity is called acceleration.

i.e., Acceleration Change in velocity

= time taken

In the above case the acceleration of the car is 20kmhr–2. It means thatfor every one hour the increase in velocity is 20km/hr.Units of Acceleration : The C.G.S unit of velocity is cm/s and time is second, hencethe C.G.S. unit of acceleration is cms–2.Similarly the S.I. unit of acceleration is ms–2.

(b) Uniform acceleration :

Observe the above car travelling on a straight road.

At the position A the velocity of the car is 10 km/hr

At the position B the velocity of the car is 15 km/hr

Change in velocity of car from A to B = 15 – 10 = 5 km/hr

The time taken for the velocity to change = 3 : 30 pm – 3 : 00 pm

acceleration of car = Change in velocity

time = km /hr 25 1012

At the position c the velocity of car is 20 km/hr

Change in velocity of car from B to C = (20 – 15) km/hr = 5 km/ hr

The time taken for the change in velocity = 4 : 00 pm – 3 : 30 pm = hr12

acceleration = Change in velocity

time = km /hr 25 1012


KINEMATICSThus acceleration of the car is same in both the cases. We say that car is movingwith uniform acceleration

(c) Variable acceleration: When the change in velocity in each unit of time is not constant,then the body is said to be moving with variable acceleration. In this case, the motionis said to as non-uniformly accelerated motion.

(d) Average Acceleration : Suppose the velocity of a particle at time t1 is 1V and at time t2

is 2V . The change produced in time interval t1 to t2 is 2 1V V

. We define the averageacceleration ava

as the change in velocity divided by the time interval. Thus,

2 1av

2 1

V Vat t

Again the average acceleration depends only on the velocities at time t1 and t2. How thevelocity changed in between t1 and t2 is not important in defining the averageacceleration.

Instantaneous Acceleration : Instantaneous acceleration of a particle at time t is

defined as a = lim v dvt dt

t 0

where v is the change in velocity between the time t and t + t. At time t the velocity

is v and at time t + t it becomes

vv v.t

is the average acceleration of the particle

in the interval t. As t apporaches zero, this average acceleration becomes theinstantaneous acceleration.

KINEMATICSWORK SHEET 7

1. A body can’t have [ ]

1) a constant speed and varying velocity

2) acceleration and a constant speed

3) constant velocity and varying speed

4) none zero speed and zero acceleration

2. Which of the following changes when a particle is moving with uniform velocity?

1) position vector 2) speed 3) velocity 4) acceleration

*3. The correct statement from the following is

1) A body having zero velocity will not necessarily have zero acceleration

2) A body having zero velocity will necessarily have zero acceleration

3) A body having non uniform velocity will have zero acceleration

4) A body having uniform speed can have only uniform acceleration


KINEMATICS*4. The position of a particle moving on X-axis is given by x = At3 + Bt2 + Ct + D.

The numeric al values of A, B, C, D are 1, 4, –2 and 5 respectively and SI units areused find the acceleration of the paritcle at t = 4s.

1) 10 m/s2 2) 32 m/s2 3) 11 m/s2 4) 28 m/s2

5. In the above problem the average acceleration during the interval t = 0 to t = 4s is

1) 20 m/s2 2) 32 m/s2 3) 11 m/s2 4) 28 m/s2

*6. A particle moves along a straight line such that its displacement at any time t isgiven by s = (t3 – 6t2 + 3t + 4) metre.

What is the velocity of the particle when its acceleration is zero?

1) –3m/s 2) 6m/s 3) –9m/s 4) 12m/s


Equations of motion:

The uniformly accelerated motion of a body is described by three equations of motion.

First equation of motion (velocity time relation) : The relation between initial velocity(u), final velocity (v), acceleration (a) and time(t) is known as the first equation ofmotion.

i.e., v = u + at

Note: It gives the velocity acquired by a body in time ‘t’.

Derivation: Consider a body having initial velocity ‘u’. Suppose it is subjected to auniform acceleration ‘a’ so that after time ‘t’ its final velocity becomes ‘v’. Now fromdefinition of acceleration we know that

Accelerationchange in velocity=

time taken

or, AccelerationFinal velocity - Initial velocity=

Time taken

av - u=

t at = v – u v = u + at

It is used to find out the velocity ‘v’ acquired by a body in time ‘t’.

Where v = final velocity of the body.

u = initial velocity of the body.

a = acceleration

and t = time taken

Notes : 1) If a body starts from rest, u = 0

v = at2) If a body moves with uniform velocity, a = 0

v = u


KINEMATICSSecond equation of motion (Displacement time relation) : The relation betweendisplacement (s), initial velocity (u), acceleration (a) and time (t) is known as the

second equation of motion. i.e., s 21ut at2

Derivation: Suppose a body has an initial velocity ‘u’ and a uniform acceleration ‘a’ fortime ‘t’ so that its final velocity becomes ‘v’. Let the displacement of the body in thistime be ‘s’. The displacement of a moving body in time ‘t’ can be found out by consideringits average velocity.

Since the initial velocity of the body is ‘u’ and its final velocity is ‘v’, the average velocity

is given by vave u v

2

.

Now displacement, s = Average velocity × time u vs time

2

u vs t2

___ (1)

But from 1st equation of motion we know that v = u + at

s u u at

t2

2u at t

2

22ut at2 2

s 21ut at2

Notes : i) If a body starts from rest, u = 0, 21s at2

ii) If a body moves with a uniform velocity, a = 0

(iii) If a body comes to rest, its final velocity, v = 0s = ut


1. An object undergoes an accelerations of 8m/s2 starting from rest. Find the distancetravelled in one second.1) 2m 2) 4m 3) 3m 4) 5m

2. A racing car has uniform acceleration of 4m/s2. What distance will it cover in10seconds after the start?1) 100m 2) 300m 3) 200m 4) 400m

3. A train starting from rest, attains a velocity of 72km/h in 5minutes. Assuming thatthe acceleration is uniform, find (i) the acceleration and (ii) the distance travelledby the train while it attained the velocity.

1) 21 ms20

, 2km 2) 21 ms15

, 3km 3) 21 ms15

, 4km 4) 21 ms20

, 1km

4. An object having a velocity 4.0 m/s is accelerated at the rate of 1.2m/s2 for 5.0s.Find the distance travelled during the period of acceleration.

1) 35m 2) 25m 3) 50m 4) 30m


KINEMATICS*5. A bus starts to move with an acceleration of 1ms–2. A man who is 48m behind the

bus runs with constant velocity of 10ms–1 to catch it. In how much time he will catchthe bus?

1) 6sec 2) 8sec 3) 10sec 4) 12sec

*6. A train starts from rest with an acceleration of 2ms–2. A man who is 9m behind thetrain runs with uniform speed and just manages to get into the train in 10sec. Thespeed of the man is

1) 9ms–1 2) 8.7ms–1 3) 4.5ms–1 4) 10.9ms–1

*7. A body moving with uniform acceleration covers 100m in the first 10 seconds and150m in the next 10seconds. The initial velocity of the body is

1) 15ms–1 2) 7.5ms–1 3) 5ms–1 4) 2.5ms–1


1. Third equation of motion (velocity displacement relation) : The relation betweeninitial velocity (u) final velocity (v), acceleration (a) and displacement (s) by the body isknown as third equation of motion.

i.e., v2 – u2 = 2as

where v = final velocity

u = initial velocity

a = acceleration

s = displacement

Derivation:Let us consider a body having initial velocity ‘u’ and moving with a uniform acceleration‘a’. If after time ‘t’, its velocity is ‘v’, then for a body moving in a straight line underuniform acceleration, the displacement ‘s’ in time ‘t’ is given by,

Displacement (s) = Average velocity × Time

But, Average velocity Initial velocity + Final velocity=

2 u v

2

Thus, we can write, s u v t

2

------- (i)

we know, v = u + at

or t v u

a

Substituting the value of t in equation (1), we get,

u v v us2 a

2 2uv u v uvs2a

2 2v us2a

v2 – u2 = 2as

This is the third equation of motion.


KINEMATICSNotes : i) It a body starts from rest, u = 0

v2 = 2as

ii) If a body moves with a uniform velocity, a = 0 v2 = u2 v = u

2. Distance travelled in nth secondConsider a body moving with uniformly accelerated motion having acceleration ‘a’. Thedistance of a particle in time ‘t’ is given by

S = ut + 21 at2 , where u = initial velocity, at time t = 0

If Sn and Sn – 1 are the distance of the particle in n and n – 1 seconds, then distance ofthe particle in nth second is, Snth = Sn – Sn–1

C BA

sn-1 snth

sn

Sn = un + 12 a n2

Sn–1 = u(n – 1) + 12 a (n – 1)2 Now

Snth = Sn – Sn–1

221 1un an u n 1 a n 12 2

2 21 aun an un u n 1 2n2 2

221 an aun an un u an

2 2 2

221 an aun an un u an

2 2 2

au an2

1u a n2

2n 1u a2

Snth

au2

(2n – 1)

Notes : If a body starts from rest, u = 0, Snth a2

(2n – 1)

Equations of kinematics are valid only for uniformly

accelerated motion i.e. when a = constant.


KINEMATICSKINEMATICS

WORK SHEET 9*1. Moving with uniform acceleration, a body covers 150m during 10sec so that it covers

24m during the tenth second. Find the initial velocity and acceleration of the body

1) 2ms–1, 5ms–2 2) 5ms–1, 2ms–2 3) 3ms–1, 4ms–2 4) 4ms–1, 3ms–1

2. A person travelling at 43.2 km/h applies the brake giving a deceleration of6.0m/s2 to his scooter. How far will it travel before stopping?

1) 6m 2) 8m 3) 12m 4) 10m

3. A bullet travelling with a velocity of 16m/s penetrates a tree trunk and comes to restin 0.4m. Find the time taken during the retardation.

1) 0.5sec 2) 0.05sec 3) 5sec 4) 0.4sec

4. A bullet going with speed 350m/s enters a concrete wall and penetrates a distanceof 5.0m before coming to rest. Find the deceleration.

1) 12.25 × 105 ms–2 2) 10.25 × 105 ms–2

3) 10.25 × 104 ms–2 4) 14 × 105 ms–2


1. First Equation of Motion :

v = u + at. It gives the velocity acquired by a body in time t.

Where v = Final velocity of the body

u = Initial velocity of the body

a = Acceleration

and t = Time taken

2. Second Equation of Motion :

s = ut + 21 at2 . It gives the displacement of a body in time t.

where s = displacement of the body in time t

u = Initial velocity of the body and a = Acceleration

3. Third Equation of Motion :

v2 = u2 + 2as. It gives the velocity acquired by a body in travelling a distance s.

where v = Final velocity

u = Initial velocity

a = Acceleration

and s = Distance travelled


KINEMATICS4. Distance travelled in ‘nth’ second

nthaS u 2n 12

where u = Initial velocity

a = uniform acceleration

snth = distance travelled in nth second

Note : If a body starts from rest, u = 0

Snth = a 2n 12

5. Notes : i) If a body starts from rest, its initial velocity, u = 0

ii) If a body comes to rest (it stops), its final velocity, v = 0

iii) If a body moves with uniform velocity, its acceleration, a = 0


1. A body A starts from rest with an acceleration a1. After two seconds , another bodyB starts from rest with an acceleration a2. If they travel equal distance in the 5th

second after the start of A, then the ratio of a1 : a2 is equal to1) 5 : 9 2) 9 : 5 3) 5 : 7 4) 7 : 5

2. A bus is beginning to move with an acceleration of 1m/s2. A boy who is 48m behindthe bus starts running at 10m/s. After what time will the boy be able to catch thebus?1) 6sec 2) 8sec 3) 10sec 4) the boy cannot catch the bus

*3. A train is moving along a straight path with uniform acceleration. Its engine passesacross a pole with a velocity of 60 km/h and the end (guard’s van) passes acrosssame pole with a velocity of 80km/h. The middle point of the train will pass acrosssame pole with a velocity1) 60km/h 2) 75km/h 3) 65km/h 4) 70.7km/h

*4. A truck starts from rest with an acceleration of 1.5metre/sec2 while a car 150metrebehind starts from rest with an acceleration of 2metre/sec2. How long will it takebefore both the truck and car side by side, and how much distance is travelled byeach?

1) 12.5sec 2) 23sec 3) 22.5sec 4) 24.5sec

5. A train starts from rest and moves with a constant acceleration of 2.0m/s2 for halfa minute. The brakes are then applied and the train comes to rest in one minute.Find (a) the total distance moved by the train, (b) the maximum speed attained bythe train and (c) the position(s) of the train at half the maximum speed.

1) 2.7km, 60m/s, 225m 2) 2.2km, 60m/s, 200m

3) 4.1km, 50m/s, 250m 4) none of these


KINEMATICS*6. A driver takes 0.20s to apply the brakes after he sees a need for it. This is called the

reaction time of the driver. If he is driving a car at a speed of 54km/h and the brakescause a deceleration of 6.0m/s2, Find the distance travelled by the car after he seesthe need to put the brakes on.1) 11m 2) 22m 3) 10m 4) 10.5m

7. A police jeep is chasing a culprit going a motorbike. The motorbike crosses a turningat a speed of 72km/h. The jeep follows it at a speed of 90km/h, crossing the turningten seconds later than the bike. Assuming that they travel at constant speeds, howfar from the turning will the jeep catch up with the bike?

1) 0.5km 2) 2km 3) 3km 4) 1km

8. A car travelling at 60km/h overtakes another car travelling at 42km/h. Assumingeach car to be 5.0m long, find the time taken during the overtake and the total roaddistance used for the overtake.

1) 27km 2) 20m 3) 38m 4) 41m


KINEMATICS

KINEMATICS

WORK SHEET – 1(SOLUTIONS)

1. Displacement has both magnitude anddirection hence displacement is a vectorquantity.

2. Since initial position and final positionare same for both player and the ball,hence they have same displacement.

3. Since initial point ad final point aresame, hence its displacement is zero.Distance covered = 40 + 40 = 80 m.

4. The numerical ratio of displacement todistance is equal to one or less than one.

Displacement is a vector quantity anddistance is a scalar quantity it is truebut it is not a correct reason for the abovestatement.

5. Consider a body moving in a circular pathof radius ‘r ’. If it completes onerevolution, then the distance covered =circumference of circle whereasdisplacement is zero, since initial pointand final point are same.

6. Distance is always greater than or equalto displacement. Hence fourth option iswrong.

7. Any moving particle returns to its initialpoint as per the definition of thedisplacement the displacement of theparticle is zero.

8. Displacement have both positive andnegative values. Hence first option is nota characteristic of displacement.

KINEMATICS


1. Let the cyclist start from point X.

O rX

radius of circle OX = r (given)

Distance covered by cyclist =Circumference of circle (once) = 2 r

Initial position of cyclist = X. After onerevolution, final position = X

Since initial position and final positionare same displacement is zero.

2. Distance covered = Half of the

circumference of circle = 12

× 2 r =

r

O rXY

r

Initial position = X, final position = Y

Magnitude of displacement XY = ?

From figure XY = XO + OY= r + r = 2r.

3. Distance covered by cyclist

O rX

r

Z


KINEMATICS

= Quadrant of circle

= 14

× Circumference of circle

= 14

× 2 r = r

2

Initial position = X Final position = Z

Magnitude of displacement = XZ = ?

Consider right triangle OXZ, we have

from pythagorus theorem,

XZ = 2 2OX OZ = 2 2r r

= 22r = 2r

4. (i) A player takes 40s to complete onerevolution of a circular path.

(ii) Here 2 minutes 20s = 140s

Let n is the number of revolution hemade

140 7n40 2

= 3.5

that means he made three and halfrevolutions in 140seconds. So, hereached the opposite of the starting pointin a circle of radius ‘r’

So, the displacement is r + r = 2r.

A BO

5. AB

represented the position vector of

the man in the East direction = 8 i m.

BC

represented the position vector of

the man in the north direction = 6 j

.

AC

represent the shortest distance i.e.,displacement of a man which is equal to

AC AB BC

2 2AC 8i 6j =10m.

toward north – east.

Since the angle between i and j

is 90°.

S

E

N

W

A 8m B

6m

C

10m

6. Let ‘O’ be the reference point (i.e. house)then

A B

C20m

50m

40m

PO

N

EW

S

total distance travelled by the man

= 50m + 40m + 20m = 110m

From the fig. OC represent thedisplacement of the man

=OC2 = OP2 + PC2 2 240 30 2500

= 50mDisplacement of man is 50m north east.

7.

( 20, 0)

PP|x axis x axis

O

(20, 0)

From the above,

OP 20 i m

OP ' 20 i m

total distance travelled

|OP PO OP = 20 + 20 + 20 = 60m

displacement = OP ' 20 i

= –20m


KINEMATICSi.e. 20 m in the negative direction.

8.

A 2km

C D

B

500m

4 km

AD 2 i 0.5 j 4 i

= 6 i 0.5 j

AD = 2 26 0.5 = 6.02 km

1/2 1tan6 12

1 1tan12

the displacement of the car = 6.02km,

along the direction tan–1 1

12

with the

positive X–axis.

9. Let P be the point on the front edge fromwhere the queen rebounds.

2-x x

2

C

2

2Q

4

PS T

R

Now, In PTC,xtan2

.... (i)

2 xIn PTC4

.... (ii)

from eq.(i) and (ii) we have4x = 4 – 2x

or 6x = 4 or 2x3

CP2 2

22 23

or CP2 4 4049 9

or CP 2 103

ft

This gives the displacement of the queenfrom the centre to the front edge.

10. (i) From the above problem

PQ presents the displacement of thequeen from the front edge to the hole.

PQ2 = 42 + 222

3

or PQ2 = 42 + 24

3

or PQ 4 103

ft.

This gives the displacement from thefront edge to the hole.

(ii) CQ = 2 2CR RQ = 2 22 2 = 2 2 ft

11.

D

A

3ft

7ft B

H

GE

C

3f

t

4ft

4ft

i

j

k

Displacement vector, r 7 i 4 j 3k

Magnitude of displacement

= 2 2 27 4 3 = 74 ft.

12. To reach opposite corner, the mosquitohas to travel 7ft along length, 4ft alongbreadth and 3ft along height.

displacement along x-axis, y-axis andz-axis are 7ft, 4ft and 3ft respectively.

13. Initial position = X Final position = Y

Magnitude of displacement XY = ?


KINEMATICS

O

r

X Z

r 60° 60°

Y

From figure, XY = XZ + ZY = 2XZ or 2ZY(according to hint given)

Consider OZX, Sin60° = XZr

XZ = r

Sin60° = r 32

XY = 2XZ = 2 × r 32

= r 3 .

14. 30 j

+ 30 i + 30 2 (Cos45° (– i

) + Sin45°

(– j

))N

S

W E

i

j

j

i

= 30 j

+ 20 i + 30 2

1 1i j2 2

= 30 j

+ 20 i – 30 i

– 30 j

= –10 i

= 10 m west.

KINEMATICS


1. Speed of the horse

distance travelled 1200m=

time taken 3 60 20 s

1200m200s

= 6ms–1

2. Distance = Speed × time

1 1

distance 1.2km 1200mtime = speed 15ms 15ms

= 80s

3. Distance = speed × time

= 20ms–1 × 25min

= 20 × 10–3km s–1 × 25 × 60s

= 500 × 10–3 × 60km

= 30km

4. Speed distance 3km 3km= time 5min 5/60hr

180km

5hr = 36km/hr.

5. Speed of train A 120km

3hr = 40kmh–1

Speed of train B 180km

4hr = 45kmh–1

So, speed of train B is faster than speedof train A.

6. Average speed 1 2

1 2

s stotal distance= total time t t

here 1

11

stv

and 2

22

stv

So, Vavg 1 2

21

1 2

s sss

v v

7. Average speed Vave

total distance travelled 100m 50m= total time taken 10s 30s

150m40s

= 3.75ms–1

8. Average speed total distance travelled=

total time taken

= 1 1 2 2 3 3

1 2 3

v t v t v tt t t

KINEMATICS


1. Average speed,

1 2 3

1 2 3

S S Stotal distance travelled= total time taken t t t


KINEMATICSLet ‘s’ the total distance travelled and lett1 time taken for 1st one third distance,with velocity of 1ms–1.

1s/3t s/31

___________ (1)

Let t2 be the time taken for 2nd one thirddistance with a velocity of 2ms–1.

2s/3t s/62

____________ (2)

Let t3 be the time taken for next onethird distance with a velocity of 3ms–1.

3s/3t s/93

_____________ (3)

ave1 2 3

s sV s s st t t3 6 9

s 1 181 1 1 11/18 11s3 6 9

= 1.79ms–1

2. Vave

total distance travelled= total time taken

Vave 1 2 3

s s 5s3 4 12=t t t

[s = distance]

s s 7s 12 7 5s s s s3 4 12 12 12

but here t1 1

1

s s/3 s hrV 10km/h 30

____(1)

time t2 2

2

s s/4 s hrV 20km/hr 80

_______ (2)

t3 3

3

5ss 5s s12 hrV 40km/hr 480 96

______ (3)

Vave s 1

s s s 1 1 130 80 96 30 80 96

480 km/hr 17.7 18km/hr27

3. Vave 1 2

s ss/2 s ' t t3 V '

2 2t ts 4.5 7.52 2 2

2 2ss 12t t

12 _______________ (1)

Vave

s ss s 1 1s6 12 6 12

1

186 12

= 4.0m/s4.

600km/h100km

100km500km/h

R

QP

S

700km/h100km

100km400km/h

average speed total distance=

total time taken100km+100km+100km+100km= 100km 100km 100km 100km

400km/hr 500km/hr 600km/hr 700km/hr

400km1 1 1 1 hr4 5 6 7

but 1 1 1 14 5 6 7

105 84 70 60 319420 420

Average speed 400 420

319

= 526.65 527km/hr

5. Vave1 2

total distance S= total time taken t t

SS/2 S/230 50

here 1S/2t

30km/h and 2

S/2t50km/h

Vave S 100 60 75

1 1 160 2S60 100


KINEMATICS

= 37.5km/h

6. Average speed of the car

1 2

1 2

S Stotal distance travelled= total time taken t t

here S = S1 + S2

60km = t t80 402 2

60km = 60t t = 1hr

Vave s 60kmt 1hr

= 60km/hr

7. a) Given, the distance travelled by theparticle in time t is S = (2.5 m/s2) t2

The distance travelled during time 0 to5.0s is s = (2.5 m/s2) (5.0s)2 = 62.5m

The average speed during this time is

ave62.5mV

5s = 12.5m/s

b) Given, S = (2.5 m/s2) t2

instantaneous speed Vave dsdt

= 2.5 × (2 × t)

= (5.0m/s2) t

At t = 5.0 s the speed is

dsVdt

= (5.0m/s2) (5.0s)

= 25m/s

8. Vave1 2

21

1 2

s s 30km 30kms 30km 30kms

40km/hr 20km/hrv v

60km 60km183 3 hrhr84 2

60 8 80km/hr 26.6km/h18 3

9. Vave total distance travelled=

total time taken

1 2 3

1 2 3

s s st t t

but here s1 = v1 × t1

= 60km/h × 0.5 h

= 30 km ____________ (1)

S2 = v2 × t2

= 30km/h × 0.2 h

= 6 km ____________ (2)

S3 = v3 × t3 = 70 × 0.7

= 49 km __________ (3)

Vave

30 6 49 km

0.5 0.2 0.7 hr

85 601.4

km/h

KINEMATICS


1. Average velocity total displacement=

total time taken

vave 1 2

1 2

s +s=t +t

s1 = v1t1 & s2 = v2t2 since t = t1 = t1

vave 1 21 2 1 2v vv t+v t v v=

t+t 2t 2

2. vave 1 2

1 2

s st t

since 1 2s ss & s2 2

and

1 21 2

s/2 s/2t & tV V

vave

1 2 1 2

s ss/2 s/2 1 1v v 2v 2v

1 2 1 2

1 2 1 2

4v v 2v v2 v v v v

or

1 2

1 2 1 2

v v2 1 1v v v v v


KINEMATICS3. In one hour, the minute hand of a clock

makes one complete revolution thatmeans it reaches the initial position.

Therefore the displacement is zero andhence average velocity is zero.

4. 1

5.

D a C

a

BaA

a

Let DB in the displacement of the

insect, DB 2 2 2a a 2a 2a

average velocity of the insect

displacement 2 a= time interval t

6. If a body starts from a point and returnsback to the same point, then itsdisplacement in zero and hence averagevelocity is zero but not average speed.

7. He simply returns to his house, i.e.,initial position hence displacement iszero and so his average velocity is zero.

8. The numerical ratio of velocity to speedof a particle is always less than (or)equal to one.

9. Vave 2 1

2 1

x x 5 50 45t t 15 12 3

= –15m/s

10. x1 = 6m, x2 = 21m

t1 = 3s, t2 = 8s

Vave 2 1

2 1

x x 21 6 15t t 8 3 5

= 3m/s

11. Vave 200km

5hr = 40km/hr towards north.

KINEMATICS


1. Average velocity of the plane

displacement of the plane= time taken

260km 260 60= km/hr30min 30

= 520 km/hr patna to ranchi

2. Average velocity of the bus

displacement of the bus= time taken

260km=8hr = 32.5 km/hr patna to ranchi.

3. Here difference of metre reading givesthe distance of the car= 12416km – 12352km = 64km

Average speed =Vave 64km2hr

= 32km/hr

Since he returns home, his initialposition and final position are same, andhence displacement is zero. Thereforeaverage velocity is zero.

4. We know,

total distance travelled by the car= total time taken

1 2 3

1 2 3

S S St t t

S 3S S SS S S 1 1 1S60 20 10 60 20 10

3 3 601 3 6 10

60

= 18km/hr

518 m/s18

= 5ms–1

5. Here the displacement is 250km –250km = 0

Hence the average velocity of the bus forthe whole journey = zero


KINEMATICS

6. (i) IIT olympiad school width = 15m, thenumber rounds made by a teacher alongthe width 15m = 10 (i.e. even)

the time period for the above duration =9.30 – 8.00 = 90min.

Average speed total distance travelled=

total time taken

15×10=90

= (5/3) m/min

(ii) Since teacher made even rounds, hisfinal position is equal to the initialposition.

hence displacement of a teacher is zero.

average velocity of a teacher is zero.

7. Position of the particle given

x = At3 + Bt2 + Ct + D

displacement at t = 0

x0 = 0 + 0 + 0 + D = D _________ (1)

displacement at t = 4s

x4 = A × 43 + B × 42 × C × 4 + D

= 64A + 16B + 4C + D

ave velocity 4 0x xt

64A 16B 4C D D 64A 16B 4C4 0 4

= 16A + 4B + C

8. the position of a stone dropped, x = 5t2

velocity of a stone 2dx dv 5tdt dt

2d5 (t )dt

= 5 × 2t = 10t

but velocity of a stone at t = 2second

v at 2s = 10 × 2 = 20m/s

9. The position of a ball rolling in a straightline, x = 2 + 6.6t – 1.1t2

Velocity of a ball dxvdt

= 0 + 6.6 – 1.1 × 2t

v = 6.6 – 2.2t

(i) Velocity of a ball at time t = 2s

v1 = 6.6 – 2.2 × 2 = 6.6 – 4.4 = 2m/s

(ii) Velocity of a boat at time t = 3s

v2 = 6.6 – 2.2 × 3 = 6.6 – 6.6 = 0

10. 1, 2 and 3

(i) Average speed of a particle is greaterthan (or) equal to the magnitude ofvelocity.

(ii) In case of uniform circular motion, speedremains constant while velocitychanges.

Hence, d v

0dt

while d v 0dt

, here

d v0

dt

represent the speed is constant

while the velocity is not constant which

represent the d v 0dt

.

KINEMATICS


1. A body can’t have constant velocity andvarying speed. The body with varyingspeed is not possible to move withconstant velocity.

2. Both speed and velocity are constant inthe case of a particle moving withuniform velocity.

A particle moving with uniform velocityhas zero acceleration.

3. A body having zero velocity will notnecessarily have zero acceleration.


KINEMATICS

4. v = 3At2 + 2Bt + C

or, dvadt

= 6 At + 2B.

At t = 4s, a = 6(1 m/s3) (4 s) + 2(4m/s2)= 32 m/s2.

5. v – 3At2 + 2Bt + C

Velocity at t = 0 is C = –2 m/s

Velocity at t = 4 is = 78 m/s (try yourself)

2 1

2 1

v vAverage accelerationt t

= 20 m/s2

6. Give that, S = t3 – 6t2 + 3t + 4 ______ (1)

dsvdt

= 3t2 – 12t + 3 ________ (2)

dva 6t 12dt

when a = 0, 6t = 12 t = 2sec

Now v = 3(2)2 – 12(2) + 3 = –9m/s.

KINEMATICS


1. Here,

u = 0, a = 8m/s2, t = 1s,

s = ? (to be calculated)

From formula, s 21ut at2

Putting values, we get,

s = 0 × 112

× 8 × (1)2

or s = 4m.

2. Here,

Initial velocity, u = 0, (starts from rest)

Uniform acceleration, a = 4m/s2

Time taken, t = 10s

Distance covered S = ?(to be calculated)

From formula, S 21ut at2


S = 0 × 1t2

× 4 × (10)2 12

× 4 × 100

S = 200m.

3. Here,

Initial velocity, u = 0

Final velocity, v = 72km/h = 20m/s

Time taken, t = 5min = 300s

(i) Acceleration a = ?(to be calculated)

(ii) Distance travelled S = ?

(to be calculated)

(i) From formula, v = u + at

We have,v ua

t


220 0 1a m/s300 15

(ii) From formula

u vS t2

Putting values, we get

0 20s 3002

= 3000m

or 3km.

4. Given velocity, v = 4.0m/s

Acceleration a = 1.2m/s2

time, t = 5.0seconds

we know, distance travelled

s 21ut at2

= 4.0 × 5.0 12

× 1.2 × (5.0)2

= 20.0 + 25.0 × 0.6 = 20.0 + 15.0 = 35.0m

5. Total distance travelled by man withuniform speed 10ms–1 is,

48 + s = 10 × t (1)

Distance travelled by bus from rest in t

s 2

21 tat (2)2 2

From equations (1) and (2)


KINEMATICS2t482

= 10t or t2 – 20t + 96 = 0

or (t – 12) (t – 8) = 0

or t = 8sec and t = 12sec

So man catches the bus in 8sec.

6. Total distance travelled by man withuniform speed v is 9 + s = vt

9 + s = v × 10 (1)

Distance travelled by train from rest in10sec is,

S = 0 × 10 12

× 2 × (10)2

= 100m (2)

From equations (1) and (2)

9 + 100 = v × 10 109v10

= 10.9 ms–1

7. From S 21ut at2

100 = 10 × 1u2

× a × (10)2

u + 5a = 10 _______ (1)

250 = 20u 12

× a × (20)2

u + 10a = 12.5 __________ (2)

multiply equation (1) with 2

2u + 10a = 20 __________ (3)

on solving equation (2) and (3)

u = 7.5ms–1

Initial velocity of the body = 7.5ms–1

KINEMATICS


1. From 21S ut at2

21150 10u a 102

150 = 10u + 50a ________ (1)

From Sn au2

(2n – 1)

24 au2

(2 × 10 –1)

48 = 2u + 19a _______ (2)

On solving equations (1) and (2) u = 5ms–1

and a = 2ms–2

2. Initial velocity of a person, u = 43.2km/h

= 12.0m/s

deceleration of a scooter, a = –6.0ms–2.

Final velocity of a person is zero

distance travelled by a person

before stopping, 2 2v uS2a

20 0 12.0 12 12S2 6.0 12

= 12m

3. Initial velocity, u = 16m/sFinal velocity, v = 0distance, s = 0.4m

retardation of the bullet, 2 2v ua2S

20 16 256 25602 0.4 0.8 8

= –320m/s2

time taken during the retardation

v u 0 16 1ta 320 20

= 0.05seconds.

4. Initial velocity, u = 350m/s.distance, S = 50cm = 0.05mfinal velocity, v = 0

deceleration, 22 2 0 350v ua2s 2 0.05

350 3500.1

= 12.25 × 105 m/s2.


KINEMATICS

KINEMATICS


1. Distance travelled by A in 5th second,

DA 1a0

2 (2 × 5 –1)

92

a1

Distance travelled by B in 3rd second,

DB 2a0

2 (2 × 3 –1)

52

a2

As per question, DA = DB

so 1 29 5a a2 2

or 1

2

a 5a 9

2. The bus moves a distance, 1x2

at2

The man moves a distance, x + 48 = v × t

or12 × 1 × t2 + 48 = 10t

or t2 – 20t + 96 = 0 or (t – 12) (t – 8)= 0

or t = 12s or t = 8s

Man will cross the bus after 8s. But againafter 12s bus will cross him due toaccelerated motion.

3. Let s = length of train,

t = time taken by full train to cross the pole,

a = uniform acc.of train.

from, v2 – u2 = 2as2 280 60 s2a

or 6400 3600 1400s2a a

The middle point of the train is cover a

distance s/2 700a

From, v2 – u2 = 2as

v2 – 602 7002aa

= 1400

v2 = 1400 + 3600 5000

= 70.7 km/h.

4. Let x be the distance travelled by thetruck when truck and car are side byside. The distance travelled by the carwill be (x + 150) as the car is 150 metrebehind the truck. Applying the formula

21s ut at2

, we have

21x 1.5 t2

___________ (1)

and (x + 150) 12

× (2) t2 _________ (2)

Here t is the common time.From eq. (1) and (2), we have

x 150 2x 1.5

Solving we get, x = 450 metre (truck) andx + 150 = 600 metre (car)

substituting the value of x in eq (1) we

get 450 21 1.5 t2

450 2t 6001.5

= 24.5sec

5. Since the train starts from rest its initialvelocity is zero.

the acceleration of the train = 2.0m/s2

time duration t1 = 30seconds

So, the distance travelled by the trainduring this period,

S1 = ut1 + 12 at1

2

S1 = 0 + 12 × 2 × (30)2 = 900m

velocity of the train, V = u + at = 0 + 2 × 30 = 60m/s

This is the maximum velocity, whenbreaks applied, the train comes to restthat means its final velocity is zero andmaximum velocity 60m/s becomes initialvelocity


KINEMATICSwhen breaks are applied, its

acceleration changes,

retardation v u 0 60

t 60

= –1ms–2.

So, the distance travelled by the train inthe second case

S2 2 2v u 0 60 60 36002a 2 1 2

= 1800m

(a) total distance moved by the train

= S1 + S2 = 900 + 1800 = 2700m = 2.7km

(b) maximum speed attained by the trainVman = 60m/s.

(c) the position of the train at half themaximum speed

22 2

3

30 0v u 900S2a 2 2 4

= 225m

6. Reaction time, t = 0.20sinitial speed of the car, u = 54km/hr

55418

= 15m/s

deceleration, a = –6.0ms–2.

distance travelled during the reactiontime, S1 = u × t1 = 15 × 0.2 = 3m

After applying the breaks, the cartravelled the distance,

222

2

0 15v u 225S2a 2 6 12

= 18.75m

total distance travelled by the can afterthe sees the need to put the brakes on

S = S1 + S2 = 3 + 18.75 = 21.75m 22m

7. Let x is the distance travelled by thepolice jeep and motorbike during t–10and t seconds respectively.

Velocity of bike = 72km/hr = u1 572

18

= 20m/sVelocity of jeep = 60km/h

u2 590

18 = 25m/s

Since the distance travelled by the both

are same,x = u1t = 20 × t ____________ (1)

x = u2 (t – 10) = 25 × (t – 10) _______ (2)

from (1) & (2)

20t = 25(t – 10)

20t = 25t – 250

5t = 250 t = 50sec.

the jeep catch up the bike after adistance

S = u1t (or) u2 (t – 10)

= 20 × 50 = 1000m = 1.0km

8. If one over takes the other, then time bythe both are same, = t

If we consider the length of the cars, thedifference of distance travelled by theboth cars = 5 + 5 = 10m

Velocity of car A & VA = 60km/h

5 506018 3

m/s

Velocity of car B, VB = 42km/hr

5 3542 m/s18 3

distance travelled by car A

A A50S V t t3

distance travelled by car B

B B35S V t t3

50 35t t 103 3

50 35t 103 3

30t 2s15

Since the fastest car travelled moredistance,

the total road distance used for theovertake

= VA × t + 5m 503

× 2 + 5 = 38.33m

= 38m

MOTION IN A STRAIGHT LINE

EXCELLENCIA JUNIOR COLLEGESSHAMI RPE T | M ADHAP UR | SU CHI T RA | EC IL

38

Motion in a straight line deals with the motion ofan object which changes its position with timealong a straight line.

The study of the motion of objects withoutconsidering the cause of motion is calledkinematics.Rest and Motion:If the position of a body does not change withtime with respect to the surroundings then it issaid to be at rest, if not it is said to be in motion.Distance and Displacement:

Distance is the actual path covered by a movingparticle in a given interval of time whiledisplacement is the change in position vector,i.e.,a vector joining initial to final position. If a particlemoves from A to B as shown in Fig. the distancetravelled is s while displacement is

f ir r r

A

R = A + BB

Distance is a scalar while displacement is a vector,both having same dimensions[L] and SI unit ismetre.

rA BOh

(a)

(b)AB

Distance = rDisplacement = 2r

Distance = 2hDisplacement = 0

MOTION IN A STRAIGHT LINESYNOPSIS

x x

hh

distance = h + 2xdisplacement = h

(d)( c)

Distance = sDisplacement = s

s

The magnitude of displacement is equal to minimumpossible distance between two positions; soDistance Displacement

For motion between two points displacement issingle valued while distance depends on actual pathand so can have many values.

For a moving particle distance can never decreasewith time while displacement can.

Decrease in displacement with time means bodyis moving towards the initial position.

For a moving particle distance can never benegative or zero while displacement can benegative.

In general, magnitude of displacement is not equalto distance. However, it can be so if the motion isalong a straight line without change in direction.

Magnitude of displacement is less than the distancetravelled in case of curvilinear motion.Ex : If an object turns through an angle along acircular path of radius r from point A to point Btheni) distance d r

ii)displacement 2 2 sin / 2x r sin2

x

r

––

O

22

rr

B Ad

x x


39 EXCELLENCIA JUNIOR COL LEGESSHAMI RPE T | M ADHAP UR | SU CHI TR A | EC IL

WE-1:An athlete completes one round of a circulartrack of radius R in 40 s. What will be hisdisplacement at the end of 2 min 20 s? (2010E)

Sol. The time = 2 min 20s = 140s

BA R R

In 40 seconds athlete completes = 1 roundIn 140 seconds athlete will completes

=140

40round =3.5 rounds

The displacement in 3 rounds = 0 So net displacement =2R

W.E-2 : If the position of a particle along Y axis isrepresented as a function of time t by theequation y(t)=t3 then find displacement of theparticle during the period t to t t

Sol. Position at time t is 3y t t

Position at time t t is 3y t t t t

displacement of the particle from t to t t is

3 3y t t y t t t t

2 33 2 33 3t t t t t t t

2 323 3t t t t t Average Speed and Average Velocity: Average speed or velocity is a measure of overall

'fastness' of motion during a specified interval of time. The average speed of a particle for a given 'interval

of time' is defined as the ratio of distance travelled tothe time taken while average velocity is defined asthe ratio of displacement to time taken.

Thus, if a particle in time interval t aftertravelling distance r is displaced by r

Average speed = Distance travelled

Time taken ,

i.e., avg

rV

t

........(i)

Average velocity = Displacement

Time ,

i.e, avg

rV

t

............(ii)

Average speed is a scalar while average velocityis a vector both having same units (m/s) and

dimensions -1LT .

For a given time interval average velocity is singlevalued while average speed can have many val-ues depending on path followed.

During the motion if the body comes back to itsinitial position.

avg = 0V

( 0r

)

but 0avgV and finite ( 0r ) For a moving body average speed can never be

negative or zero while average velocity can benegative.

If a graph is plotted between distance (or displace-ment) and time, the slope of chord during a giveninterval of time gives average speed (or ) averagevelocity

avg

Δr= = tan = slope of chord

ΔtV

Instantaneous speed and Instantaneous velocity : Instantaneous velocity is defined as rate of change

of position of the particle with time. If the posi-tion r

of a particle at an instant t changes by

r in a small time interval t

limt o

r drV

t dt

The magnitude of velocity is called speed,i.e

Speed= velocity i.e, VV

Velocity is a vector while speed is a scalar , both

having same units (m/s) and dimensions -1LT . If during motion velocity remains constant

throughout a given interval of time, then themotion is said to be uniform and for uniform

motion, V=constant=Vavg

If velocity is constant, speed will also be constant.However , the converse may or may not be true,i.e,if speed=constant,velocity may or may not beconstant as velocity has a direction in addition tomagnitude which may or may not change,e.g, incase of uniform rectilinear motion

constantV and constantV

While in case of uniform circular motion

V=constant but V is not constant

, due tochange in direction.

velocity can be positive or negative as it is a vec-tor but speed can never be negative as it is magni-

tude of velocity, i.e, V V

As by definition dr

Vdt

, the slope of displace-

ment-versus time graph gives velocity , i.e,



40

tandr

V slope of r t curvedt

Time

v

(b)

dt

dA

Timet(a)

velo

city

As by definition dr

Vdt

; . ,i e dr Vdt

and from fig. Vdt dA

So, dA dr i.e, r dA Vdt i.e, area under velocity-time graph gives displace-ment while without sign gives distance.

Average speed is the total distance divided by total time

avg

Total distance travelledv =

Total time taken

If a body travels a distance 1s in time 1t , 2s in

time 2t and 3s in time 3t then the average speed

is 1 2 3

avg1 2 3

+ s + sv =

t + t + t

s

If an object travels distances s1, s

2, s

3 etc. with speeds

v1, v

2, v

3 respectively in the same direction. Then

Average speed =

1 2 3

31 2

1 2 3

s s sss s

v v v

If an object travels first half of the total journeywith a speed 1v and next half with a speed 2vthen its average speed is

1 2avg

1 2

1 2 1 2 1 2

2v vs + s 2s 2v = = = =

s s s s 1 1 v + v+ + +v v v v v v

If a body travels first 1/3 rd of the distance with aspeed v

1 and second 1/3rd of the distance with a

speed v2 and last 1/3rd of the distance with a speed

v3, then the average speed

avg

1 2 3

s s s+ +

3 3 3v =s s s

+ +3v 3v 3v

1 2 3avg

1 2 2 3 3 1

3v v vv

v v v v v v

If an object travels with speeds v1, v

2, v

3 etc.,

during time intervals t1, t

2, t

3 etc.,

then its average speed 1 1 2 2 3 3

1 2 3

v t v t v t ....

t t t ....

If t

1 = t

2 = t

3 = .... = t, then

1 2 3 1 2avg

v t v t v t ..... v v ...v

nt n

i.e.The average speed is equal to the arithmeticmean of individual speeds.

The actual path length traversed by a body iscalled distance.

Note : If x y zv = v i + v j+ v k

varies with time t then

drdr = vdt v =

dt

Integrating on both sides thenf 2

i 1

r t

r tdr = vdt

displacement of the particle from time 1t to 2tis given by

2 2 2

1 1 1

t t t

f i x y zt t t

ˆ ˆ ˆs = r - r = v idt + v jdt + v kdt

WE-3: A particle starting from point A, travellingupto B with a speed S, then upto point ‘C’with a speed 2S and finally upto ‘A’ with aspeed 3S, then find its average speed.

120O

B

C

A

Sol. Average speed = Total dis tan ce travelled

Total time taken

B

C

A

2S t2

2 /35 /6 /2

t13S

St3

Total distance travelled= AB + BC + CA = 2 r



Total time taken is

1 2 31 2 3

AB BC CAT t t t

V V V

r 2 r 5 r

2S 6S 18S

(arc length = radius angle)

avg

2 rV 1.8S

r 2 r 5 r2S 6S 18S

WE-4.For a man who walks 720 m at a uniformspeed of 2 m/s, then runs at a uniform speedof 4 m/s for 5 minute and then again walksat a speed of 1 m/s for 3 minutes. His averagespeed is

Sol. Where s1 = 720 m and

11

1

st 360s 6 min .

v

s2 = (4)(5)(60) = 1200m, t

2 = 300 s

s3 = (1)(3)(60) = 180 m, t

3 = 180 s

1 2 3avg

1 2 3

s s sv

t t t

=720 1200 180

360 300 180

avgv 2.5m / s WE-5. A particle is at x = +5 m at t = 0, x = -7m at

t = 6s and x = +2m at t = 10s. Find the averagevelocity of the particle during the interval (a)t = 0 to t = 6s; (b) t = 6s to t = 10s, (c) t = 0 tot = 10s.

Sol. x1 = +5m, t

1 = 0, x

2 = -7m; t

2 = 6s, x

3 =+2m, t

3=

10sa) The average velocity between the times t = 0to t = 6s is

2 11

2 1

x x 7 5v 2m / s

t t 6 0

b) The average velocity between the times t2 = 6s

to t3 = 10s is

3 22

3 2

2 7x x 9v 2.25m / s

t t 10 6 4

c) The average velocity between times t1 = 0 to t

3= 10s is

3 13

3 1

x x 2 5v 0.3m / s

t t 10 0

WE-6. A particle traversed one third of the distancewith a velocity v

0. The remaining part of the

distance was covered with velocity v1 for half

the time and with a velocity v2 for the

remaining half of time. Assuming motion tobe rectilinear, find the mean velocity of theparticle averaged over the whole time ofmotion.

Sol.t2/2t2/2t1

C DS

v0 v1 v2

A B

For AC; 0 1

Sv t

3 1

0

St

3v ---(1)

For CB; 2S

CD DB3

2 21 2

2S t tv v

3 2 2

21 2

4St

3 v v

Since, average velocity is defined as

avg2 2

1

S 2STotal Displacement 3 3v

t tTotal Time t2 2

1 2

S=

t + t

0 1 2avg

0 1 2

3v v vv

4v v v

Acceleration

The rate of change of velocity is equal toacceleration.Average and Instantaneous acceleration

If the velocity of a particle changes ( either in mag-nitude or direction or both) with time the motionis said to be accelerated or non-uniform. In caseof non-uniform motion if change in velocity is V

in time interval t , then

2 1avg

2 1

v - vΔva = =

Δt t - t

.......(1)

Instantaneous acceleration or simply accelerationis defined as rate of change of velocity with timeat a given instant. So if the velocity of a particle v

at time t changes by V

in a small time interval

t then

Δt 0

ΔV dVa = lim =

Δt dt

.......... (2)

Regarding acceleration it is worth noting that:

It is a vector with dimensions -2LT and SI units

2m / s . If acceleration is zero, velocity will be constant

and the motion will be uniform. However, ifacceleration is constant(uniform), motion is non-uniform and if acceleration is not constant thenboth motion and acceleration are non-uniform.

As by definition /V ds dt



42

So, 2

2

dv d ds d sa = = =

dt dt dt dt

.........(3)

i.e., if s

is given as a function of time, secondderivative of displacement w.r.t. time gives accel-eration.

If velocity is given as a function of position, thenby chain rule

dva =

dt

dv dx= .

dx dt or

dva = v

dx

dxas = v

dt

As acceleration a = dv / dt

, the slope of ve-

locity-time graph gives acceleration.Velocity - Time Graph

1t 2t

1V

2V

V

t

V

B

O

A

t

avg

Va tan

t

avga

= slope of the line joining two points in v-t graph.

The slope of a versus t curve, i.e, da

dt

is a

measure of rate of non-uniformity ofacceleration(usually it is known as JERK).

Acceleration can be positive or negative. Positiveacceleration means velocity is increasing with timewhile negative acceleration called retardation meansvelocity is decreasing with time.Equations of motion :

If a particle starts with an initial velocity u,acceleration a and it gains velocity v in time t then

dva =

dt or dv = a dt or

v t

u 0dv = a dt or V t

u 0V = a t

or v = u + at .......(1A)In vector form v = u + a t

........(1B) Now again by definition of velocity, Eqn. (1A)

reduces tods

= u + atdt

or s t

0 0ds = u + at dt or

21

s = ut + at2

........(2A)

In vector form 21

s = ut + at2

.............(2B)

From eqns. (1A) and (2A), we get

2v - u v - u1

s = u + aa 2 a

or 2 2 22as = 2uv - 2u + v + u - 2uvi.e., 2 2v = u + 2as .......(3)

In scalar form

v.v = u.u + 2a.s

or 2 2v = u + 2a.s

Distance travelled = average speed x time

2

u vs t

.............(4)

Distance travelled in nth second

n

1s = u + a n -

2

..... (5)

If acceleration and velocity are not collinear, v canbe calculated using

1/222v = u + at + 2uat cosθ

with at sinθ

tan =u + atcosθ

.......(6)

atv

u

GRAPHSCharacteristics of s-t and v-t graphs

Slope of displacement time graph gives velocity. Slope of velocity-time graph gives acceleration. Area under velocity-time graph gives displacement

Let us plot v-t and s-t graphs of some standardresults. To draw the following graphs assumethat the particle has got either a one-dimensionalmotion with uniform velocity or with constantacceleration.



S.No Situation v-t graph s-t graph Interpretation

v-t graph

v= constant

v

v= at

v

1

2

3

4

5

6

Uniform motion i)Slope of s-t graph = v = constant.ii) In s-t- graph s = 0 at t = 0

Uniformly acceleratedmotion with u=0 ands=0 at t = 0

Uniformly acceleratedmotion with

Uniformly acceleratedmotion with

Uniformly retarded motion till velocity becomes zero

Uniformly retarded then accelerated in opposite direction

I) u=0, i.e.,v=0 and t=0ii) u=0, i.e., slope of s-t graph at t=0, should be zeroiii) a or slope of v-t graph is constant

I) s = s at t=00

I)Slope of s-t graph at t=0 gives u. ii) Slope of s-t graph at t=t becomes zeroiii) In this case u can’t be zero.

0

I)At time t=t , v=0 orslope of s-t graph is zero.ii) In s-t graph slope or velocity first decreasesthen increases with opposite sign.

0

s-t graph

s=vt

t

s

t

s

v=u+at

v

us=ut+1/2 at

2s

v=u+at

v

u

s

s0

v=u-at

v

u

s

t0

t0

s

t0

0

0

0

u and

s s at t

0

0

0

u and

s s at t

0u I)slope of s-t graph at t=0 is not zero

, i.e., v or

ii) v or slope of s-t graph gradually goes on increasing

_

v

u

t0

O



44

If a particle starts from rest and moves with uniformacceleration ‘a’ such that it travels distances

ms and ns in the thm and thn seconds

then n

aS = 0 + 2n -1

2 1 (u=0)

m

aS = 0 + 2m -1

2 2

subtracting eq (2) from eq (1) n ms s

an m

.

A particle starts from rest and moves along astraight line with uniform acceleration. If ‘s’ is the

distance travelled by it in n seconds and ns is the

distance travelled in the thn second, then

n2

2n 1s

s n

(fraction of distance fallen in nth

second during free fall ) Moving with uniform acceleration, a body crosses

a point 'x' with a velocity 'u' and another point ‘y’with a velocity ‘v’. Then it will cross the mid pointof ‘x’ and ‘y’ with velocity

S S

v1 v

yxu

In this case acceleration 2 2 2 21 1v - u v - v

a = =2S 2S

on solving, 1v 2

uv 22 .

If a bullet loses (1/n)th of its velocity while passingthrough a plank, then the minimum no. of suchplanks required to just stop the bullet is .

u uu

n v = 0

x x x

Let m be the no of planks required to stop thebullet

2

2uu - - u = 2a

nx

1

2 20 - u = 2amx 2Dividing eq (2) with eq (1)

2 2

22

0 - u 2am=

2auu - - u

n

x

x

2

2 22

1

11 1

um

uu u

n n

2

2 2 2

1

1 211

nm

n n nn

n

2

2 1

nm

n

The velocity of a bullet becomes 1

n of the initial

velocity while penetrating a plank. The number ofsuch planks required to stop the bullet.

u uu

n v = 0

x x x

2

2u- u = 2a

nx

1

2 20 - u = 2amx 2

From eq (1) and eq (2) ; 2

2 1

nm

n

A bullet loses 1

n of its velocity while penetrating

a distance x into the target. The further distancetravelled before coming to rest.

u

x y

n

uu v=0

Let x is the distance covered by the bullet to lose

the 1

thn

of initial velocity

2 2v - u = 2as2

2uu - - u = 2a

nx

1

Let y is the further distance covered by the bulletto come to rest

2 20 - u = 2a x + y 2 From eq (1) and (2)

21

2

ny x

n



If the velocity of a body becomes

th1

n

of its initial

velocity after a displacement of ‘x’ then it will cometo rest after a further displacement of

u

x y

n

uv=0

2

2u- u = 2a

nx

1

2

2 u0 - = 2a

ny

2

From eq. (1) and eq (2) ; 2 1

xy

n

WE-7. A body covers 100cm in first 2 seconds and128cm in the next four seconds moving withconstant acceleration. Find the velocity ofthe body at the end of 8sec?

Sol. distance in first two seconds is

21 1 1

1s = ut + at

2S1

t1 t2

S2

1100 = 2u + a 4

2......(1)

distance in (2+4)sec from starting point is

2

1 2 1 2 1 2

1s + s = u t + t + a t + t

2

1228 = 6u + a 36

2......(2)

from eq (1) and (2)We get a = -6 cm/s2 , sub a=-6 in eq - (1)

1100 2u 6 4

2

2u = 112 u = 56 cm/sv = u + at = 56 - 6 x 8 v 8cm / s

WE-8. A car starts from rest and moves withuniform acceleration ‘a’. At the same instantfrom the same point a bike crosses with auniform velocity ‘u’. When and where willthey meet ? What is the velocity of car withrespect to the bike at the time of meeting ?

Sol.2

car

1s = at

2 1 ,

bikes = ut 2

if they meet at the same point

car bikes = s

21at = ut

2

2ut

a

2

bike

2u 2us ut u

a a

carv at 2u

carv w.r.t. bike at the time of meeting,

cb car bikev = v - v

= 2u - u = u

W.E-9: What does d v /dt

and v /

d dt repre-

sent? can these be equal ? can:

(a) d v /dt=0

while v / 0d dt

(b) v / 0d dt

while v / 0

d dt ?

Sol. v /d dt

represents time rate of change of speed

as v u

, while v /d dt

represents magnitude

of acceleration.If the motion of a particle has translationalacceleration (without change in direction)

as vv v , v

d dn n

dt dt

or vv

d dn

dt dt

[as n is constant]

or vv 0

d d

dt dt

In this case both these will be equal and notequal to zero.(a) The given condition implies that:

v / 0,d dt

i.e., . 0acc while v / 0d dt

,

i.e., speed =constant.This actually is the case of uniform circularmotion. In case of uniform circular motion

2dv u= a = = constant

dt r

while v =constant.

i.e., v 0d

dt

(b) v / 0d dt

means a = 0

, i.e., a = 0

or v / 0d dt

or v = constant

And when velocity v

is constant speed will be

constant,



46

i.e., speed v

=constant or v 0d

dt

So, it is not possible to have v

0d

dt

while v 0d

dt

.

WE-10:In a car race, car A takes time t less thancar B and passes the finishing point with avelocity v more than the velocity withwhich car B passes the point . Assumingthat the cars start from rest and travel with

constant accelerations 1a and 2a respec-

tively , the value of V

tis

Sol. The distance covered by both cars is sameThus, s

1=s

2=s

If the cars take time t1 and t

2 for the race and their

velocities at the end of race be v1 and v

2 , then it is

given that

1 2 2 1v v v and t t t

Now,

1 21 2

2 1

2 1

2a s 2a sv vv

t t t 2s 2s

a a

1 21 2

2 1

a a va a

t1 1a a

WE-11 : A drunkard walking in a narrow lane takes5 steps forward and 3 steps backward, followedagain by 5 steps forward and 3 steps backward,and so on. Each step is 1m long and requires1sec. How long the drunkard takes to fall in apit 13m away from the start?

Sol. Distance of the pit from the start =13-5=8mTime taken to move first 5m=5sec5 steps (i.e., 5m) forward and 3steps(i.e.,3m)backward means that net distance moved =5-3=2mand time taken during this process =5+3=8sec

Time taken in moving 8×8

8m = = 32sec2

Total time taken to fall in the pit =32+5=37sec

WE-12 : An particle travels inside a straighthollow tube 2m long of a particle acceleratorunder uniform acceleration. How long is theparticle in the tube if it enters at a speed of1000 m/s and leaves at 9000 m/s. What is itsacceleration during this interval ?

Sol. Let ‘a’ be the uniform acceleration of -particle.According to given problem s = 2m, v = 9000 m/s and u = 1000 m/s.Since v2 - u2 = 2aS,

2 29000 1000 2a 2 a = 2 x 107 m/s2

Let the particle remains in the tube for time ‘t’,then v u at

47

v u 9000 1000t 4 10 s

a 2 10

WE-13:A car starts from rest and moves withuniform acceleration of 5 m/s2 for 8 sec. If theacceleration ceases after 8 seconds then findthe distance covered in 12s starting from rest.

Sol. The velocity after 8 sec v = 0 + 5 x 8 = 40 m/sDistance covered in 8 sec

0

1s 0 5 64 160m

2

After 8s the body moves with uniform velocityand distance covered in next 4s with uniform ve-locity.s = vt = 40 x 4 = 160 mThe distance covered in12 s =160 +160 =320m.

WE-14 : A car is moving with a velocity of 40 m/s.The driver sees a stationary truck ahead at adistance of 200 m. After some reaction time

t the breaks are applied producing a

(reaction) retardation of 8 2/m s . What is themaximum reaction time to avoid collision ?

Sol. The car before coming to rest (v = 0)

22 2v u 2as 0 40 2 8 s

s 100m The distance travelled by the car is 100mTo avoid the clash, the remaining distance 200 -100 = 100m must be covered by the car with uni-form velocity 40 m/s during the reaction time t .

10040 t 2.5s

t

The maximum reaction time t 2.5s



WE-15 : Two trains one travelling at 54 kmph andthe other at 72 kmph are headed towards oneanother along a straight track. When theyare 1/2 km apart, both drivers simultaneouslysee the other train and apply their brakes. Ifeach train is decelerated at the rate of 1 m/s2,will there be a collision ?

Sol. Velocity of the first train is 54 kmph = 15 m/sDistance travelled by the first train before comingto rest

2

1

u 225s 112.5m

2a 2

Velocity of the second train is 72 kmph= 20 m/sDistance travelled by the second train before com-ing to rest

2

2

u 400s 200m

2a 2

Total distance travelled by the two trains beforecoming to rest = s

1 + s

2 = 112.5 + 200 = 312.5m

Because the initial distance of separation is 500mwhich is greater than 312.5m, there will be nocollision between the two trains.

WE-16. A bus accelerates from rest at a constant rate‘ ’ for some time, after which it deceleratesat a constant rate ‘ ’ to come to rest. If thetotal time elapsed is t seconds. Then evalu-ate following parameters from the given grapha) the maximum velocity achievedb) the total distance travelled graphically andc) Average velocity

Vmax

O

A

Bt1 t2

V

t

Sol. a) = Slope of line OA = max max

11

v vt

t

= Slope of line AB = max max

22

v vt

t

max max1 2

v vt t t

maxv

maxv t

b) Displacement = area under the v-t graph= area of OAB

= 1 2 max max

1 1 1base height t t v tv

2 2 2

= 21

t2

max

avg

vtotaldisplacement 1v t

total time 2 2

WE-17: A body starts from rest and travels a distanceS with uniform acceleration, then movesuniformly a distance 2S and finally comes torest after moving further 5S under uniformretardation. Find the ratio of averagevelocity to maximum velocity

Sol.

Vmax

O t1t1 t2 t3

v

t

area of ΔOAC max 1 1

max

1 2SS V t (or) t ;

2 V

area of ABCD max 2 2max

2S2S V t (or) t

V

area of ΔBDE max 3 3

max

1 10S5S V t (or) t

2 V

avg

TotaldisplacementV ;

Total time

avg

max max max

S 2S 5SV

2S 2S 10SV V V

m a x

81 4

SS

V

a v g

m a x

V 8 S 4

V 1 4 S 7

WE-18: The acceleration-displacement (a - x) graphof a particle moving in a straight line is asshown. If the particle starts from rest,thenfind the velocity of the particle when displace-ment of the particle is 12m.



48

2

2 8 10 12 x(m)

4

2a /m s

GFE

C

D

BA

OSol. 2 2v - u = 2ax

2 2v - uax =

2

2 2

2 2

v u ( u=0 )

v 2 area under a x graph Area under a-x graph =Area of OAE + Area of rectangle ABEF+ Area of trapezium BFGC + Area of CGD

Area 1 1 12 2 6 2 2 4 2 2 4 24

2 2 2

v 2 24 4 3m / s

WE-19: Velocity-time graph for the motion of acertain body is shown in Fig. Explain thenature of this motion. Find the initialvelocity and acceleration and write theequation for the variation of displacementwith time. What happens to the moving bodyat point B ? How will the body move afterthis moment ?

A711 15

tB

C

v

Sol. The velocity -time graph is a straight line with-ve slope, the motion is uniformly retarding oneupto point B and uniformly accelerated afterwith-ve side of velocity axis .At point B the body stops and then its direction ofvelocity reversed.

The initial velocity at point A is -10v = 7ms

The acceleration

-2 -2t 0v - vΔv 0-7a = = = = -0.6364ms -0.64ms

Δt Δt 11

The equation of motion for the variation ofdisplacement with time is

2 21s = 7t - 0.64t 7 0.32

2t t

WE-20: A particle starts from rest and acceleratesas shown in the graph. Determinea) the particle’s speed at t = 10s and at t = 20sb) the distance travelled in the first 20s.

-1

1

2

-2

-3

5 1015 20

a m/s2

t sec

Sol. a) Upto 10 sec the particle moves with uniformacceleration, hence the velocity at t = 10s,v

1 = u + at = 0 + 2 x 10 = 20m/s

From t = 10s to t = 15s the acceleration is zero,soThe velocity of the particle at t=15s is 20m/sVelocity at t = 20sec, v

2 = v

1 + at

= 20 + (-3)5 = 5 m/sb) Distance travelled in first 10 sec

221 1

1 1s ut a t 0 10 2 10 100m

2 2

Distance travelled when t = 10 sec to t = 15 sec

2s vt 20 5 100m Distance travelled when t = 15 sec to t = 20 sec

23

1s 20 5 3 5 62.5m

2

Total distance travelled in 20 sec = s1 + s

2 + s

3 =

100 + 100 + 62.5 = 262.5 m

W.E-21: Velocity (v) versus displacement (x) plotof a body moving along a straight line is asshown in the graph. The corresponding plotof acceleration (a) as a function of displace-ment (x) is (EAM-2014)

v

0 100 200

Vel

ocit

y

xDisplacement



0

a

1)x

x

x

Displacement100 200

O100 200

Displacement

3)

a

O

2)x100 200

Displacement

4)

100 200ODisplacement

a

Sol. From the given graph equation for velocity v = kxon differentiationdv

kvdt

------ (i)

dv

dt= k(kx) = k2x;

a = k2x and v = -kx + v0

on differentiation 0

dvkv k kx v

dt

a = k2x - kv0

-------(ii)So, according to the eq. (ii) the shape of a-x graphis as below

0 x

Equations of Motion for VariableAcceleration : When acceleration ‘a’ of the particle is a function

of time i.e., a = f(t)

dvf t

dt dv f t dt Integrating both sides within suitable limits, we have

v t

u 0

dv f t dt t

0

v u f t dt When acceleration ‘a’ of the particle is a function

of distance a = f(x)

dvf x

dt dv dx

f xdx dt

0

v x

u x

vdv f x dx 0

x2 2

x

v u 2 f x dx

WE-22: A particle located at x = 0, at time t = 0,starts moving along the positive x-directionwith a velocity v that varies as v = α x . Thedisplacement of particle varies with time as

Sol.dx dx

x dtdt x

On integrating, we get

x t

0 0

dxdt

x

[ at t = 0, x = 0 and let at any time t, particle be at x]x1/2

0

xt

1/ 2

or

1/2x t2

2x t

WE-23 : The acceleration (a) of a particle movingin a straight line varies with its displacement(S) as a = 2S. The velocity of the particle iszero at zero displacement. Find thecorresponding velocity-displacementequation.

Sol. a = 2S dv

v 2Sds

vdv 2SdS v s

0 0

vdv 2SdS v S2 2

0 0

v S2

2 2

2

2vS

2 v 2S

WE-24: An object moving with a speed of 6.25m/s, is decelerated at a rate given by

dv= -2.5 v

dt, where v is instantaneous

speed. The time taken by the object, to cometo rest, would be (AIEEE-2011)

Sol.dv

2.5 vdt

(or) 1

dv 2.5dtv

On integrating, within limits(v

1 = 6.25 m/s to v

2 = 0)

2

1

v 0 t1/2

v 6.25m/s 0

v dv 2.5 dt

01/2

6.252 v 2.5 t (or)

1/22 6.25

t 2s2.5



50

WE-25:The velocity of a particle moving in thepositive direction of the X-axis varies as

V K S where K is a positive constant.

Draw V-t graph.

Sol. V K s

0 0

S tdS dSK S K dt

dt S

2 212 and S=

4S Kt K t

2 21 12

4 2

dSV K t K t

dt

V t

O

V

tV t

The V-t graph is a straight line passing throughthe origin.

ACCELERATION DUE TO GRAVITY The uniform acceleration of a freely falling body

towards the centre of earth due to earth’s gravitationalforce is called acceleration due to gravity

It is denoted by ‘g’ Its value is constant for all bodies at a given place.

It is independent of size, shape, material, natureof the body.

Its value changes from place to place on thesurface of the earth.

It has maximum value at the poles of the earth.The value is nearly 9.83 m/s2.

It has minimum value at equator of the earth. Thevalue is nearly 9.78 m/s2.

On the surface of the moon, earthmoon

gg

6

The acceleration due to gravity of a body alwaysdirected downwards towards the centre of theearth, whether a body is projected upwards ordownwards.

When a body is falling towards the earth, itsvelocity increases, g is positive.

The acceleration due to gravity at the centre ofearth is zero.

MOTION UNDER GRAVITYEquation of motion for a body projectedvertically downwards : When a body is projected vertically downwards

with an initial velocity u from a height hthen a = g, s = h

a) v = u + gt b) h = ut + 2

1gt2

c) v2 - u2 = 2gh d) Sn = u +

2

g(2n-1)

In case of freely falling body u = 0, a = +g

a) v = gt b) 21

S gt2

c) 2v 2gS d) n

1S g n

2

For a freely falling body, the ratio of distancestravelled in 1 second, 2seconds, 3 seconds, .... =1 : 4 : 9 : 16....

For a freely falling body, the ratio of distancestravelled in successive seconds = 1 : 3 : 5 : 9 ....

A freely falling body passes through two points Aand B in time intervals of t

1 and t

2 from the start,

then the distance between the two points A and

B is = 21

22 tt

2

g

A freely falling body passes through two points Aand B at distances h

1 and h

2 from the start, then

the time taken by it to move from A to B is

t = 1212 hh

g

2

g

h2

g

h2

Two bodies are dropped from heights h1 and h

2simultaneously. Then after any time the distancebetween them is equal to (h

2~ h

1).

A stone is dropped into a well of depth ‘h’, thenthe sound of splash is heard after a time of ‘t’.

Time taken by the body to reach water, 1

2ht

g

Time taken by sound to travel a distance ‘h’,

2sound

ht

V

Time to hear splash of sound is

1 2t t t =sound

2h h+

g V A stone is dropped into a river from the bridge

and after ‘x’ seconds another stone is projected



down into the river from the same point with avelocity of ‘u’. If both the stones reach the water

simultaneously, then 1 t 2 t xS S

221 1gt u t x g t x

2 2

A body dropped freely from a multistoried buildingcan reach the ground in t

1 sec. It is stopped in its

path after t2 sec and again dropped freely from

the point. The further time taken by it to reach the

ground is 2 23 1 2t t t .

H1 H2

H3

t2

t1 t3=?

We know that H1 = H

2 + H

3

2 2 21 2 3

1 1 1gt gt gt

2 2 2

2 2 21 2 3t t t 2 2

3 1 2t t t

Equations of motion of a body ProjectedVertically up : Acceleration (a) = -g

a) v = u - gt b) s = ut - 2

1gt2

c) v2 - u2 = -2gh d) sn = u -

2

g(2n-1)

Angle between velocity vector and accelerationvector is 1800 until the body reaches the highestpoint.

At maximum height, v = 0 and a = g

2

max

2

max uHg2

uH (independent of mass

of the body) A body is projected vertically up with a velocity

‘u’ from ground in the absence of air resistance

‘R’. then. a dt t

i) a d

ut t

g

ii) Time of flight a d

2uT t t

g

A body is projected vertically up with a velocity‘u’ from ground in the presence of constant airresistance ‘R’. If it reaches the ground with avelocity ‘v’, thena) Height of ascent = Height of descent

b) Time of ascent a

mut

mg R

c) Time of descent d

mvt

mg R

d) a dt t

e) v mg Rv u

u mg R

f) For a body projected vertically up under airresistance, retardation during its motion is > g

At any point of the journey, a body possess thesame speed while moving up and while movingdown.

Irrespective of velocity of projection, all the

bodies pass through a height 2

g in the last second

of ascent. Distance traveled in the last second of

its journey 2

gu .

The change in velocity over the complete journeyis ‘2u’ (downwards)

If a vertically projected body rises through a height‘h’ in nth second, then in (n-1)th second it will risethrough a height (h+g) and in (n+1)thsecond itwill rise through height (h-g).

If velocity of body in nth second is ‘v’ then in

thn 1 second it is (v + g) and that in

thn 1 second is (v - g) while ascending

A body is dropped from the top edge of a towerof height ‘h’ and at the same time another body isprojected vertically up from the foot of the towerwith a velocity ‘u’.

h

u=0

a) The separation between them after ‘t’ seconds

is = h ut



52

b) The time after which they meet h

tu

c) The height at which they meet above the

ground = 2

2

ghh

2u

d) The time after which their velocities are equal

in magnitudes is u

t2g

e) If they meet at mid point, then the velocity ofthrown

body is u gh and its velocity of meeting is zerof) Ratio of the distances covered when themagnitudes of their velocities are equal is 1 : 3.

A body projected vertically up crosses a point Pat a height ‘h’ above the ground at time ‘ 1t ’

seconds and at time 2t seconds ( 1 2t and t aremeasured from the instant of projection) to samepoint while coming down.

h1t

t2

21/ 2h ut gt ; 2 2 2 0gt ut h This is quadratic equation in t

Sum of the roots, 1 2

2ut t

g (time of flight)

Velocity of projection , 1 2

1u g t t

2

Product of the roots, 1 2

2ht t

g

Height of P is 1 2

1h gt t

2

Maximum height reached above the ground

2

1 2

1H g t t

8

Magnitude of velocity while crossing P is

2 1g t t

2

A body is projected vertically up with velocity 1uand after ‘t’ seconds another body is projected

vertically up with a velocity 2u .

a) If 2 1,u u the time after which both the bodieswill meet with each other is

1 2h h

221 2

1 1

2 2u x gx u x t g x t

22

2 1

1u t gt

2xu u gt

for the first body..

b) If 1 2u u u , the time after which they meet

is u t

g 2

for the first body and

u t

g 2

for the second body.

Height at which they meet = 2 2

2 8

u gt

g

A rocket moves up with a resultant accelerationa. If its fuel exhausts completely after time ' 'tseconds, the maximum height reached by therocket above the ground is

1 2h h h

=2 2 21 1

at + a t2 2g

2 2 2

2

v a th = =

2g 2g

, v at

21 a

h at 12 g

An elevator is accelerating upwards with an

acceleration a. If a person inside the elevatorthrows a particle vertically up with a velocity u

relative to the elevator, time of flight is 2u

tg a

In the above case if elevator accelerates down,

time of flight is 2u

tg a

The zero velocity of a particle at any instant doesnot necessarily imply zero acceleration at thatinstant. A particle may be momentarily at rest andyet have non-zero acceleration.For example, a particle thrown up has zero velocityat its uppermost point but the acceleration at thatinstant continues to be the acceleration due togravity.



WE-26: Drops of water fall at regular intervals fromthe roof of a building of height h = 16m.The first drop striking the ground at the samemoment as the fifth drop is ready to leave fromthe roof. Find the distance between thesuccessive drops.

Sol. Step-I : Time taken by the first drop to touch the

ground = 2h

tg

For h = 16m, t 16 2 2

4g g

Time interval between two successive drops is

1 1 2t t t

n 1 4 g

Where n = number of dropsStep-II :Distance travelled by 1st drop

2

1

1 1 24 16 16

2 2S g t g m

g

Distance travelled by 2nd drop

2

2

1 1 23 9 9

2 2S g t g m

g

Distance travelled by 3rd drop

2

3

1 1 22 4 4

2 2S g t g m

g

Distance travelled by 4th drop

2

4

1 1 21

2 2S g t g m

g

Distance between 1st and 2nd drops=

1 2 16 9 7S S m Distance between 2nd and 3rd drops=

2 3 9 4 5S S m Distance between 3rd and 4th drops=

3 4 4 1 3S S m Distance between 4th and 5th drops=

4 5 1 0 1S S m WE-27: A body falls freely from a height of 125m

(g = 10 m/s2). After 2 sec gravity ceases toact Find time taken by it to reach theground ?

Sol. 1) Distance covered in 2s under gravity

221

1 1S gt 10 2 20m

2 2

Velocity at the end of 2s

V = gt = 10 2 = 20 m/sNow at this instant gravity ceases to act, there after velocity becomes constant.The remaining distance which is 125 - 20 = 105mis covered by body with constant velocity20 m/sTime taken to cover 105 m with constant velocityis given by

1 1

S 105t t 5.25s

V 20

Total time taken = 2 + 5.25 = 7.25 sWE-28: A parachutist drops freely from an

aeroplane for 10 seconds before the parachuteopens out. Then he descends with a netretardation of 2 m/s2. His velocity when hereaches the ground is 8 m/s. Find the heightat which he gets out of the aeroplane ?

Sol. Distance he falls before the parachute opens is

1S 1

100 4902

g m

Then his velocity , u=gt = 98.0 m/sVelocity on reaching ground = v=8 /m sretardation = 2 2/m s

2 22v u 2aS 2 2385S m

Total distance 1 2S S S = 2385 + 490 =2875 m( height of aeroplane)

WE-29: If a freely falling body covers half of itstotal distance in the last second of its journey.Find its time of fall

Sol. Suppose t is the time of free fall

2 12n

gS n and 2

2

gS n

nS

S

2

2 1 122

2

gn

gn

2 4 2 0n n and 2 2n sec

WE-30:A body is projected vertically up with a velocityu. Its velocity at half of its maximum height andat 3/4th of its maximum height are

Sol. From v2 - u2 = 2aS, here a = –g; when S = H/2, then

2

2 2 u uv u 2 g v

4g 2

When S = 3H/4, then

2

2 2 3u uv u 2 g v

4 2g 2



54

WE-31 : A stone is allowed to fall from the top of atower 300m height and at the same time an-other stone is projected vertically up from theground with a velocity 100 m/s. Find whenand where the two stones meet ?

Sol.300 m

300-x

x

height of the tower, h= 300mSuppose the two stones meet at a height x fromground after t seconds

r

r

St

u , 0ru u u , rS h

ht

u

3003sec

100

height of the stone from the point of projection is

21X = ut- g t

2

1100 3 9.8 9

2 255.9m

WE-32: A stone is dropped from certain heightabove the ground. After 5s a ball passesthrough a pane of glass held horizontally andinstantaneously loses 20% of its velocity. Ifthe ball takes 2 more seconds to reach theground, the height of the glass above theground is

Sol. In 5s velocity gained v = gt = 50 m/s. Velocityafter passing through the glass pane

8050 40m / s

100

Height of the glass pane above the ground is

21h ut gt

2 = 21

40 2 10 22

= 100m

Body Projected Vertically up from a Tower A body projected vertically up from a tower of

height ‘h’ with a velocity ‘u’ (or) a body droppedfrom a rising balloon (or) a body dropped froman helicopter rising up vertically with constantvelocity ‘u’ reaches the ground exactly below thepoint of projection after a time ‘t’. Then

TOWER

h-h

+ X - X

(a) Height of the tower is 2gt2

1uth

(b) Time taken by the body to reach the ground

t = g

gh2uu 2

(c) The velocity of the body at the foot of the tower

v = gh2u 2 (d) Velocity of the body after ‘t’ sec. is

v u gt (e) Distance between the body and balloon after

this time = 21

2gt

WE-33:A ball is thrown vertically upwards fromthe top of a tower. Velocity at a point ‘h’ mvertically below the point of projection istwice the downward velocity at a point ‘h’ mvertically above the point of projection. Themaximum height reached by the ball abovethe top of the tower is (MED-2012)

Sol. If AB is the tower then according to the problem,velocity at ‘P’ is given as twice the velocity at ‘Q’

2 2PV u gh ; 2 2QV u gh ; P QV 2 V

TOWER

A

B

C

Ph

hQH



2 22u 2gh u 2gh 2 10ghu

3

From the top of the tower maximum height reached

2uH

2g

10gh5h3

H2g 3

WE-34: From a tower of height H, a particle isthrown vertically upwards with a speed u.The time taken by the particle to hit theground is n times that taken by it to reachthe highest point of its path. The relationbetween H, u and n is ( jee main- 2014)

Sol. Time taken to reach the maximum height 1

ut

g

If t2 is the time taken to hit ground

i.e., 2

2

1H ut gt

2

But t2 = nt

1 ; So,

2 2

2

nu 1 n uH u g

g 2 g

2 2 2nu 1 n u

Hg 2 g

22 2gH nu n

WE-35 : A balloon starts from rest, moves verti-cally upwards with an acceleration g/8 ms-2.A stone falls from the balloon after 8 s fromthe start. Further time taken by the stone toreach the ground (g = 9.8 ms-2) is

Sol. The distance of the stone above the ground aboutwhich it begins to fall from the balloon is

21 gh 8 4g

2 8

The velocity of the balloon at this height can be

obtained from v = u + at; gv 0 8 g

8

This becomes the initial velocity (u) of the stoneas the stone falls from the balloon at the height h.

u ' g

For the total motion of the stone 1 21

h = -u t + gt2

214g gt gt

2 and 2 2 8 0t t

Solving for ‘t’ we get t = 4 and -2s. Ignoring nega-tive value of time, t = 4s.

Three bodies are projected from towers of sameheight as shown. 1st one is projected vertically upwith a velocity ‘u’. The second one is thrown downvertically with the same velocity and the third oneis dropped as a freely falling body. If t

1, t

2, t

3 are

the times taken by them to reach ground, then,

hh h

u=u u=0

1 2 3

For 1st body, 2

1 1

1

2h ut gt 1

For 2nd body, 2

2 2

1

2h ut gt 2

For 3rd body, 23

1

2h gt 3

from 21 t + 12 t

1 2 1 2 1 2

1

2h t t gt t t t

i) Height of the tower 1 2

1

2h gt t 4

ii) From eq (3) & (4),

3 1 2t t t 5iii) Equating R.H.S of (1) & (2),

velocity of projection 1 2

1u g t t

2 6

iv) Time difference between first two bodies to

reach the ground 2u

tg

7

Relative Motion in one dimension Velocity of one moving body with respect to other

moving body is called Relative velocity. A and B are two objects moving uniformly with

average velocities vA and v

B in one dimension, say

along x-axis having the positions xA(0) and x

B(0)

at t = 0. If x

A(t) and x

B(t) are positions of objects A and

B at time t then

A A Ax t x 0 v t ; B B Bx t x 0 v t



56

The displacement from object A to B is given by

BA B Ax t x t x t

B A B Ax 0 x 0 v v t

BA B Ax 0 v v t

Velocity of A w.r.t B is AB A Bv = v - v

2 2AB A B A Bv = v + v - 2v v cosθ

Two bodies are moving in a straight line in same

direction then, AB A Bv = v - v

( 00 )

Two bodies are moving in a straight line in opposite

direction then, AB A Bv = v + v

, ( 0180 )

If two bodies move with same velocity and in samedirection, then position between them does notvary with time.

If two bodies move with unequal velocity and insame direction, then position between them firstdecreases to minimum and then increases.

If the particles are located at the sides of n sidedsymmetrical polygon with each side a and eachparticle moves towards the other, then time afterwhich they meet is

Initial separation

TRe lative velocity of approach

aT

v vcosn

aT

v 1 cosn

and 2

aT

2v sinn

Shortcut to solve the problems

For Triangle n = 3 2a

T3v

;

For Square n = 4 a

Tv

For hexagon, n = 6 2a

Tv

WE.36: A passenger is at a distance ‘d’ from a bus,when the bus begins to move with a con-stant acceleration a. Then find the minimumconstant velocity with which the passengershould run towards the bus so as to catch is

Sol. passenger busS d S ;21

vt d at2

22 v v 2ad

at 2vt 2d 0 ta

for minimum velocity 2v 2ad 0 v 2ad WE.37. Two trains, each travelling with a speed of

137.5kmh , are approaching each other onthe same straight track. A bird that can fly at60kmph flies off from one train when theyare 90 km apart and heads directly for theother train. On reaching the other train, itflies back to the first and so on. Totaldistance covered by the bird before trainscollide is

Sol. Relative speed of trains 137.5 37.5 75kmh

Time taken by them to meet r

r

St

u 90 6

h75 5

Distance travelled by the bird, birdx V t

6

60 72km5

( birdV 160kmh ) ,

WE.38: On a two-lane road, car A is travelling witha speed of 36 kmph. Two cars B and Capproach car A in opposite directions withspeed of 54 kmph each. At a certain instant,when the distance AB is equal to AC, bothbeing 1 km, B decides to overtake A beforeC does. What minimum acceleration of carB is required to avoid an accident?

Sol. Velocity of a car A, AV 36km / h 10m / s

Velocity of car B, BV 54km / h 15m / s

Velocity of car C, CV 54km / h 15m / s Relative velocity of car B with respect to car A,

BA B AV V V 15 10 5m / s Relative velocity of car C with respect to car A,

CA C AV V V 15 10 25 /m s At a certain instance, both cars B and C are at thesame distance from car A i.e., s 1km 1000m Time taken (t) by car C to cover 1000m is

1000

t 40s25

The acceleration produced by car B is

211000 5 40 at

2

21600a 1m / s

1600



WE.39: A body is at rest at x = 0. At t = 0, it startsmoving in the positive x-direction with aconstant acceleration. At the same instantanother body passes through x = 0 moving inthe positive x-direction with a constant speed.

The position of the first body is given by 1x t

after time ‘t’ and that of the second body by

2x t after the same time interval. Which of

the following graphs correctly describes

1 2x x as a function of time ‘t’ ?[AIEEE -

2008]

1) 2)

1 2x x 1 2x x

O Ot t

O

3) 4)

1 2x x 1 2x x

O t t

Sol. As, 21

1x t = at

2 and 2x t vt

2

1 2

1x - x = at - vt

2 (parabola)

Clearly, graph (2) represents it correctly.

WE.40.A particle has an initial velocity ˆ ˆ3i 4 j and

an acceleration of ˆ ˆ0.4i 0.3j . It speed after10s is (AIEEE-2009)

Sol. v = u + at

ˆ ˆ ˆ ˆ3 4 0.4 0.3 10i j i j

ˆ ˆ ˆ ˆ ˆ ˆ3 4 4 3 7 7i j i j i j

v = 49 + 49

7 2

C.U.Q

DISTANCE AND DISPLACEMENT1. The numerical ratio of displacement to

distance is1) always less than 1 2) always greater than 13) always equal to 14) may be less than 1 or equal to one

2. The location of a particle is changed. Whatcan we say about the displacement anddistance covered by the particle?1) Both cannot be zero2) One of the two may be zero3) Both must be zero 4) Both must be equal

3. Consider the motion of the tip of the minutehand of a clock. In one houra) the displacement is zerob) the distance covered is zeroc) the average speed is zerod) the average velocity is zero1) a & b are correct 2) a,b & c are correct3) a & d are correct 4) b,c & d are correct

SPEED AND VELOCITY

4. The numerical value of the ratio of averagevelocity to average speed is1) always less than one 2) always equal to one3) always more than one4) equal to or less than one.

5. If a particle moves in a circle describing equalangles in equal intervals of time, then thevelocity vector1) remains constant. 2) changes in magnitude.3) changes in direction.4) changes both in magnitude and direction.

6. In which of the following examples of motion,can the body be considered approximately apoint objecta) a railway carriage moving without jerksbetween two stations.b) a monkey sitting on top of a man cyclingsmoothly on a circular trackc) a spinning cricket ball that turns sharplyon hitting the groundd) a trembling beaker that has slipped off theedge of a table1) a,b 2) b,c 3)a,c 4)b,d



58

7. An object may havea) varying speed without having varying ve-locityb) varying velocity without having varyingspeedc) non zero acceleration without having vary-ing velocityd) non zero acceleration without having vary-ing speed.1) a,b & c are correct 2) b & d are correct3) a,b & d are correct 4) a & d are correct

8. The distance travelled by a particle in astraight line motion is directly proportionalto t1/2, where t = time elapsed. What is thenature of motion ?1) Increasing acceleration2) Decreasing acceleration3) Increasing retardation4) Decreasing retardation

ACCELERATION9. If a body starts from rest, then the time in

which it covers a particular displacement withuniform acceleration is1) inversely proportional to the square root of thedisplacement2) inversely proportional to the displacement3) directly proportional to the displacement4) directly proportional to the square root of thedisplacement

10. Check up only the correct statement in thefollowing.1) A body has a constant velocity and still it canhave a varying speed2) A body has a constant speed but it can have avarying velocity3) A body having constant speed cannot have anyacceleration.4) None of these.

11. When the speed of a car is u, the minimumdistance over which it can be stopped is s. Ifthe speed becomes nu, what will be theminimum distance over which it can bestopped during the same time ?1) s/n 2) ns 3) s/n2 4) n2s.

12. The distance covered by a moving body isdirectly proportional to the square of the time.The acceleration of the body is1) increasing 2) decreasing3) zero 4) constant

13. Mark the incorrect statement for a particlegoing on a straight line.1) If the velocity and acceleration have oppositesign, then the object is slowing down.2) If the position and velocity have opposite sign,then the particle is moving towards the origin.3) If the velocity is zero at an instant, then theacceleration should also be zero at that instant.4) If the velocity is zero for a time interval,thenthe acceleration is zero at any instant within thetime interval.

MOTION UNDER GRAVITY14. B

1, B

2 and B

3 are three balloons ascending

with velocities v,2v and 3v, respectively. Ifa bomb is dropped from each when they areat the same height, then1) bomb from 1B reaches ground first

2) bomb from 2B reaches ground first

3) bomb from 3B reaches ground first4) they reach the ground simultaneously

15. The distances moved by a freely falling bodydur ing 1st, 2nd, 3rd,......nth second of its motionare proportional to1) even numbers 2) odd numbers3) all integral numbers4) squares of integral numbers

16. To reach the same height on the moon as onthe earth, a body must be projected up with

1) higher velocity on the moon.

2) lower velocity on the moon.

3) same velocity on the moon and earth.

4) it depends on the mass of the body.

17. At the maximum height of a body thrownvertically up

1) velocity is not zero but acceleration is zero.

2) acceleration is not zero but velocity is zero.

3) both acceleration and velocity are zero.

4) both acceleration and velocity are not zero.

18. A ball is dropped freely while another isthrown vertically downward with an initialvelocity ‘v’ from the same pointsimultaneously. After ‘t’ second they areseparated by a distance of

1) 2

vt 2)

2gt2

1 3) vt 4)

2gt2

1vt



19. The average velocity of a freely falling bodyis numerically equal to half of the accelerationdue to gravity. The velocity of the body as itreaches the ground is

1) g 2) 2

g3)

2

g4) g2

20. Two bodies of different masses are droppedsimultaneously from the top of a tower. If airresistance is proportional to the mass of thebody, then,1) the heavier body reaches the ground earlier.2) the lighter body reaches the ground earlier.3)both the bodies reach the ground simultaneously.4) cannot be decided.

21. A man standing in a lift falling under gravityreleases a ball from his hand. As seen byhim, the ball1) falls down 2) remains stationary3) goes up 4) executes SHM

22. A particle is dropped from certain height. Thetime taken by it to fall through successivedistances of 1 m each will be

1) all equal, being equal to 2 / g second2) in the ratio of the square roots of the integers 1,2, 3, ...........3) in the ratio of the difference in the square roots of the integers, i.e.,

1, 2 1 , 3 2 , 4 3 ,......

4) in the ratio of the reciprocals of the square roots

of the integers, i.e., 1 1 1

, , ,......1 2 3

23. A body, freely falling under gravity will haveuniform1)speed 2)velocity 3)momentum 4)acceleration

24. A person standing near the edge of the top ofa building throws two balls A and B. The ballA is thrown vertically upward and B is thrownvertically downward with the same speed, The

ball A hits the ground with a speed AV and

the ball B hits the ground with a speed BV .then1) A BV V 2) A BV V 3) A BV V

4) the relation between AV and BV depends onheight of the building above the ground

25. A lift is coming from 8th floor and is just aboutto reach 4th floor. Taking ground floor asorigin and positive direction upwards for all

quantities, which one of the following iscorrect?1) x<0, v<0, a>0 2) x>0, v<0, a<03) x>0, v<0, a>0 4) x>0, v>0, a<0

GRAPHS26. Choose the correct statement :

1) The area of displacement - time graph givesvelocity.2) The slope of velocity - time graph givesacceleration.3) The slope of displacement - time graph givesacceleration.4) The area of velocity - time graph gives averagevelocity.

27. Velocity-time graph of a body thrownvertically up is 1) a straight line 2) a parabola 3) a hyperbola 4) circle

28. Velocity - displacement graph of a freelyfalling body is1) straight line passing through the origin2) straight line intersecting ‘x’ and ‘y’ axes3) parabola 4) hyperbola

29. Displacement - time graph of a body projectedvertically up is1) a straight line 2) a parabola3) a hyperbola 4) a circle

30. The displacement - time graphs of two bodiesA and B are OP and OQ respectively. IfPOX is 600 and QOX is 450, the ratio ofthe velocity of A to that of B is

Y

d

O t X

Q

P

600

450

1) 2:3 2) 3 : 1 3) 1: 3 4) 3:131. If the distance travelled by a particle and

corresponding time be laid off along y and xaxes respectively, then the correct statementof the following is1) the curve may lie in fourth quadrant2) the curve lies in first quadrant3) the curve exhibits peaks corresponding tomaxima4) the curve may drop as time passes



60

32. In relation to a velocity - time graph1) the curve can be a circle2) the area under the curve and above the timeaxis between any two instants gives the averageacceleration3) the slope at any instant gives the rate of changeof acceleration at that instant4) the area under the curve and above the timeaxis gives the displacement

33. The displacement - time graph of a particlemoving with respect to a reference point is astraight line1)the reference point is stationary with zero velocity2) the acceleration of the object is zero3) body moves with uniform velocity4) all the above

34. For a uniform motion1) the velocity - time graph is a straight line parallelto time axis2) the position - time graph is a parabola3) the acceleration - time graph is a straight lineinclined with time axis4) the position - time graph is a straight line

35. Figure shows the displacement- time graphof a particle moving on the x-axis

x

t0t

O

1) the particle is continuously going in positive Xdirection2) the particle is at rest3) the velocity increases up to a time 0t and thenbecomes constant.4) the particle moves at constant velocity up to atime 0t and then stops.

36. The variation of quantity A with quantity B.plotted in the Fig. describes the motion of aparticle in a straight line.a) Quantity B may represent time.b) Quantity A is velocity if motion is uniform.c) Quantity A is displacement if motion is uniform.d) Quantity A is velocity if motion is uniformlyaccelerated.

A

BO

v

t

1) a,c,d 2) b,c,d 3) a,b 4) c,d37. The displacement-time graph of a moving

particle is shown in Fig. The instantaneousvelocity of the particle is negative at the point

Time

Dis

plac

emen

t

CD

E F

O1) D 2) F 3) C 4) E

38. Which of the following option is correct forhaving a straight line motion represented bydisplacement-time graph.

sOA

B

C

D

t

1) The object moves with constantly increasingvelocity from O to A then it moves with constantvelocity.2) Velocity of the object increases uniformly.3) Average velocity is zero.4) The graph shown is impossible.

39. The displacement of a particle as a function oftime is shown in the figure. The figure shows that

1 2 3 4Time in seconds

1

2

Dis

plac

emen

t

1) the particle starts with certain velocity but themotion is retarded and finally the particle stops2) the velocity of the particle is constant through out3) the acceleration of the particle is constant throughout4) the particle starts with constant velocity, thenmotion is accelerated and finally the particle moveswith another constant velocity.



40. A uniform moving cricket ball is turned backby hitting it with a bat for a very short timeinterval.Show the variation of its accelerationwith time. (Take acceleration in the back warddirection as positive)

1.

3.

2.

4.

t t

tt

a

a

a

a

RELATIVE VELOCITY41. A small body is dropped from a rising balloon.

A person A stands on ground, while anotherperson B is on the balloon. Choose the correctstatement : Immediately, after the body isreleased.1) A and B, both feel that the body is coming(going) down.2) A and B, both feel that body is coming up.3) A feels that the body is coming down, while Bfeels that the body is going up4) A feels that the body is going up, while B feelsthat the body is going down.

42. Seeta is moving due east with a velocity of

1v m / s and Geeta is moving due west with a

velocity of 2v m / s . The velocity of Seeta withrespect to Geeta is

1) 1 2v v due east 2) 1 2v v due east

3) 1 2v v due west 4) 1 2v v due west

C.U.Q - KEY1) 4 2) 1 3) 3 4) 4 5) 3 6) 17) 2 8) 4 9) 4 10) 2 11) 4 12) 413) 3 14) 1 15) 2 16) 2 17) 2 18) 319) 1 20) 3 21) 2 22) 3 23) 4 24) 325) 1 26) 2 27) 1 28) 3 29) 2 30) 231) 2 32) 4 33) 4 34) 1 35) 4 36) 137) 4 38) 3 39) 1 40) 1 41) 4 42) 1

LEVEL - I (C.W)

DISPLACEMENT AND DISTANCE1. A body is moving along the circumference of

a circle of radius ‘R’ and completes half ofthe revolution. Then, the ratio of itsdisplacement to distance is1) : 2 2) 2:1 3) 2 : 4) 1:2

2. A body completes one round of a circle ofradius ‘R’ in 20 second. The displacement ofthe body after 45 second is

1) 2

R2) 2 R 3) R2 4) 2R

SPEED AND VELOCITY3. If a body covers first half of its journey with

uniform speed v1 and the second half of the

journey with uniform speed v2, then the

average speed is

1) 1 2v v 2) 1 2

1 2

2 v v

v v

3) 1 2

1 2

v v

v v 4) 1 2v v4. A car is moving along a straight line, say OP

in figure. It moves from O to P in 18 s andreturn from P to Q in 6 s. What are the aver-age velocity and average speed of the car ingoing from O to P and back to Q?

1) 1 110m s , 20m s 2) 1 120m s ,10m s

3) 1 110m s ,10m s 4) 1 120m s , 20m s

5. For a body moving with uniform acceleration‘a’, initial and final velocities in a time inter-val ‘t’ are ‘u’ and ‘v’ respectively. Then, itsaverage velocity in the time interval ‘t’ is

1) v at 2)at

v2

3) v at 4)at

u2

ACCELERATION6. A body moves with a velocity of 3m/s due

east and then turns due north to travel withthe same velocity. If the total time of travelis 6s, the acceleration of the body is

1) 3 m/s2 towards north west

2) 2

1 m/s2 towards north west

3) 2 m/s2 towards north east 4) all the above



62

7. If a body travels 30m in an interval of 2s and50m in the next interval of 2s, then theacceleration of the body is 1) 10 m/s2 2) 5 m/s2 3) 20 m/s2 4) 25 m/s2

8. A bullet travelling horizontally looses 1/20th

of its velocity while piercing a wooden plank.Then the number of such planks required tostop the bullet is1) 6 2) 9 3) 11 4) 13

9. If 2 0.4nS n find initial velocity and

acceleration1) 2.2 units, 0.4 units 2) 2.1 units, 0.3 units3) 1.2 units, 0.4 units 4) 2.2 units, 0.3 units

10. A particle starts moving from rest underuniform acceleration. It travels a distance ‘x’in the first two seconds and a distance ‘y’ inthe next two seconds. If y = nx, then n=(1993 E)1) 1 2) 2 3) 3 4) 4

11. A particle is moving in a straight line with ini-tial velocity ‘u’ and uniform acceleration ‘a’.If the sum of the distances travelled in tth and(t+1)th second is 100cm, then its velocity af-ter ‘t’ seconds in cm/s is1) 20 2) 30 3) 80 4) 50

12. A particle is moving with uniform accelera-tion along a straight line ABC. Its velocity at‘A’ and ‘B’ are 6 m/s and 9 m/s respectively.If AB : BC = 5 : 16 then its velocity at ‘C’ is1) 9.6 m/s 2) 12 m/s 3) 15 m/s 4) 21.5 m/s

13. A car moving on a straight road acceleratesfrom a speed of 4.1 m/s to a speed of 6.9 m/sin 5.0 s. Then its average acceleration is1) 0.5m/s2 2) 0.6m/s2 3) 0.56m/s2 4) 0.65m/s2

MOTION UNDER GRAVITY14. A body projected vertically upwards with a

velocity of 19.6 m/s reaches a height of 19.8m on earth. If it is projected vertically up withthe same velocity on moon, then the maxi-mum height reached by it is1) 19.18 m2) 3.3 m 3) 9.9 m 4) 118.8 m

15. A ball is thrown straight upward with a speedv from a point h meter above the ground. Thetime taken for the ball to strike the ground is

1) 2

v 2hg1 1

g v

2) 2

v 2hg1 1

g v

3) 2

v 2hg1 1

g v

4) 2

v 2hg2

g v

16. A ball is dropped on the floor from a height of10m. It rebounds to a height of 2.5m. If theball is in contact with the floor for 0.01 s, thenthe average acceleration during contact isnearly

1) 2500 2m / s upwards

2) 21800 2m / s downwards

3) 21500 2m / s upwards

4) 21500 2m / s downwards17. A body falling from rest has a velocity ‘v’ af-

ter it falls through a distance ‘h’. The distanceit has to fall down further, for its velocity tobecome double, is ..... times ‘h’.1) 5 2) 1 3) 2 4) 3

RELATIVE VELOCITY18. A ball is dropped from a building of height

45m. Simultaneously another ball is thrownup with a speed 40m/s. The rate of change ofrelative speed of the balls is

1) 20 1ms 2) 140 ms 3) 130 ms 4) 10 ms

19. Two cars 1 & 2 starting from rest are moving

with speeds 1 2v and v m/s 1 2> v )(v . Car 2 isahead of car ‘1’ by s meter when the driverof the car ‘1’ sees car ‘2’. What minimum re-tardation should be given to car ‘1’ to avoidcollision. (2002 A)

1) 1 2v v

s

2) 1 2v v

s

3) 2

1 2v v

2s

4)

2

1 2v v

2s

20. Two cars are travelling towards each otheron a straight road at velocities 15 m/s and 16m/s respectively. When they are 150m apart,both the drivers apply the brakes and the carsdecelerate at 3 m/s2 and 4 m/s2 until they stop.Separation between the cars when they cometo rest is1) 86.5 m 2) 89.5 m 3) 85.5 m 4) 80.5 m

LEVEL - I (C.W) - KEY01) 3 02) 2 03) 2 04) 1 05) 2 06) 207) 2 08) 3 09) 1 10) 3 11) 4 12) 313) 3 14) 4 15) 1 16) 3 17) 4 18) 419) 4 20) 4



LEVEL - I (C.W) - HINTS

1.

A B

C

Displacement : Distance : 2R R2. In 40sec body completes two revolutions.

In 5 sec it covers 1/4 th of the circle and angle

traced is 2

. So displacement 2 sin

2s R

3. Average speed = 1 2

1 2

s s

t t

1 2

1 2

2v vv =

v + v

4. avgv =totaldisplacement

total time = 1 2

1 2

s s

t t

5. avg

v u v v atv

2 2

6.

WW E

N

S

-3i3i

4j

v ua

t

; i f1 2v = v i, v = v j

f iΔv = v - v 2 1= v j- v i ; 2

1

vtan θ =

v

7.s1

t1

s2

t2 ;

21 1 1

1

2s ut t

2

1 2 1 2 1 2

1

2s s u t t a t t

8.1 1

2020

nn ; no. of planks =

1n2

n2

9.1

2ns u an a 1

2u a an

..........(1)

2 0.4ns n ..........(2)

from (1) and (2) 0.4n an ; 1

22

u a

10. 21x a 2

2 ; 21

x y a 42

11. 2 12t

ts u t ; 1 2 1

2t

ts u t

1 100t ts s ; v u ft

12.

A B C

5x 16x1v 2v

3v

s1 s2

;2 22 23 22 1

1 22 2

v vv va

s s

13.v u

at

14.2

2

uh

g ;

1h

g E M

M E

h g

h g

15.21

h vt gt2

; 2 2 2 0gt vt h

16. a= 2 12gh 2gh

t

17. 2v 2gh ; 24v 2gx

18. Relative acceleration is zero as ‘g’ is downwardsfor the both the bodies.

19. 1 2relu v v ; 0relv ; 2 2 2rel relv u as

20.

u1

s1 s2s

u2v1=0 v2=0

s2 21 1 1 12v u a s ; 2 2

2 2 2 22v u a s

1 2s s s s .

LEVEL - I (H.W)

DISPLACEMENT AND DISTANCE1. A body moves from one corner of an

equilateral triangle of side 10 cm to the samecorner along the sides. Then the distance anddisplacement are respectively1) 30 cm & 10 cm 2) 30 cm & 0 cm3) 0 cm & 30 cm 4) 30 cm & 30 cm.

SPEED AND VELOCITY2. For a train that travels from one station to

another at a unifor m speed of 40 kmh–1 andreturns to final station at speed of 60 kmh–1,then its average speed is1) 98 km/hr 2) 0 km/hr3) 50 km/hr 4) 48 km/hr

3. If the distance between the sun and the earthis 1.5x1011 m and velocity of light is 3x108 m/s, then the time taken by a light ray to reachthe earth from the sun is1) 500 s 2) 500 minute 3) 50 s 4) 35 10 s



64

ACCELERATION

4. A body is moving with velocity 130mstowards east. After 10s its velocity becomes

140ms towards north. The averageacceleration of the body is [AIPMT 2011]

1) 27 ms 2) 27 ms 3) 25ms 4) 21ms

5. A body starting with a velocity ‘v’ returns toits initial position after ‘t’ second with thesame speed, along the same line.Acceleration of the particle is

1) t

v22) zero 3)

t2

v4)

v2

t

6. A body starting from rest moving with uniformacceleration has a displacement of 16 m infirst 4 s and 9 m in first 3 s. The accelerationof the body is1) 1 ms–2 2) 2 ms–2 3) 3 ms–2 4) 4 ms–2

7. A body starts from rest and moves with anuniform acceleration. The ratio of distancecovered in the nth second to the distancecovered in ‘n’ second is

1)

2n

1

n

2 2)

n

1

n

12 3)

n

1

n

22 4) 2n

1

n

2

8. A bus accelerates uniformly from rest andacquires a speed of 36kmph in 10s. Theacceleration is1) 1 m/s2 2) 2 m/s2 3) 1/2 m/s2 4) 3 m/s2

9. Speeds of two identical cars are U and 4U ata specific instant. The ratio of the respectivedistances in which the two cars are stoppedfrom that instant is1) 1:1 2) 1:4 3) 1:8 4) 1:16

10. A car moving along a straight highway withspeed of 1126Kmh is brought to a stop within a distance of 200m. what is the retardationof the car1) -23.06ms 2) -24ms 3) -25.06ms 4) -26ms

MOTION UNDER GRAVITY11. Two balls are projected simultaneously with

the same velocity ‘u’ from the top of a tower,one vertically upwards and the othervertically downwards. Their respective timesof the journeys are t

1 and t

2. At the time of

reaching the ground, the ratio of their finalvelocities is1) 1:1 2) 1:2 3) 2:3 4) 2:1

12. Two bodies are projected simultaneously withthe same velocity of 19.6 m/s from the top ofa tower, one vertically upwards and the othervertically downwards. As they reach theground, the time gap is 1) 0 s 2) 2 s 3) 4 s 4) 6 s

13. Two bodies begin to fall freely from the sameheight. The second one begins to fall s af-ter the first. The time after which the 1st bodybegins to fall, the distance between the bod-ies equals to l is

1) 2

l

g

2)

g

l

3) 2

lg

4) 2

g

l

14. A balloon is going upwards with velocity 12m/sec. It releases a packet when it is at aheight of 65 m from the ground. How muchtime the packet will take to reach the ground

2g 10m / sec

1) 5 sec 2) 6 sec 3) 7 sec 4) 8 sec15. A body thrown up with some initial velocity

reaches a maximum height of 50m. Anotherbody with double the mass thrown up withdouble the initial velocity will reach amaximum height of1) 100m 2) 200m 3) 400m 4) 50m

16. The distance moved by a freely falling body(starting from rest) during the 1st, 2nd and3rd ... nth second of its motion, areproportional to1) (n-1) 2) (2n-1) 3) (n2-1) 4) (2n-1)/n2

17. A ball released from a height ‘h’ touches theground in ‘t’s. After t/2s since dropping, theheight of the body from the ground

1) 2

h2)

4

h3)

4

h34)

3

2

h

18. A boy standing at the top of a tower of 20 m

height drops a stone Assuming 210 g ms ,

the velocity with which it hits the ground is [AIPMT 2011]

1) 120 ms 2) 140 ms 3) 15ms 4) 110ms

19. A ball thrown vertically upwards with an ini-tial velocity of 1.4 m/s returns in 2s. The to-



tal displacement of the ball is1) 22.4 cm 2) zero 3) 44.8 m 4) 33.6m

20. A stone is dropped from a certain height whichcan reach the ground in 5s. It is stopped mo-mentarily after 3s and then it is again re-leased. The total time taken by the stone toreach the ground will be1) 6s 2) 6.5s 3) 7s 4) 7.5s

RELATIVE VELOCITY21. What are the speeds of two objects if, when

they move uniformly towards each other, theyget 4 m closer in each second and when theymove uniformly in the same direction with theoriginal speeds, they get 4 m closer each 10s?1) 2.8 m/s and 12 m/s2) 5.2 m/s and 4.6 m/s3) 3.2 m/s and 2.1 m/s4) 2.2 m/s and 1.8 m/s

22. Two trains are each 50m long movingparallel towards each other at speeds10 m/s and 15 m/s respectively, at what timewill they pass each other?1) 8 s 2) 4 s 3) 2 s 4) 6 s

23. A ball is dropped from the top of a building100 m high. At the same instant another ball

is thrown upwards with a velocity of 140 ms

form the bottom of the building. The two ballswill meet after.1) 5 s 2) 2.5 s 3) 2s 4) 3 s

LEVEL - I (H.W) - KEY01) 2 02) 4 03) 1 04) 3 05) 1 06) 207) 1 08) 1 09) 4 10) 1 11) 1 12) 313) 1 14) 1 15) 2 16) 2 17) 3 18) 119) 2 20) 3 21) 4 22) 2 23) 2

LEVEL - I (H.W) -HINTS1. Displacement = shortest distance between initial

point and final point

2.1 2

avg1 2

2v vv

v v

3.s = vt ,s

t =v

4.

WW E

N

S

-30i

30i

40j

i 1v = v i ;f 2v = v j ; f iΔv = v - v =

2 1v j - v i

2 21 2Δv = v + v ;

Δva =

Δt

5.v u

at

; 6. n

1s a n

2

;21

s an2

7. 2 12n

as n :

2

2

as n ; 2

2 1ns n

s n

8. v = u + at ;9. 2 2v u 2as ; v = 0 both the cases

2u s ;

2

1 1

2 2

u s

u s

10. 2 2v u 2as

11. 2v u 2gh is same for both the bodies.

12.2u

tg

13.2

1

gtH

2 ;

2

2

g tH

2

; 1 2-l H H

14.21

h ut gt2

15.2

max

uH

2g ( independent of mass)

16. n

1s g n

2

;3 5

Ratio = : :2 2 2

g g g..... 2 1

2

gn

ns 2 1n

17. 21

h gt2

18. v 2gh19. Since the ball returns back to its initial position,

the displacement is zero.

20. 1 2h h h , 2 2 21 2

1 1 1gt gt gt

2 2 2 ; tot 1 2t t t

21. A Bv v 4m/s ; A B

4v v

10m/s

22.1 2 1 2

r 1 2

l l l lt

v v v

23. rr

r 1 2

d ht a 0

v u u

; 1 0u

LEVEL - II (C.W)

DISPLACEMENT AND DISTANCE1. A person moves 30m north and then 20m

towards east and finally 30 2 m in south-west direction. The displacement of theperson from the origin will be



66

1) 10m along north 2) 10 m along south3) 10m along west 4) zero

SPEED AND VELOCITY

2. If a car covers 2/5th of the total distance with

1v speed and 3/5th distance with 2v thenaverage speed is

1) 1 2

1v v

2 2) 1 2v v

2

3)

1 2

1 2

2v v

v v 4)1 2

1 2

5v v

3v 2v3. Four persons A, B,C and D initially at the

corners of a square of side length ‘d’. If everyperson starts moving with same speed v suchthat each one faces the other always, theperson will meet after time

1) d

v2)

2d

v3)

d

2v4)

d

2v4. A man walks on a straight road from his home

to a market 2.5 km away with a speed of 5km/h. Finding the market closed, he instantlyturns and walks back home with a speed of7.5 km/h. What is the (a) magnitude of aver-age velocity and (b) average speed of the manover the time interval 0 to 50 min (in kmph).1) 0,4 2) 0,6 3) 0,8 4)0,12

ACCELERATION5. A starts from rest and moves with acceleration

a1. Two seconds later, B starts from rest and

moves with an acceleration a2. If the

displacement of A in the 5th second is thesame as that of B in the same interval, theratio of a

1 to a

2 is

1) 9:5 2) 5:9 3) 1:1 4) 1:36. A body travels 200cm in the first two seconds

and 220cm in the next 4 seconds withdeceleration. The velocity of the body at theend of the 7th second is1) 20 cm/s 2) 15 cm/s 3) 10 cm/s 4) 0 cm/s

7. A bullet moving at 20 m/sec. It strikes awooden plank and penetrates 4 cm beforecoming to stop. The time taken to stop is1) 0.008 sec 2) 0.016 sec3) 0.004 sec 4) 0.002 sec

8. An automobile travelling with a speed of60km/h can brake to stop within a distanceof 20m. If the car is going twice as fast i.e.,120km/h the stopping distance will be1) 20 m 2) 40 m 3) 60 m 4) 80 m

9. A police party is moving in a jeep at a constantspeed v. They saw a thief at a distance x on amotorcycle which is at rest. The moment thepolice saw the thief, the thief started atconstant acceleration . Which of thefollowing relations is true if the police is ableto catch the thief? [2011-E]

1) 2v x 2) 2v 2 x 3) 2v 2 x 4) 2v x

10. Velocity of a body moving with uniformacceleration of 3m/s2 is changed through30m/s in certain time. Average velocity ofbody during this time is 30m/s. Distancecovered by it during this time is1) 300 m 2) 200 m 3) 400 m 4) 250 m

11. A person is running at his maximum speed of4 m/s to catch a train. When he is 6m fromthe door of the compartment the train startsto leave the station at a constant accelera-

tion of 21 /m s . Find how long it takes him tocatch up the train1. 2sec 2. 3 sec 3. 4 sec 4. none

12. A body is moving along the +ve x-axis with

uniform acceleration of 24ms . Its velocity

at x=0 is 110 .ms The time taken by the bodyto reach a point at x=12m is

1) 2 ,3s s 2) 3 ,4s s 3) 4 ,8s s 4) 1 , 2s s

MOTION UNDER GRAVITY13. A freely falling body takes ‘t’ second to travel

first (1/x)th distance. Then, time of descent is

1) x

t 2) xt 3)

t

x4)

xt

1

14. The distance travelled by a body during lastsecond of its upward journey is ‘d’, when thebody is projected with certain velocity verticallyup. If the velocity of projection is doubled, thedistance travelled by the body during the lastsecond of its upward journey is

1) 2d 2) 4d 3) d/2 4) d15. A rocket is fired and ascends with constant

vertical acceleration of 10m/s2 for 1 minute.Its fuel is exhausted and it continues as a freeparticle. The maximum altitude reached is(g=10m/s2)1) 18 km 2) 36 km 3) 72 km 4) 108km



16. A parachutist after bailing out falls 50mwithout friction. When parachute opens, it

decelerates at 22m / s . He reaches the

ground with a speed of 3 /m s . At whatheight, did he bail out ?

1) 91m 2) 182m 3) 293m 4) 111m

17. A body is thrown vertically upwards with aninitial velocity ‘u’ reaches a maximum heightin 6s. The ratio of the distance travelled bythe body in the first second to the seventhsecond is

1) 1:1 2) 11:1 3) 1:2 4) 1:1118. A body is thrown vertically up to reach its

maximum height in t seconds. The total timefrom the time of projection to reach a pointat half of its maximum height while return-ing( in seconds ) is (2008 E)

1) 2 t 2)1

1 t2

3)3t

24)

t

219. Water drops fall from a tap on to the floor 5.0m

below at regular intervals of time. The first dropstrikes the floor when the fifth drop beings tofall. The height at which the third drop will befrom ground, at the instant when the first dropstrikes the ground is (Take g = 10ms-2)1) 1.25m 2) 2.15m 3) 2.75m 4) 3.75m

20. A boy throws n balls per second at regulartime intervals. When the first ball reaches themaximum height he throws the second onevertically up. The maximum height reachedby each ball is

1) 22 1

g

n 2) 22

g

n 3) 2

g

n 4)

g

n21. A body is thrown vertically upward from a

point ‘A’ 125m above the ground. It goes upto a maximum height of 250 m above theground and passes through ‘A’ on itsdownward journey. The velocity of the bodywhen it is at a height of 70m above the groundis (g = 10 m/s2) (MED-2013)1) 20 m/s 2) 50 m/s 3) 60 m/s 4) 80 m/s

22. A body is released from the top of a tower ofheight H m. After 2s it is stopped and theninstantaneously released. What will be itsheight after next 2s (in metres)?1)H-5 2) H-10 3)H-20 4)H-40

23. A ball dropped from 9th stair of a multistoriedbuilding reaches the ground in 3 sec. In the

first second of its free fall, it passes through‘n’ stare then ‘n’ equal to1) 1 2) 2 3) 3 4) 4

RELATIVE VELOCITY24. Two particles P and Q simultaneously start

moving from point A with velocities 15m/sand 20 m/s respectively. The two particlesmove with accelerations equal in magnitudebut opposite in direction. When P overtakesQ at B then its velocity is 30m/s. The veloc-ity of Q at point B will be1) 30m / s 2) 5m / s 3) 10m / s 4) 15m / s

25. Two trains A and B, 100m and 60m long, aremoving in opposite directions on paralleltracks. The velocity of the shorter train is 3times that of the longer one. If the trains take4s to cross each other, the velocities of thetrains are

1) 1 1A Bv 10ms , v 30ms

2) 1 1A Bv 2.5ms , v 7.5ms

3) 1 1A Bv 20ms , v 60ms

4) 1 1A Bv 5ms , v 15ms

LEVEL - II (C.W) - KEY01) 3 02) 4 03) 1 04) 2 05) 2 06) 307) 3 08) 4 09) 3 10) 1 11) 1 12) 113) 2 14) 4 15) 2 16) 3 17) 2 18) 219) 4 20) 2 21) 3 22) 4 23) 1 24) 225) 1

LEVEL - II (C.W) - HINTS

1.C

B

-30i O

A

-30 j

30j

20 i

W E

N

S

Total displacement from the origin,20 30 30 30s i j j i =-10i

2.total distance

avg speed=total time

1 2

1 2

1 2

s ss s

v v



68

3. 2

aT

2vsinn

, (n=4)

4. (a) 1 2

avg1 2

s sv =

t t

; (b)

1 2avg

1 2

s sv =

t t

5. 1

as 2n 1

2 ; 2

as 2n 1

2

According to given prob. 1 2S S

6.s1 s2

t2t1

21 1 1

1

2s ut t

2

1 2 1 2 1 2

1

2s s u t t a t t

7. 2 2v - u = 2as ; v - u

t =a

; u + v

s = t2

8. 2 2v - u = 2as ; as, v=0

2u s ;

2

1 1

2 2

u s

u s

9. Distance travelled by the police party in ‘t’ sec. is vt.

Distance travelled by thief 21

x t2

21x t vt

2

2αt- vt + x = 0

2

2αt - 2vt + 2x = 0 t=2v 2± 4v - 8αx24v > 8αx ; 2v > 2αx

10. avg

v ua ;s v t

t

11.

21d at vt

2

12.21

s ut at2

13.2 2

d

h 1 1gt h gt

x 2 2

14. distance covered in the last second of upwardjourney = distance covered in 1st second of down-ward journey. This is independent of velocity ofprojection.

15. 21h at , v at

2 : max.height = H =

2v

2gtotal distance from the ground = (H + h)

21

12

aat

g

16. u = 2g H (g = 9.8 m/s2) v2 – u2 = 2as ; height at which he bails out= (H + s)

17. , , 6a a

ut u gt u g

g

height in the first second, 1

gh u

2

height covered in the first second of downward

journey, 2

gh

2

18. H = 21

2gt ;

2H 1g(t )

2 2 ; find t = ?

Total time = t + t

19.h

5th

4th

3rd

2nd

1st

x

3x

5x

7x

3 5 7 16h x x x x x ;5

16 16

hx

Distance of 3rd drop from the ground = 12x20. Time interval between two balls = Time of ascend

=1

n=

u

gg

un

;2u

h2g

21. The body is freely falling from a height of 250m. Itsvelocity at a height of 70m from the ground meansvelocity of freely falling body after travelling 180m.

v = 2gh .

22.

s1 s2

t2t1 2

1 1

12sec

2h gt t ; 2

2 2

12sec

2h gt t

11 2h H h h

23.21

2nh gt , h= height of each story (constant)

2n t24. relv at ; For P , 30 = 15 + at

For Q, v = 20 - at25. A B3v v ; rel A Bv = v + v

A B

rel

l + lt =

v ; 1 2 A Bs s v v t



LEVEL - II(H.W)

1. Two particles move along x-axis in the samedirection with uniform velocities 8 m/s and4 m/s. Initially the first particle is 21m to theleft of the origin and the second one is 7m tothe right of the origin. The two particles meetfrom the origin at a distance of 1) 35 m 2) 32 m 3) 28 m 4) 56 m

2. A moving car possesses average velocities of1 15ms ,10ms and 115ms in the first, second,

and third seconds respectively. What is thetotal distance covered by the car in these 3s?1) 15m 2) 30m 3) 55m 4) 45m

ACCELERATION3. The average velocity of a body moving with

uniform acceleration after travelling adistance of 3.06 m is 0.34 m/s. If the changein velocity of the body is 0.18 ms-1 during thistime, its uniform acceleration is ( in ms-2 )1) 0.01 2) 0.02 3) 0.03 4) 0.04

4. If a body looses half of its velocity on penetrating3cm in a wooden block, then how much will itpenetrate more before coming to rest1) 1 cm 2) 2cm 3) 3cm 4) 4cm

5. A car moving with a speed of 50km/hr can bestopped by brakes after atleast 6m. If thesame car is moving at a speed of 100km/hr,the minimum stopping distance is1) 12m 2) 18m 3) 24m 4) 6m

6. A particle moving with a constant accelera-tion describes in the last second of its motion36% of the whole distance. If it starts fromrest,how long is the particle in motion andthrough what distance does it moves if it de-scribes 6 cm in the first sec.?1) 5 ;150s cm 2) 10 ;150s cm

3) 15 ;100s cm 4) 20s ; 200cm7. A bus starts from rest with a constant accel-

eration of 25 /m s .At the same time a cartravelling with a constant velocity 50 m/s overtakes and passes the bus. How fast is the bustravelling when they are side by side?1) 10 m/s 2) 50 m/s 3) 100 m/s 4) 150m/s

8. A particle moving with uniform retardationcovers distances 18m, 14m and 10m insuccessive seconds . It comes to rest aftertravelling a further distance of1)50 m 2) 8 m 3) 12 m 4) 42 m

MOTION UNDER GRAVITY9. The splash of sound was heard 5.35s after

dropping a stone into a well 122.5m deep.Velocity of sound in air is

1) 350 cm/s 2) 350 m/s 3) 392 cm/s 4) 0 cm/s

10. Two stones are thrown vertically upwards withthe same velocity of 49m/s. If they are thrownone after the other with a time lapse of 3second, height at which they collide is

1) 58.8 m 2) 111.5 m 3) 117.6 m 4) 122.5 m

11. A stone projected upwards with a velocity ‘u’reaches two points ‘P’ and ‘Q’ separated bya distance ‘h’ with velocities u/2 and u/3. Themaximum height reached by it is

1) 5

h92)

5

h18 3)

5

h364)

5

h72

12. A ball is dropped from the top of a building.The ball takes 0.5s to fall past the 3m lengthof a window at certain distance from the topof the building. Speed of the ball as it crossesthe top edge of the window is (g=10m/s2)

1) 3.5 ms-1 2) 8.5 ms-1 3) 5 ms-1 4) 12 ms-1

13. A body thrown vertically up with a velocity ‘u’reaches the maximum height ‘h’after‘T’ second.Correct statement among the following is1) at a height h/2 from the ground its velocity is u/22) at a time ‘T’ its velocity is ‘u’3) at a time ‘2T’ its velocity is ‘-u’4) at a time ‘2T’ its velocity is ‘-6u’

14. A ball is projected vertically upwards with avelocity of 25 ms-1 from the bottom of a tower.A boy who is standing at the top of a tower isunable to catch the ball when it passes him inthe upward direction. But the ball againreaches him after 3 sec when it is falling. Nowthe boy catches it Then the height of the toweris (g=10ms-2)1) 5 m 2) 10 m 3) 15 m 4) 20 m

15. A person sitting on the top of a tall building isdropping balls at regular intervals of onesecond. When the 6th ball is being dropped,the positions of the 3rd, 4th, 5th balls fromthe top of the building are respectively1) 4.9m, 19.6m, 44.1m 2) 4.9m, 14.7m, 24.5m3) 44.1m, 19.6m, 4.9m 4) 24.5m, 14.7m, 4.9m



70

16. A stone projected vertically up from theground reaches a height y in its path at 1t

seconds and after further 2t seconds reachesthe ground. The height y is equal to

1) 21 ttg2

1 2) 221 ttg

2

1

3) 21ttg2

14) 21ttg

17. A person standing on the edge of a well throwsa stone vertically upwards with an initialvelocity 5 ms-1. The stone gone up, comes downand falls in the well making a sound. If theperson hears the sound 3 second afterthrowing, then the depth of water (neglect timetravel for the sound and take g = 10ms-2)1) 1.25 m 2) 21.25 m3) 30m 4) 32.5 m

18. A ball is thrown vertically upwards with aspeed of 10 m/s from the top of a tower 200mheight and another is thrown verticallydownwards with the same speedsimultaneously. The time difference betweenthem on reaching the ground is (g=10m/s2)1) 12s 2) 6s 3) 2s 4) 1s

19. A body is projected vertically upwards with avelocity ' 'u . It crosses a point in its journey

at a height ' 'h twice , just after 1 and 7 sec-

onds .The value of u in 1 210ms is g ms

1) 50 2)40 3) 30 4) 2020. A stone thrown vertically up from the ground

reaches a maximum height of 50m in 10s.Time taken by the stone to reach the groundfrom maximum height is1) 5s 2) 10s 3) 20s 4) 25s

21. A freely falling body travels-- of total distancein 5th second1) 8% 2) 12% 3) 25% 4) 36%

22. A body is projected with a velocity u. It passesthrough a certain point above the ground in t

1sec. The time after which the body passes throughthe same point during the return journey

1)2

1

ut

g

2) 12

ut

g

3)2

13u

tg

4)

2

123

ut

g

23. A boy throws a ball in air in such a mannerthat when the ball is at its maximum heighthe throws another ball. If the balls are thrownwith the time difference 1 second, the maxi-mum height attained by each ball is1) 9.8 m 2) 19.6 m 3) 4.9 m 4) 2.45 m

RELATIVE VELOCITY24. Two cars are travelling in the same direction

with a velocity of 60 kmph. They areseparated by a distance of 5 km. A truckmoving in opposite direction meets the twocars in a time interval of 3 minute. Thevelocity of the truck is (in kmph)1) 20 2) 30 3) 40 4) 60

25. A police van moving on a highway with aspeed of 30 kmph fires a bullet at a thief’scar speeding away in the same direction witha speed of 192 kmph. If the muzzle speed ofthe bullet is 150 m/s, with what speed doesthe bullet hit the thief’s car? (Note: Obtainthat speed which is relevant for damaging thethief’s car).1) 25m/s 2) 50m/s 3) 75m/s 4) 105m/s

26. Two cars are moving in same direction withspeed of 30kmph. They are separated by adistance of 5km. What is the speed of a carmoving in opposite direction if it meets thetwo cars at an interval of 4 min?1) 60 kmph 2) 5 kmph3) 30 kmph 4) 45 kmph

LEVEL - II(H.W)-KEY01) 1 02) 2 03) 2 04) 1 05) 3 06) 107) 3 08) 2 09) 2 10) 2 11) 3 12) 113) 3 14) 4 15) 3 16) 3 17) 3 18) 319) 2 20) 2 21) 4 22) 2 23) 3 24) 325) 4 26) 4

LEVEL - II(H.W)-HINTS1. 121 x u t ; 27 x u t

2. 1 1 2 2 3 3s u t u t u t

3.ave

distancet =

v , a = changein velocity v u

total time t

4. 2

1y =

2 1

x n

n

5. 2 2v u 2as , v=0 ; 2u s

6.21

2s at ; 2

2 1 36

100

n

n



7.21

s = at , s = vt2

, v=at

8.1

s u a n2

9.sound

2h ht

g v

10.2 2

2 8

u gth

g 11. 2 2v u 2as ;

2

2

uH

g

12.21

S ut at2

213 v 0.5 10 0.5

2

13. After a time 2T, body returns to the point of projec-tion with the same velocity in opposite direction.

14.2v

tg

; 2 2v u 2gh

15.

6

5

4

3

2

1

g

2

3g

2

5g

2

7g

2

9g

2

16. h = y = 1 2

1gt t

2 17. h = –ut +

21

2gt

18.2u

tg

19. 2 1

1u g(t t )

2

20. a dt t 21. 2

2 1ns n

h n

22.ta

t1

P

1u

tg

; a

ut

g

23. 2

gh

2n

24. d

u u

v ;

rel

d dt

v u v

25. The relative velocity of bullet with respect to thief

bt b p tv v v v

26. d

u u

v ;

rel

d dt = =

v u + v

LEVEL-III

SPEED AND VELOCITY1. A motorist drives north for 35.0 minutes at

85.0 km/h and then stops for 15.0 minutes.He next continues north, travelling 130 kmin 2.00 hours. What is his total displacement1) 85 km 2)179.6 km 3)20 km 4)140 km

2. The coordinates of a moving particle at any

time ‘t’ are given by 3x t and 3y t . Thespeed of the particle at time ‘t’ is given by

1) 2 2 2) 2 23t

3) 2 2 23t 4) 2 2 2t

ACCELERATION3. The relation between time t and distance x is

2t ax bx where a and b are constants. Theacceleration is1) 32av 2) 22av 3) 22abv 4) 32bv

4. Two cars start in a race with velocities u1 and

u2 and travel in a straight line with

accelerations ‘ ’ and . If both reach thefinish line at the same time, the range of therace is

1) )uu()(

)uu(2212

21

2) )uu()uu(2

2121

3) 2

221

)(

)uu(2

4) 21uu2

5. A point moves with uniform accelerationv

1 ,v

2 and v

3 denote the average velocities in

three successive intervals of time t1, t

2 and

t3. Correct relation among the following is

1) (v1-v

2) : (v

2-v

3) = (t

1-t

2) : (t

2-t

3)

2) (v1-v

2) : (v

2-v

3) = (t

1+t

2) : (t

2+t

3)

3) (v1-v

2) : (v

2-v

3) = (t

1-t

2) : (t

2+t

3)

4) (v1-v

2) : (v

2-v

3) = (t

1+t

2) : (t

2-t

3)

6. A train starts from rest and moves with uniformacceleration for some time and acquires avelocity ‘v’. It then moves with constantvelocity for some time and then decelerates



72

at rate and finally comes to rest at the nextstation. If ‘L’ is distance between two stationsthen total time of travel is

1) L v 1 1

v 2

2) L v 1 1

v 2

3) L v 1 1

v 2

4) L v 1 1

v 2

7. A car, starting from rest, accelerates at therate f through a distance S, then continues atconstant speed for time t and then decelerateat the rate (f/2) to come to rest. If the totaldistance travelled is 15S, then

1) S ft 2) 21

S ft6

3) 21S ft

72 4) 21

S ft4

8. An express train moving at 30 m/s reducesits speed to 10 m/s in a distance of 240 m. Ifthe breaking force is increased by 12.5% inthe beginning find the distance that it travelsbefore coming to rest1) 270 m 2) 240 m 3) 210 m 4) 195 m

9. For motion of an object along the x-axis, thevelocity v depends on the displacement x as

2v 3x 2x, then what is the acceleration at

x 2m .

1) 248ms 2) 280ms 3) 218ms 4) 210ms

MOTION UNDER GRAVITY10. The friction of the air causes a vertical

retardation equal to10% of the acceleration

due to gravity. Take 2g 10m / s . Themaximum height and time to reach themaximum height will be decreased by1) 9%, 9% 2) 11%, 11%3) 9%, 10% 4) 11%, 9%

11. A parachutist after bailing out falls for 10swithout friction. When the parachute openshe descends with an acceleration of 2 m/s2

against his direction and reached the groundwith 4 m/s. From what height he has droppedhimself ? (g = 10m/s2)1) 500m 2) 2496m 3) 2996m 4) 4296m

12. A body is dropped from the roof of a multi-storied building. It passes the ceiling of the

15th storey at a speed of 120 ms . If the

height of each storey is 4m, the number of

storeys in the building is (take 210g msand neglect air resistance)1) 20 2) 25 3) 30 4) 35

13. A body is projected vertically up with veloc-ity 198ms . After 2 s if the acceleration dueto gravity of earth disappears , the velocityof the body at the end of next 3 s is1)49ms-1 2)49.6ms-1 3)78.4ms-1 4)94.7ms-1

14. The velocity of a particle is 20v v gt ft .

If its position is x = 0 at t = 0, then its dis-placement after unit time (t = 1) is (AIE-2007)

1) 0v 2g 3f 2) 0

g fv

2 3

3) 0v g f 4) 0

gv f

2

15. A ball A is dropped from the top of a buildingand at the same time an identical ball B isthrown vertically upward from the ground.When the balls collide the speed of A is twicethat of B. At what fraction of the height of thebuilding did the collision occur?

1) 1

32)

2

33)

5

34)

7

316. An object falls from a bridge that is 45m

above the water.It falls directly into a smallrow-boat moving with constant velocity thatwas 12m from the point of impact when theobject was released. What was the speed of

the boat? 2g 10 ms

1. 2 /m s 2. 3 /m s 3. 5 /m s 4. 4 /m s

GRAPHS17. An elevator is going up. The variation in the

velocity of the elevator is as given in thegraph. What is the height to which the eleva-tor takes the passengers

3.6

velocity

Time2 10 12O

1) 3.6 m 2) 28.8 m 3) 36.0 m 4)72.0 m



18. The velocity time graph of a body moving ina straight line is shown in the figure. The dis-placement and distance travelled by the bodyin 6 sec are respectively (in metres) .

t(sec)

v(m

/s)

12 4

56

1) 8,16 2) 16,8 3)16,16 4) 8,819. The velocity-time graph of a stone thrown

vertically upward with an initial velocity of130ms is shown in the figure. The velocity in

the upward direction is taken as positive andthat in the downward direction as negative.What is the maximum height to which thestone rises?

3020100

102030

Time (s)

velo

city

(m

s)

-1

1 2 3 4 5

B

A

C

1) 30 m 2) 45 m 3) 60 m 4) 90 m20. The variation of velocity of particle moving

along a straight line is shown in the figure.The distance travelled by the particle in 4s is

20

10

V(m

s)

-1

21 3 4t(s)

1) 55 m 2) 30 m 3) 25 m 4) 60 m21. Figure shows the displacement-time (x-t)

graph of a body moving in a straight line.Which one of the graphs shown in figurerepresents the velocity-time (v-t) graph of themotion of the body.

0 5 10 15 20 t

x

1)5 10 15 20

v

t0 2)5 10 15 20

v

t0

5 10 15 20t3) 0

vv

05 10 15 20 t4)

22. Velocity -time (v-t) graph for a moving objectis shown in the figure . Total displacement ofthe object during the time interval when thereis non-zero acceleration and retardation is

10 20 30 40 50 60

v(m

/s) 4

3

2

10

t(sec)

1) 60m 2) 50m 3) 30m 4) 40m



74

RELATIVE VELOCITY23. An armored car 2m long and 3 m wide is

moving at 110ms when a bullet hits it in a

direction making an angle 1 3

tan4

with the

length of the car as seen by a stationaryobserver. The bullet enters one edge of thecar at the corner and passes out at thediagonally opposite corner. Neglecting anyinteraction between the car and the bullet andeffect of gravity, the time for the bullet tocross the car is1) 0.20 s 2) 0.15 s 3) 0.10 s 4) 0.50 s

24. Two particles start simultaneously from thesame point and move along two straight lines.One with uniform velocity v and other with auniform acceleration a. If is the angle be-tween the lines of motion of two particles thenthe least value of relative velocity will be attime given by

1)v

sina

2)v

cosa

3)v

tana

4)v

cota

25. A jet airplane travelling at the speed of 500km/h ejects its products of combustion at thespeed of 1500 km/h relative to the jet plane.What is the speed of the later with respect toan observer on ground?

1) 100kmph 2) 1000kmph

3) 10kmph 4) 11kmphLEVEL - III - KEY

1) 2 2) 3 3) 1 4) 1 5) 2 6) 17) 3 8) 1 9) 2 10) 1 11) 3 12) 113) 3 14) 2 15) 2 16) 4 17) 3 18) 119) 2 20) 1 21) 4 22) 2 23) 1 24) 225) 2

LEVEL - III - HINTS

1. 1 1s v t ; 1 2s s s

2. x y

dx dyv , v

dt dt ; 2 2

x yv v v

3.dx dv

v , adt dt

4. Range = dist. covered before they meet

1 2

1 1s t u t t u t

2 2

solve for t and substitute in the above equation.

5.change in average velocity

aaverage time

6.

v

Time

t1t3t2

O

v

1

v

t=

3

v

t=

1 2 3T t t t ; weknow 1 2 3

1 1L = vt + vt + vt

2 2

312

vtvtL = + vt +

2 2 ; 2 1 3

L 1t = - t + t

v 2 ......(1)

7.

2 2

32 23

s

v - o = 2fs® s = 2sf

o - v = -2 .s2

2s = 12s ; 12s = vt also v = 2fs

12s = 2fs.t ; 2144s = 2fs.t ; 21

s = ft72

8. 2 2v u 2as

9. Diff. eq. (1) w.r.t x is dv

= 6x - 2dx

;dv

a vdx

10.1

hg

;2 1

1 2

h g

h g ;

2 1 1 2

1 2

100 100h h g g

h g

1

11

10 10

g gg g

11.2

1

1

2s gt , v = gt ; 2 2

1 2v - v = 2as ; 1 2s = s + s

12. Distance traveled when it gains velocity of 20m/s



2v - 0 = 2gh ; 2v

h =2g

= 400

20=20m

Number of stories =20

54

Total no.of stories 15 5 20 13. v = u - gt . after g disappears body moves up

with uniform velocity.

14. Given velocity of particle ; 20v = v + gt + ft

displacement s = vdt

15.

2

B

2 2A B

1ut gth 2

1 1h h gt ut gt2 2

16.2h

tg

; s = v × t

17. area of v t graph; 18.area of v t graph

19.2

2

uh

a ; From graph u=30m/s;

23010 /

3a m s

20. area of under v-t graph21. o to 5s - velocity is +ve and constant

5 to 15s - slope is zero15 to 20s - velocity is -ve and constant

22. Area under trapezium gives displacement.

RELATIVE VELOCITY23.

3m v sin 37b

0

vb

2m

2m=370

=tan-1 3

4

10m/s

cos 370

bv

Relative velocity along x-axis is 0bv cos37 -10

Distance travelled by bullet along x- axis is

0bv cos37 -10 t = 2 .........(1)

Distance travelled by bullet along y-axis is

0bv sin37 t = 3 ..........(2)

Solving equation (1) and (2) we get t=0.20s24. At any time velocity of first car is V and that of

second car is v = v + at = 0 + at

22relv = v + at - 2vat cosα

relv is minimum if 2r

dv = 0

dt ;

vcosαt =

a

25. Velocity of jet aeroplane = 500 j

velocity of fuel w.r.to plane 1500 j

f pv - v = -1500 j

; f pv = v -1500 j

500 1500j j 1000 j speed of fuel w.r.to ground is -1000km/hr..

LEVEL-IV

Matching Type Questions1. Match the following

Column I Column II

a) d v

dt

p) Acceleration

b) dv

dt

q) Magnitude of acceleration

c) dr

dt

r) Velocity

d) d r

dt

s) magnitude of velocity

2. A particle moves in a straight line with zeroinitial velocity. Then, match the followingterms :Column I Column II

a) 1

2(slope of 2v - x ) graph p) Speed

b) slope of v - t graph q) Accelerationc) Slope of x t graph r) Area under v-t

graph

d) 2 Area under a x graph s) Velocity

Statements Type QuestionsMark your answer as1) If Statement I is true, Statement II is true, State-ment II is a correct explanation for Statement I2)If Statement I is true, Statement II is true, State-ment II is not a correct explanation for Statement I3) If Statement I is true, Statement II is false4) If Statement I is false, Statement II is true



76

3. Statement I : An object may have varying speedwithout having varying velocity.Statement II : If the velocity is zero at an instant,the acceleration may not be zero at that instant.

4. Statement I : A body is momentarily at rest atthe instant it reverses the direction.Statement II : A body cannot have accelerationif its velocity is zero at a given instant of time.

5. Statement I : On a curved path average speedof a particle can never be equal to average veloc-ity.Statement II : Average speed is total distancetravelled divided by total time. Whereas averagevelocity is, final velocity plus initial velocity dividedby two.

6. Statement I : Particle A is moving eastwards andparticle B northwards with same speed. then,velocity of A with respect to B is in south -eastdirection.Statement II : Relative velocity between them iszero as their speeds are same.

7. Statement I : A lift is ascending with decreasingspeed means acceleration of lift is downwards.Statement II : A body always moves in thedirection of its acceleration.

8. Statement I : Two balls of different masses arethrown vertically upwards with the same speed.They will pass through their point of projection inthe downward direction with the same speed.Statement II : The height and the downwardvelocity attained at the point of projection areindependent of the mass of ball.

9. Statement I: The v-t graph perpendicular to timeaxis is not possible in practice.Statement II: Infinite acceleration can’t berealized in practice.

10. Statement I: Magnitude of average velocity isequal to average speed, if velocity is constant.Statement II: If velocity is constant , then therein no change in the direction of motion.

11. Statement I: The average velocity of a particle

having initial and final velocity v1 and v

2 is 1 2v v

2

Statement II: If r1 and r

2 be the initial and final

displacement in time t, then 1 2avg

r - rv =

t.

12. Statement I: If a particle is thrown upwards thendistance travelled in last second of upward journeyis independent of the velocity of projection.

Statement II: In last second distance travelled is4.9m. (take 29.8 /g m s )

13. Statement I: Distance between two particlesmoving with constant velocities always remainsconstant.

Statement II: In the above case relative motionbetween them is uniform.

14. Statement I: For one dimensional motion, the anglebetween acceleration and velocity should be zero.

Statement II: One dimensional motion is on astraight line.

15. Statement I: If displacement is a linear functionof time, its average and instantaneous velocity willbe same.

Statement II: If the acceleration of a movingparticle is zero, the particle moves linearly.

16. Statement I: If two particles are moving withsame speed but their velocity vectors are oppo-site, then the distance between the particles mustbe changing.

Statement II:If the velocity of one particle w.r.t.other is zero, then separation between the par-ticle must be constant.

17. Statement I: If the magnitude of acceleration ofa particle is constant then speed must change.

Statement II: In uniform circular motion speedof the particle is constant but it has some accel-eration.

18. Statement I: A bus moving due north takes a turnand starts moving towards east with same speed.There will be no change in the velocity of bus.

Statement II:Velocity is a vector quantity.

19. Statement I:When a particle is moving on astraight line, its velocity is constant.

Statement II: The net acceleration of a movingparticle may change its direction of motion ormagnitude of velocity or both.

20. Statement I:The direction of velocity vector isalways along the tangent to the path therefore itsmagnitude may be given by its slope.

Statement II:The slope of tangent to path alwaysmeasure the magnitude of velocity at that point.



LEVEL - IV - KEYMatching type

1) a -p; b-q; c-r; d-s2) a-q; b-q; c-s; d-p

Statement Type3) 4 4)3 5)3 6)3 7)3 8) 19) 1 10)1 11)4 12)1 13)4 14) 415)2 16)2 17)4 18)4 19)4 20)2

LEVEL - IV - HINTS

3. If speed varies, then velocity will definitely vary.

At highest point of a particle thrown upwardsa 0 but v=0.

4. Any body momentarily at rest must change its di-rection, when a particle is released from rest,v 0 but a 0

5. On a curved path, distance>displacement

average speed > average velocity

Further,

average velocity displacement

=Total time

6.A B AB A Bv = ui, v = u j, v = v - v = ui - u j

0tan 1 , 45 in S-E direction.

7. Ascending means velocity is upwards. Speed is de-creasing. It means acceleration is downwards. Fur-ther, the body moves in the direction of velocity.

8. Both maximum height and time of flight are inde-pendent on masses.

9.

v

to,

vSlope = = = a

t

10. In one dimensional motion vavg

is numerically equalto v

avg velocity, since there is no change in direction.

11. If acceleration is not constant 1 2avg

v + vv ¹

2

2 1avg

r - rsv = =

t t

12. If a particle is thrown upwards then distance trav-elled in last second of upward journey is equal todistance travelled in first second of a freely falling

body.g

s = = 4.9m2

It is independent of velocity of projection.

13. If distance between two particles moving with con-

stant velocities always remains constant v 0r

14. In one dimension motion angle between velocityand acceleration becomes zero and 1800 in dif-ferent situation. These are two possible value only.

15. If displacement is linear function of time then ve-locity is constant that is motion will be of uniformnature.

16. Correct speed same but direction of velocity op-posite than distance between particle will bechanging. Statement-(1) correct statement-(2)correct but (2) is not the correct explanation ofStatement-(1).

17. acceleration = constant ; velocity changes

speed must changes statement-(1)-false

In uniform circular motion

speed = constant but it has some accelerationstatement-(2) correct.

18. Velocity change either by changing its magnitudeor by direction

Statement-(1)-false. ; Statement-(2) - correct.

19. In st. line motion particle may move with constantvelocity but it is not necessary acceleration velocity changes either by magnitude or by direc-tion or both.

20. Statement -(1) correct

Yes it may not have variable speed when haveconstant velocity

Statement-(2) correct

But Statement-(2) is not the correct explanationfor Statement -(1).

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