Iit Jee Study

KINEMATICS

DHANALAKSHMI NAGAR

NEAR ANNAMAIAH CIRCLE,

TIRUPATI.

PH NO. 9440025125

KINEMATICS

www.physicsashok.in 1 KINEMATICS

THEORY OF KINEMATICSKinematics : The study of the motion of an object without taking into consideration cause of its motion is called

kinematics.

NOTE : The word kinematics comes from the Greek word Kinema which means

motion. The word dynamics comes from the Greek word dynamics which means power.

BASIC DEFINITIONSDistance and Displacement

Suppose an insect is at a point A(x1, y1, z1) at t = t1. It reaches at point B(x2, y2, z2) at t = t2 through path ACBwith respect to the frame shown in fig. The actual length of curved path ACB is the distance travelled by theinsect in time t = t2 t1.

A B

C

X

Y

Z

Ar Br

If we connect point A (initial position) and point B (final position) by a straight line, then the length of straight lineAB gives the magnitude of displacement of insect in time interval t = t2 t1.The direction of displacement is directed from A to B through the straight line AB from the concept of vector,the position vector of A

is A 1 1 1 r x i y j z k= + +

and that of B is B 2 2 2 r x i y j z k= + +

.According to addition law of vectors,

A Br AB r+ =

B AAB r r= -

( ) ( ) ( )2 1 2 1 2 1 AB x x i y y j z z k= - + - + -

The magnitude of displacement is

( ) ( ) ( )2 2 22 1 2 1 2 1| AB | x x y y z z= - + - + -

NOTE : Distance covered by the body is always equal to or greater

than its magnitude of displacement.

Example 1. A man walks 3m in east direction, then 4m in north direction. Find distance covered and thedisplacement covered by man.

Sol. The distance covered by man is the length of path = 3m + 4m = 7m.Let the man starts from O and reaches finally at B (shown in figure).

OB

represents the displacement of man. From figure,

( ) ( )2 2| OB | OA AB= +

( ) ( )2 2| OB | 3m 4 m 5 m= + = W E

N

S

O 3 m A

B

4 m

KINEMATICS


and4m 4

tan3m 3

q = =

1 4tan

3- q=

The displacement is directed at an angle 14

tan3

- north of east.

Average Speed and Average VelocitySuppose we wish to calculate the average speed and average velocity of the insect (in section (i)) betweent = t1 and t = t2. From the path (shown in fig.) we see that at t = t1, the position of the insect is A(x1, y1, z1) andat t = t2, the position of the insect is B(x2, y2, z2).

A B

C

X

Y

Z

Ar Br

The average speed is defined as total distance travelled by a body in a particular time interval divided by thetime interval. Thus, the average speed of the insect is

av2 1

The length of curve ACBv

t t=

-The average velocity is defined as total displacement of the body in a particular time interval divided by the timeinterval.Thus, the average velocity of the insect in the time interval t2 t1 is

av2 1

ABv

t t=

-

B Aav

2 1

r rv

t t-

=-

( ) ( ) ( )2 1 2 1 2 1av

2 1

x x i y y j z z kv

t t- + - + -

=-

Example 2. In one second a particle goes from point A to point B moving in a semicircle (fig.).Find the magnitude of average velocity.

Sol. avAB

| v |t

=D

A

B

1 0m.

av2.0

| v | m / s1.0

=

av| v | 2 m / s=

Example 3. A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h. If AB =6BC the average speed in km/h between A and C is _______

Sol. AB = 40t1 ...(1)BC = 60t2 ...(2)

KINEMATICS


total distance travelledAverage speed =time taken

av1 2

AB BCVt t

+=+

From eqn. (1) and (2) A B C1 2

av1 2

40t 60tVt t

+=+ ...(3)

According to questionAB = 6BC40t1 = 6 60t2 From eqn (1) and (2)t1 = 9t2

From eqn (3)2 2

av2 2

40 9t 60tV9t t

from eqn (3)

2av

2

420tV10t

= avV 42 km / h=

Instantaneous VelocityInstantaneous velocity is defined as the average velocity over smaller and smaller interval of time.Suppose position of a particle at t is r

and at t + t is r r+ D . The average velocity of the particle for time

interval t is avr

vt

D=

D

.

From our definition of instantaneous velocity, t should be smaller and smaller. Thus, instantaneous velocity is

t 0

r drv lim

t dtD D

= =D

Example 4. Let at any time t, the position vector of a particle is r x i y j z k= + +

. Find the velocity of the particle.

Sol. x-component of velocity, xdx

vdt

=

y-component of velocity, ydy

vdt

=

z-component of velocity, zdz

vdt

=

Thus, velocity of particle

x y 2 v v i v j v k= + +

dx dy dz v i j kdt dt dt

= + +

Average and Instantaneous AccelerationIn general, when a body is moving, its velocity is not always the same. A body whose velocity is increasing issaid to be accelerated.Average acceleration is defined as change in velocity divided by the time interval.Let us consider the motion of a particle. Suppose that the particle has velocity 1v

at t = t1 and at a later time

KINEMATICS


t = t2 it has velocity 2v

. Thus, the average acceleration during time interval t = t2 t1 is

2 1av

2 1

v v va

t t t- D

= =- D

If the time interval approaches to zero, average acceleration is known as instantaneous acceleration.Mathematically,

t 0

v dva lim

t dtD D

= =D

Example 5. The velocity of a point depends on time t, as v c t b= +

where c

and b

are constant vectors.Find the instantaneous acceleration at any time t.

Sol. Acceleration at any time t,

( )dv da c t bdt dt

= = +

a 0 b b= + =

IMPORTANT FEATURES1. If a body is moving continuously in a given direction on a straight line, then the magnitude of displacement is

equal to distance.2. Generally, the magnitude of displacement is less or equal to distance.3. Many paths are possible between two points. For different paths between two points, distances are different

but magnitudes of displacement are same.4. The slope of distance-time graph is always greater or equal to zero.5. The slope of displacement-time graph may be negative.6. If a particle travels equal distances at speeds v1, v2, v3, ...... etc. respectively, then the average speed is

harmonic mean of individual speeds.

7. If a particle moves a distance at speed v1 and comes back with speed v2, then 1 2

av1 2

2v vv

v v=

+but avv 0=

8. If a particle moves in two equal intervals of time at different speeds v1 and v2 respectively, then1 2

avv v

v2+

=

9. The average velocity between two points in a time interval can be obtained from a position versus time graphby calculating the slope of the straight line joining the co-ordinates of the two points.

x2x1

t1 t2(a)

x2x1

t1 t2(b)

( )x2 x1( )t2 t1

BA

The graph [shown in fig.], describes the motion of a particle moving along x-axis (along a straight line).Suppose we wish to calculate the average velocity between t = t1 and t = t2. the slope of chord AB [shown infig.(b)] gives the average velocity.

KINEMATICS


Mathematically, 2 1av2 1

x xv tan

t t-

= q=-

10. If a body moves with constant velocity, the instantaneous velocity is equal to average velocity. The instantaneousspeed is equal to modulus of instantaneous velocity.

11. x-component of displacement is xx v dtD = y-component of displacement is yy v dtD = z-component of displacement is zz v dtD = Thus, displacement of particle is

r x i y j z kD = D + D + D

12. If particle moves on a straight line, (along x-axis), then dx

v .dt

=

13. The area of velocity-time graph gives displacement.14. The area of speed-time graph gives distance.15. The slope of tangent at position-time graph at a particular instant gives instantaneous velocity at that instant.16. The slope of velocity-time graph gives acceleration.17. The area of acceleration-time graph in a particular time interval gives change in velocity in that time interval.

ONE, TWO AND THREE DIMENSIONAL MOTIONOne Dimensional Motion

If only one of the three co-ordinates is required to specify the position of an object in space changing w.r.t.time, then the motion of the object is called one dimensional motion. Motion of a particle in a straight line canbe described by only one component of its velocity and acceleration. For example, motion of a block in astraight line, motion of a train along a straight track, a man walking on a level and narrow road, an object fallingunder gravity, etc.

Two Dimensional MotionIf two of the three co-ordinates are required to spacify the position of an object on space changing w.r.t. time,then the motion of the object is called two dimensional motion. The motion of a particle through its verticalplane at some angle with horizontal. ( 90) is an example of two dimensional (2D) motion.This is a projectile motion. Similarly, a circular motion is an example of 2D motion. A 2D motion takes placein a plane and its velocity (or acceleration) can be described by two components in any two mutuallyperpendicular directions (vx and vy).

Three Dimensional MotionIf all the three co-ordinates are required to specify the position of an object in space changing w.r.t. time, thenthe motion of an object is called three dimensional motion. Such a motion is not restricted to a straight line orplane but takes place in space. In a 3D motion velocity and acceleration of a particle can be resolved in threecomponents (vx, vy, vz, ax, ay, az). A few examples of 3D motion are a flying bird, a flying kite, a flyingaeroplane, the random motion of gas molecules, etc.

UNIFORMLY ACCELERATED MOTIONA motion, in which change in velocity in each unit of time is constant, is called uniformly accelerated motion. So,for uniformly accelerated motion, acceleration is constant or approximately. So for uniformly accelerated motion( a

= constant) , equations of motion are as under,,

v u a t= +

...(i)

KINEMATICS


21s u t a t2

= +

...(ii)

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

...(iii)

where u

= initial velocity of particle

v

= velocity at time t

s

= displacement of particle at time tIf motion is described in one dimension, so vector sign () need not to be used.Normally, vertical upward motion is taken negative and vertical downward motion is taken positive. Similarly,for horizontal rightward motion is taken positive and leftward motion is taken negative.

Sign convention : Sign convention for (a) motion in vertical (b) motion in horizontal, is shown in Fig.

+ve ve ve+ve(a) (b)

Example 6. At a distance L = 400m from the traffic light brakes are applied to a locomotivemoving at a velocity v = 54 km/hr. Determine the position of the locomotive relative to the traffic light 1 min afterthe application of the brakes if its acceleration is 0.3 m/sec2.

Sol.5u 54 15 m / s

18

a = 0.3 m/s2 v = u + at

0 = 15 0.3 t0

015t 50 sec0.3

= =

After 50 second, locomotive comes in rest permanently. v2 = u2 + 2as

O2 = 152 2 0.3 S0

0225 2250S 375m0.6 6

= = =

the distance of the locomotive from traffic light= 400 375 = 25 metre

Example 7. A car moves in the xy plane with acceleration ( ) 2 3i 4 j m / s+ .(a) Assuming that the car is at rest at the origin at t = 0, derive expression for the velocity as function of time.(b) Find the equation of path of car and find the position vector as function of time.

Sol. Here, ux = 0, uy = 0, uz = 0ax = 3 m/s

2 , ay = 4 m/s2

(a) vx = ux + axtor vx = 3tand vy = uy + ayt

KINEMATICS


or vy = 4t

x y v v i v j= +

( ) v 3ti 4tj= +

(b) 2x x1

x u t a t2

= +

or 2 21 3

x 3t t2 2

= =

and 2y y1

y u t a t2

= +

or 2 21

y (4)t 2t2

= =

2 4

y 2 x x3 3

= = 23

x t2

=

4

y x3

=

Hence, the path is straight line.The position of car is

2 23 r x i y j t i 2t j2

= + = +

Motion Under GravityThe most familiar example of motion with costant acceleration on a straight line is motion in a vertical directionnear the surface of earth. If air resistance is neglected, the acceleration of such type of particle is gravitationalacceleration which is nearly constant for a height negligible with respect to the radius of earth. The magnitude ofgravitational acceleration near surface of earth is g = 9.8 m/s2 = 32 ft/s2.

Discussion :Case-I : If particle is moving upwards :

In this case applicable kinematics relations are :v = u gt ...(i)

21h ut gt2

= - ...(ii)

g

u

v2 = u2 2gh ...(iii)Here, h is the vertical height of the particle in upward direction.

NOTE : For maximum height attained by projectileh = hmax, v = 0

i.e., (0)2 = u2 2ghmax

2

maxu

h =2g

Case-II : If particle is moving vertically downwards :In this case :

v = u + gt ...(i)v2 = u2 + 2gh ...(ii)

g

u

KINEMATICS


21h ut gt2

= + ...(iii)

Here, h is the vertical height of the particle in downward direction.

Example 8. A particle is projected vertically upwards with velocity 40 m/s. Find the displacement and distancetravelled by the particle in

(a) 2s (b) 4s (c) 6s [Take g = 10 m/s2]Sol. Here, u is positive (upwards) and a is negative (downwards). So, first we will find t0, the time when velocity

becomes zero.

0u 40

t 4sa 10

= = =

(a) t < t0. Therefore, distance and displacement are equal.

21d s ut at2

= = +

1d 40 2 10 4 60 m

2= - =

(b) t = t0. So, again distance and displacement are equal.21d s ut at

2= = +

1d 40 4 10 16 80m

2= - =

(c) t > t0. Hence, d > s1

s 40 6 10 36 60 m2

= - =

While ( )2

20

u 1d a t t

2a 2= + -

22(40) 1d 10 (6 4) 100 m

2 10 2= + - =

Example 9. A ball is thrown upwards from the ground with an initial speed of u. The ball is at a height of 80 m at twotimes, the time interval being 6s. Find u. Take g = 10 m/s2.

Sol. Here, u = u m/s, a = g = 10 m/s2 and s = 80 m.

Substituting the value in 21

s ut at2

= + , we have

80 = ut 5t2

or 5t2 ut + 80 = 0

or2u u 1600

t10

+ -=

and2u u 1600

t10

- -=

Now, it is given thatve

+ve s = 80 m

u

KINEMATICS


2 2u u 1600 u u 16006

10 10+ - - -

- =

or2

2u 1600 6 or u 1600 305

-= - =

or u2 1600 = 900 u2 = 2500or u = 50 m/sIgnoring the negative sign, we have

u = 50 m/s

Example 10. A disc arranged in a vertical plane has two groves of same length directedalong the vertical chord AB and CD as shown in the fig. The same particles slidedown along AB and CD. The ratio of the time tAB/tCD is :

60A

B

D

C

(A) 1 : 2 (B) 1 : 2 (C) 2 : 1 (D) 2 : 1

Sol. 2AB AB1S g t2

=

2CD CD

1S g cos 60 t2

=

But SAB = SCD

2AB

AB

2CDCD

1 g tS 21S g cos 60 t2

=

60

A

B

D

C60

g

g

gcos60

or2AB2CD

t1 2

t=

AB

CD

t 1 : 2t

=

Example 11. A stone is dropped from a height h. Simultaneously another stone is thrownup from the ground with such a velocity that it can reach a height of 4h. Find the time when two stones crosseach other.

Sol. For second stone,2 2

0v v 2g (4h)= -2 2

00 v 8gh= - 0v 8ghbut they meth at height H in time t0. Displacement of 1st stone is

20

1h H gt2

...(1)and that of second stone is h

4h(1)(2)

v0

v = 0

20 0 0

1H v t gt2

= - ...(2)After solving eqn (1) and (2)

or 2 20 0 0 01 1h v t g t g t2 2

h = v0t0

KINEMATICS


00

h htv 8gh

= = 0h

t8g

=

Example 12. A rocket is launched at an angle 53 to the horizontal with an initial speed of 100 ms1. Itmoves along its initial line of motion with an acceleration of 30 ms2 for 3 seconds. At this time its enginefails & the rocket proceeds like a free body. Find :(i) the maximum altitude reached by the rocket(ii) total time of flight.

Sol. S0 = ut + at2= 100 3 + 30 9= 300 + 135 = 435 m

In OAB

0

hsin53S

=

04h S sin 53 435 87 4 348 m5

v0 = u + at = 100 + 30 3 = 190 m/s (velocity at the time of switch off)After engin switch off

0x 03v v sin 37 190 114 m / s5

0y 04v v cos37 190 152 m / s5

ay = 10 m/s2, ax = 0

(i) At maximum altitudes vy = 02 2y 0y y 0v v 2a h= +

20y y 00 v 2a h= +

20y

0y

vh

2a

53

37u

O

h

B

Engin fail

37A

v0xS0

0152 152h2 10

011552h

10= 0h 1155.2 m=

The maximum altitude reached by the rocket is= h0 + h

= (1155.2 + 348) m 1503.2 m

(ii)Total time of flight.2

0y y1y v t a t2

= +

20y 0 y 0

1h v t a t2

- = +

20 0

11155.2 152t 10 t2

KINEMATICS


20 01155.2 152t 5t- = -

20 05 t 152 t 1155.2 0- - =

20 0t 30.4 t 231.04 0- - =

0t 35.54 sec.=

NON-UNIFORMLY ACCELERATED MOTIONWhen motion of a particle is not uniform i.e., acceleration of particle is not constant or acceleration is a functionof time, then following relations hold for one dimensional motion :

(i)ds

vdt

=

(ii)dv dv

a vdt ds

= =

(iii) ds = vdt and(iv) dv = adt or vdv = adsSuch problems can be solved either by differentiation or integration applying some boundary conditions.

Example 13. A particle is moving with a velocity of v = (3 + 6t + 9t2) cm/s. Find out :(a) the acceleration of the particle at t = 3 s.(b) the displacement of the particle in the interval t = 5 s to t = 8 s.

Sol. (a) Acceleration of particle

( ) 2dva 6 18t cm / sdt

= = +

At t = 3 s,a = (6 + 18 3) cm/s2

a = 60 cm/s2

(b) Given, v = (3 + 6t + 9t2) cm/s

or ( )2ds 3 6t 9tdt

= + +

or ds = (3 + 6t + 9t2)dt

( )8 8 25 5

ds 3 6t 9t dt= + +

82 35

s 3t 3t 3t = + + or s = 1287 cm

Example 14 : The motion of a particle along a straight line is described by the function x = (2t 3)2 where x is inmetres and t is in seconds.(a) Find the position, velocity and acceleration at t = 2 s.(b) Find the velocity of the particle at origin.

Sol. (a) Position, x = (2t 3)2

Velocity, dxv 4 2t 3 m / sdt

and acceleration, 2dv

a 8 m / sdt

= =

KINEMATICS


At t = 2 s,x = (2 2 3)2 = 1.0 mv = 4(2 2 3) = 4 m/s and a = 8 m/s2

(b) At origin, x = 0or (2t 3) = 0 v = 4 0 = 0

EQUATIONS OF MOTION FOR VARIABLE ACCELERATIONCase-I : When acceleration a of the particle is a function of time :

Since, acceleration of a particle is a function of time, i.e., a = f(t)

dv

f (t)dt

=

dv = f(t)dtIntegrating within the proper limits, we get

t

0v u f (t)dt= +

Case-II : When acceleration a of the particle is a function of distance :Since, acceleration of a particle is a function of distance, i.e., a = f(x)

dv

f (x)dt

=

dv dx

. f (x)dx dt

=

vdv = f(x)dxIntegrating with in proper limits, we get

0

x2 2x

v u 2 f (x)dx= + Case-III : When acceleration a of the particle is a function of velocity :

Since, acceleration of a particle is a function of velocity, i.e., a = f(v)

dv

f (v)dt

=

dv

dtf (v)

=

Integrating within proper limits, we getv

u

dvt

f (v)=

Therefore, we get v as a function of t i.e., v(t).

Againdv

f (v)dt

=

dv dx

. f (v)dx dt

=

dv

v f (v)dx

=

vdv

dxf (v)

=

KINEMATICS


0

x v

x u

vdvdx

f (v)=

v0 u

vdvx x

f (v)- =

v

0 u

vdvx x

f (v)= +

Therefore, we get x as a function of v i.e., x(v).

GRAPHS IN ONE DIMENSIONAL MOTIONThe tabular forms of st and vt graphs are given for one-dimensional motion with uniform velocity or withconstant acceleration.

Table-1 : Displacement-Time GraphS.No. Different Cases st Graph The main Features of Graph

1. At rest

s

t

Slope = v = 0

2. Uniform motion

s

t

s = ut s = 0 at t = 0

3. Uniformly accelerated motion

s

t

s = at212 u = 0 i.e., slope of st graph at t = 0,with u = 0, s = 0 at t = 0 should be zero.


s

t

s = ut + at212 Slope of st graph gradually goes onwith u 0 but s = 0 at t = 0 increasing.


s

t

s = s + ut + at21200s

s = s0 at t = 0with u 0 but s = s0 at t = 0

6. Uniformly retarded motion

s

t0t

At t = t0, slope of st graph becomeszero.

Table-2 : Velocity-Time GraphS.No. Different Cases vt Graph The main Features of Graph

1. Uniform motion

v

t

v = constanta = 0

(i) Slope of vt graph = a = 0(ii) v = constant

KINEMATICS


2. Uniformly accelerated motiont

v = at

v

(i) u = 0 i.e., v = 0 at t = 0with u = 0, s = 0 at t = 0 (ii) a or slope of vt graph is constant.


v = u + atvu

At t = 0, v 0 and slope of vt graphwith u 0 but s = 0 at t = 0 is not zero.


v = u + atvu

v = u at t = 0with u 0 but s = s0 at t = 0

5. Uniformly decelerated motiont

v = u at

v

u

0t

Slope of vt graph = a (retardation)

6. Non-uniformly accelerated motiont

v

Slope of vt graph increases withtime

Acceleration-Time Graph(i) When at graph is a straight line parallel to time-axis, then acceleration = a = constant.

a

t

Slope = 0

(a)

(ii) When at graph is a straight line passing through origin, then acceleration of particle is increasing uniformly.

t

a

Slope

= +ve

(b)

(iii) When at graph is a straight line of negative slope, then acceleration is decreasing uniformly.

t

aSlope = ve

(c)

KINEMATICS


Example 15. The fig. shows the vt graph of a particle moving in straight line.Find the time when particle returns to the starting point.

v

t252010

10

20

Sol. When the particle comes at initial position, total displacement is zero.Since, the area of v t graph gives displacement. In this case area ofv t graph should be zero.

01 1S 20 25 t 25 v2 2

or 0v0 250 t 252

or0

500t 25v

...(1)

Also 0v20tan

25 20 t 25

from figure

or 0v4

t 25

0v 4 t 25 ...(2)v0

t

v20

20 25From eqn (1) and (2)

500 125t 25

4 t 25 t 25

2t 25 125

t 25 125 5 5 t = (25 + 5 5 ) sec. t 36.2 sec=

Example 16. From the velocity-time plot shown in figure, find the distancetravelled by the particle during the first 40 seconds. Also find the averagevelocity during this period.

5m/sV

5m/s20 40

t(s)

Sol. The distance fravelled by the particle during the first 20 second.

11S v t2

11S 5 202

1S 50 m=

The distance travelled by the partile during next 20 second is

21S 5 202

S2 = 50 mSince distance is a sclar quantity therefore

total distance = S1 + S2 = 50 + 50S 100 m=

Example 17. A ball is dropped from a height of 80m on a floor. At each collision the ball losses half of its speed.Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor,[Take g = 10 ms2].

KINEMATICS


Sol. The time in first collision21h ut gt

2= + (during downward motion)

or 21

80 0 10 t2

= + ( )u 0=

or2 80

t 4 s10

= =

Final speed just before first collisionv = 0 + 10 4 = 40 m/s

It now loses half of initial speed after the collision i.e., when it first bounces its initial speed is 40

20 m / s.2

=

So, the time is loosing half of its speed.0 = 20 10 t (during upward motion)

20t 2 s

10= = (final speed = 0)

In 2 s, it attains height21h 20 2 10 (2)

2= -

h = 40 20 = 20 mNow, it is dropped again from 20m with zero initial speed. Time taken in reaching the ground

2120 0 10(t)2

= +

t = 2 sAlso final speed v2 = 0 + 2 10 20 (from v2 = u2 + 2gh) v = 20 m/sThus, with the above data, we can draw the speed-time graph.

Speed (m/s)

4020

4 6 8 Time(s)

Since, velocity is a vector quantity so from the above graph, we can now draw the velocity-time graph.[Take downward motion positive and upward motion negative in case of vt graph]

4020

4 6 820

Velocity (m/s)

Time(s)

Example 18. The velocity-time graph of an object moving along a straight line is as shown in the fig.Calculate the distance covered by the object :

v

20 ms1

O0 2 5 10

C

BA

A B

t (s)

KINEMATICS


(a) between t = 0 to t = 5 s and(b) between t = 0 to t = 10 s.

Sol. (a) Let x1 be the distance covered in the time interval between t = 0 to t = 5 s. Then, x1 = area of the trapeziumOABB

AB OB 3 5AA 20 80 m

2 2 + +

= = = (b) Let x2 be the distance covered in the time interval between t = 0 to t = 10 s. Then, x2 = area of thetrapezium OABC.

AB OC 3 10AA 20 130 m

2 2+ +

= = =

IMPORTANT FEATURES1. For a particle having zero initial velocity if v t, s t2 and v2 s then acceleration of particle must be

constant i.e., particle is moving rectilinearly with uniform acceleration.2. For a particle having zero initial velocity if s t, where > 2, then particles acceleration increases with time.3. For a particle having zero initial velocity if s t, where < 0, then particles acceleration decreases with time.4. When a body is non-uniformly accelerated, then problem can be solved either be differentiation or integration

(applying some boundary conditions).Differentiation : s t v t a tIntegration : a t v t s t

By boundary condition we mean that velocity or displacement at some time (usually at t = 0) should be knownto us. Otherwise we cannot find constant of integration.

5. Equation dv

a vds

= or vdv = ads is useful when acceleration displacement equation is known and velocitydisplacement equation is required.

6. When either u = 0 or u a

, motion is only accelerated.

7. It can be observed when either u = 0, u a

or u a

.

8. When u a

motion is first retarded (till the velocity becomes zero) and then accelerated in opposite direction.9. Following are few points we may conclude in case of a one dimensional motion :

(a) slope of displacement-time graph gives velocity ds

as vdt

= .

(b) slope of velocity-time graph gives acceleration dv

as adt

= .

(c) area under velocity-time graph gives displacement as (ds = vdt).(d) area under acceleration-time graph gives change in velocity (as dv = adt).(e) displacement-time graph in uniform motion is a straight line passing through origin, if displacement is zero attime t = 0 (as s = vt).(f) velocity-time graph is a straight line passing through origin in a uniformly accelerated motion if initial velocityu = 0 and a straight line not passing through origin if initial velocity u 0 (as v = u + at).

(g) displacement-time graph in uniformly accelerated or retarded motion is a parabola 21

as s ut at2

= .

10. Slopes of vt or st graphs can never be infinite at any point, because infinite slope of vt graph means infiniteacceleration. Similarly, infinite slope of st graph means infinite velocity. Hence, the following graphs are notpossible :

KINEMATICS


v

t t

s

11. At one time, two values of velocity or displacement are not possible. Hence, the following graphs are notacceptable :

v

0t t

v

1

v2s

0t t

s

1

s2

RELATIVE MOTIONIf the velocities of two bodies are known w.r.t a common frame of reference, then the velocity of a body can bemeasured w.r.t. the second body. Therefore, if the velocities of two bodies A and B w.r.t. the ground are Av

and Bv

then the relative velocity of A w.r.t. B is

AB A Bv v v= -

Similarly, we see thatAB BAv v= -

Also, relative acceleration of A w.r.t. B isand AB A Ba a a= -

AB BAa a= -

Example 19. Car A has an acceleration of 6 m/s2 due east and car B, 8 m/s2 due north. What is the acceleration ofcar B w.r.t. to car A ?

Sol. It is a two dimensional motion.Therefore, BAa

= acceleration of car B w.r.t. car AA

B Aa a= -

2 2 2BAa (6) (8) 10 m / s= - =

BAaN

S

EW

Ba = 8 m/s2

Aa = 6 m/s2and 1 18 4

tan tan6 3

- - a = =

Thus, BAa

is 10 m/s2 at an angle of -14

= tan3

from west towards north.

Discussion(a) If a satellite is moving in equatorial plane with velocity v

and a point on the surface of earth with velocity u

relative to the centre of earth, the velocity of satellite relative to the surface of earth

SEv v u

= -(b) If a car is moving at equator on the earths surface with a velocity CEv

relative to earths surface and a point on

the surface of earth with velocity vE relative to its centre, then

CE C Ev v v

= -(c) If the car moves from west to east (the direction of motion of earth)

KINEMATICS


vC = vCE + vEand if the car moves from east to west (opposite to the motion to earth)

vC = vCE vE(d) For crossing the river in shortest time, the boat should sail perpendicular to

the flow. If the width of river is d and v the velocity of boat in still water,then,

dt

v=

O

rvrv

v

B C

The position of boat at the other bank is C (not B).The displacement of the boat OC OB BC= = +

2 2OC (OB) (BC)= +2

2 2 2r r

dOC d (v t) d v

v

= + = + (e) For crossing the river in shortest distance, the boat moves as such its horizontal component of velocity balances

the speed of flow.OB = the shortest path = d

vr = v sin rvsin =

v

O

rv

rv

v

BA

2cos = 1sin 2 2 2

r rv v vcos = 1v v

- =

2 2

r

d dt

v cos v v vv

= =q -

2 2r

dt

v v=

-In this case, the magnitude of displacement = d.

(f) If boat crosses the river along the shortest path, then time is not least.

Example 20. A police van moving on a highway with a speed of 30 km/h fires a bullet at a thiefs car speeding awayin the same direction with a speed of 192 km/h. If the muzzle speed of the bullet is 150 m/s, with what speeddoes the bullet hits the thiefs car ?

Sol. Let p bv , v

and tv

represent velocity of the police van, muzzle velocity of bullet and the velocity of the thiefsscar respectively. This is a one dimensional motion.

p25

v 30km / h m / s3

= =

bv 150 m / s=

and t160

v 192 km / h m / s3

= =

Since, bullet is fired from the moving police van, the effective velocity of the bullet will be

t b p25 475

v v v 150 m / s3 3

= + = + =

KINEMATICS


if the relative velocity of the bullet w.r.t the thiefs car is btv ,

then

bt b tv v v= -

bt475 160 315

v 105 m / s3 3 3

= - = =

Example 21. A, B & C are three objects each moving wit constant velocity. As speed is 10m/sec in adirection PQ

. The velocity of B relative to A is 6 m/sec at an angle of cos1(15/24) to PQ. The velocity

of C relative to B is 12 m/sec in a direction QP

, then find the magnitude of the velocity of C.Sol. A v 10 i=

B v 6cos i 6sin j

Here15 351cos , sin24 24

And CB v 12 i= -

BA B Av v v= -

or B 6cos i 6sin j v 10 i

B v cos i 6sin j 10 i

15 351 6 i 6 j 10 i24 24

=15 351 10 i j4 4

55 351 i j4 4

But CB C Bv v v= -

C CB B55 351 v v v 12 i i j4 4

= + = - + +

7 351 i j4 4

= +

22

C7 351v4 4

49 351 400 204 4 4+= = = Ans. 5 m / s

Example 22. A man crosses a river in a boat. If he crosses the river in minimum time he takes 10 min with a drift120 m. If he crosses the river taking shortest path, he takes 12.5 min, find :(a) width of the river(b) velocity of the boat with respect to water,(c) speed of the current.

Sol. Let vr = velocity of riverv = velocity of river in still water andd = width of river

A

B

d

For minimum time

v

KINEMATICS


Given, tmin = 10 min

ord

10v

= ...(i)

Drift in this case will be,x = vrt

120 = 10vr ...(ii)Shortest path is taken when vb is along AB. In this case,

2 2b rv v v= -

Now, 2 2b r

d d125

v v v= =

- ...(iii)

Solving these three equations we get,A

Brv

v

Shortest path

bv

v = 20 m/min, vr = 12 m/min and d = 200 m.

Example 23. A man with some passengers in his boat, starts perpendicular to flow of river 200m wide andflowing with 2m/s. boat speed in still water is 4m/s. When he reaches half the width of river the passengersasked him they want to reach the just opposite end from where they have started.(a) Find the direction due to which he must row to reach the required end.(b) How many times more total time, it would take to that if he would have denied the passengers.

Sol. (a) Event (1) From A to B,Time taken by boat to recent from A to B is

y

100 100t 25sec.v 4

= = =

Also, AD = vx t = 2t = 50 mEvent (2) From B to C,Actual velocity of boat should be along BC. This actual velocity is found by resultant of vrel and vr.

rel b rv v v= -

xcomput of actual velocity isvx = vr vrel sin = 2 4sin

and y comput of actual velocity isvy = vrel cos

vbvrel 200mB

E

vrvrel

Cy

A DThe time taken to go from B to C is

0y

BEtv

=

rel

100 100v cos 4cos

25cos

Also, EC = vx t0 = (2 4 sin ) t0

254sin 2

cos

But EC = AD = 50 m

KINEMATICS


25EC 4sin 2

cos

2550 4 sin 2cos

50 cos = 100 sin 50cos = 2 sin 11 + cos = 2 sin

22cos 4sin cos2 2 2

1tan2 2a =

1 1tan2 2

1 12 tan

2

(b) If boat crosses the river with initial condition

1y rel

200 200tv v

= =

1200t 50sec.

4= =

If boat cross the river with final condition,2 AB BCt t t= +

t2 = t + t0

225t 25

cos

1tan

2 2

2cos2 5

2cos 2cos 12

42 15

8 315 5

= - =

2

25 125 200t 25 253 3 35

= + = + =

2

1

t 200 4t 3 50 3

2

1

t 4t 3

=

Example 24. To a person going west wards with a speed of 6 km/h rain appears to fall vertically downwards witha speed of 8 km/h. Find the actual direction of rain.

Sol. Let Mv

= velocity of man = 6 km/h

v

= relative velocity of rain w.r.t. man = 8 km/h

KINEMATICS


Rv

= actual velocity of rainIn this case

( )R Mv v v= + -

OEW

Vertical

S

N

SB

A

C

v

Mv

RvR Mv v v= -

or R Mv v v= +

2 2Rv (6) (8) 10 km / h= + =

The velocity of rain ( )Rv

is given by the vector OC

, the resultant of vectors OA

and OB

as shown in figure.

If is the angle that Rv

makes with the vertical, then

MBC | v | 6tan 0.75OB 8| v |

or = 36 52 (east of vertical)

Example 25. Rain is falling vertically with a speed of 20 ms1 relative to air. A person is running in the rainwith a velocity of 5 ms1 and a wind is also blowing with a speed of 15 ms1 (both towards east). Find theangle with the vertical at which the person should hold his umbrella so that he may not get drenched.

Sol. ra v 20 k= -

m v 5 i=

a v 15 i=

x

Nz

EW

Sra r av v v= -

r v 20 k 15i= - +

rm r mv v v= -

20k 15i 5i= - + - 20 k 10 i= - +10 1tan20 2

1 1tan2

vrmvertically

Example 26. An aircraft flies at 400 km/h in still air. A wind of 200 2 km / h is blowing from the south. The pilotwishes to travel from A to a point B north east of A. Find the direction he must steer and time of his journey ifAB = 1000 km.

Sol. Given that vw = 200 2 km / h . vaw = 400 km/h and av

should be along AB or in north-east direction. Thus,

the direction of awv

should be such as the resultant of wv

and awv

is along AB or in north-east direction.

A

NB

Cav

45

45

awv = 400 km/h

wv = 200 2 km/h

E

KINEMATICS


Let awv

makes an angle with AB as shown in fig. Applying sine law in triangle ABC, we getAC BC

sin 45 sin

orBCsin sin 45AC

200 2 1 1sin400 22

= 30therefore, the pilot should steer in a direction at an angle of (45 + ) or 75 from north towards east.

Further,a| v | 400

sin(180 45 30 ) sin 45

or asin105 km| v | (400)sin 45 h

acos15 km 0.9659 km| v | (400) (400)sin 45 h 0.707 h

akm| v | 546.47h

The time of journey from A to B is

a

AB 1000t h546.67| v |

t = 1.83 h

Example 27. A glass wind screen whose inclination with the vertical canbe changed, is mounted on a cart as shown in figure. The cart movesuniformly along the horizontal path with a speed of 6 m/s. At what v=6m/smaximum angle to the vertical can the wind screen be placed sothat the rain drops falling vertically downwards with velocity 2 m/s,do not enter the cart ?

Sol. c v 6 i=

r v 2 j=

rc r cv v v= - vr

vrc y

D vc x

c

r

v 6tan 3v 2

AEcos

AE cos ED sintanBE cos

2

sin32sin / 2

sincos

vr

E

A

B

D

C

2

2sin cos2 23 cot

22sin2

KINEMATICS


or cot 32

or1tan

2 3

or 11tan

2 3

1 12 tan

3

MOTION IN TWO AND THREE DIMENSIONSINTRODUCTION

A body is free to move in space. In this case, the initial position of body is taken as origin.Any convenient co-ordinate system is chosen. Let us suppose that at an instant t, the body is at point P(x, y, z).The position vector of the body is r x i y j z k= + +

. Thus, velocity

dr dx dy dz v i j kdt dt dt dt

= = + +

The velocity along x-axis is, xdx

vdt

=

and acceleration along x-axis is xxdv

adt

= .

The velocity along y-axis is ydy

vdt

=

and the acceleration along y-axis is yydv

adt

=

Similarly, zdz

vdt

= and zzdv

adt

=

The acceleration of the body x y z a a i a j a k= + +

.Discussion(a) If ax is constant,

vx = ux + axt2

x x1

x u t a t2

= +2 2x x xv u 2a x= +

If ax is a variable,

xx v dx= x xdv a dt=

(b) If ay is constant2

y y1

y u t a t2

= +

vy = uy + ayt2 2y y yv u 2a y= +

KINEMATICS


If ay is variable,

yy v dt= y ydv a dt=

(c) If az is constant,vz = uz + azt

2z z

1z u t a t

2= +

2 2z z zv u 2a z= +

If az is variable,

zz v dt= z zdv a dt=

If the motion of the body takes place in xy plane, then az = 0, vz = 0, uz = 0

Example 28. A bird flies in the xy plane with a velocity 2 v t i 3t j= +

. At t = 0, bird is at origin.Calculate position and acceleration of bird as function of time.

Sol. We have given2 v t i 3t j= +

Here, vx = t2, vy = 3t and vz = 0

Since, vx = t2

or 2dx tdt

orx t 20 0

dx t dt

or3tx3

Also, vy = 3t

ordy 3tdt

ory t

0 0dy 3t dt

23ty

2

Thus, position of bird is r x i y j

3 2t 3t r i j3 2

vx = t2

2

xx

dv d(t)a 2tdt dt

and vy = 3t

KINEMATICS


or ydv dt

3dt dt

ay = 3 unit

Thus, acceleration of bird is x y a a i a j

a 2t i 3 j

PROJECTILE MOTIONWe next consider a special case of two dimensional motion : A particle moves in a vertical plane with initialvelocity 0v

but its acceleration is always the free fall acceleration g

, which is downward. Such a particle is

referred to as a projectile (meaning that is projected or launched) and its motion is called projectile motion. Aprojectile might be a baseball in flight or a golf ball, but it is not air plane or a duck in flight. Or goal here is toanalyze projectile motion using the tools for two-dimensional motion and making the assumption that air has noeffect on the projectile.Now let us consider a projectile launched so that its initial velocity v0 makes an angle with with the horizontal(shown in figure). For discussion of motion, we take origin at the point of projection. Horizontal direction as x-axis and vertical direction as y-axis is taken.

g 0v

A x

y

O

The initial velocity of projectile along x-axis is ux = u cos .The component of gravitational acceleration along x-axis is ax = g cos 90 = 0.The component of initial velocity along y-axis is uy = u sin .The acceleration along y-axis is ay = g

NOTE : In projectile motion, the horizontal and the vertical motions areindependent of each other, that is, neither motion affects the other.

Discussion(a) The instantaneous velocity of the projectile as function of time : Let projectile reaches at point (x, y)

after time t [Fig.]. vx = ux = u cos and vy = uy gt = u sin gt

x y v v i v j

gu

A x

y

O

yv

xvP(x,y)

v u cos i (u sin gt) j

The instantaneous speed2 2| v | (u cos ) (u sin gt)

Also, x = uxt = (u cos ) t = ut cos

KINEMATICS


2y

1y u t gt2

or 21y u sin t gt2

The position of the projectile is r x i y j

21 r ut cos i ut sin gt j2

(b) Trajectory of projectile : The yx graph gives the path or trajectory of the projectile.

From discussion of instantaneous velocity of projectile.x = ut cos ...(i)

and 21y ut sin gt2

...(ii)

xt

u cos

...(iii)

Putting the value of t from Eq. (iii) into Eq. (ii).2x 1 xy u sin g

u cos 2 u cos

or2

2 2gxy x tan

2u cos

...(iv)

This is the required path or projectile.

Multiplying the eq. (iv) by 2 22u cos

g

to both sides, we get

2 2 22 2u sin cos 2u cosx x y

g g

Adding 22u sin cos

g

to both sides, we get

22 2 2 2 2u sin cos 2u cos u sinx yg g 2g

This is of the form,(x a)2 = c(y b) ...(v)

which is the equation of a parabola. Hence, the equation of the path of the projectile is a parabola.

NOTE : The trajectory of a projectile will be parabolic when direction of velocity ofprojectile is different from direction of acceleration and its acceleration is

constant both in magnitude and direction.

(c) Time of flight : In Fig., the time taken by projectile to reach at point A from point O is known as time of flight.Here, OA = vxT, where T is time of flight. The total displacement along y-axis during motion of projectile fromO to A is zero. So, y = 0

KINEMATICS


but 2y1y u T gT2

or 210 u sin T gT2

2u sinT

g

...(vi)

(d) Range of projectile : Distance OA is known as range of projectile [Fig.]The time taken to reach point A from point O is

2u sinTg

The range x2u sinR u T u cos

g

2sin cosR ug

2u sin 2R

g

...(vii)

Caution : This equation does not give the horizontal distance travelled by a projectile when the final height is not thelaunch height.Regarding range of projectile, two cases are discussed below :

Case-I : Range will be maximum if sin 2 = 1 or = 45

2

max.u

Rg

(at = 45)

Case-II : For given velue of v, the range of projectile will be same for angle and 90 though their times of flightand maximum heights are different.

290

u sin 2 90R

g

290

u sin 180 2R

g

2

90u sin 2

R Rg

Thus, for the case as shown in fig,x

yu

u

3060

OR30 = R60

(e) Height attained by projectile : At the maximum height (at point B) the vertical component of velocity is zero.

2 2y yv u 2gH or (0)2 = (u sin )2 2gH

2 2u sin

H2g

x

y yv = 0

xu

A

H

B

O

KINEMATICS


NOTE : If air resists or opposes the projectile motion [Fig.], then

0

uy

x

I

II

0

> 0ITrajectory in vacuumIITrajectory in presence of air resistance

O

Time taken by projectile during upward motion < Time taken during downward motion. The values of height attained and of range of a projectile decrease. The projectile returns to the ground with less speed. At its trajectory its horizontal velocity also decreases. Time of flight also decreases. At which angle, the projectile returns to the ground, increases.

Example 29. Prove that the maximum horizontal range is four times the maximum height attained by the projectile;when fired at an inclination so as to have maximum horizontal range.

Sol. For = 45, the horizontal range is maximum and is given by2

maxu

Rg

Maximum height attained2 2 2

maxmax

Ru sin 45 uH

2g 4g 4

or Rmax = 4 Hmax

Example 30. There are two angles of projection for which the horizontal range is the same. Show that the sum ofthe maximum heights for these two angles is independent of the angle of projection.

Sol. There are two angles of projection and 90 for which the horizontal range R is same.

Now,2 2

1u sin

H2g

and 2 2

2u sin 90

H2g

2 2

2u cos

H2g

Therefore, 2 2

2 21 2

u uH H sin cos

2g 2g

Clearly, the sum of the heights for the two angles of projection is independent of the angle of projection.

Example 31. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60 with the vertical.Find the time when the particle will move perpendicular to its initial direction, taking g = 10 m/sec2.

Sol. Here ax = 0ay = gux = 100 sin60 = 50 3uy = 100 cos60 = 50

KINEMATICS


0 x y u u i u j 50 3 i 50 j= + = +

y y yv u a t 50 gt= + =

x xv u 50 3 m / s= =

y

x60 ( ) v 50 3 i 50 gt j= + -

But u

and v

are perpendicular..

u v 0

or 50 3 i 50 j 50 3 i 50 gt j 0 or 7500 + 2500 500 t = 0

10000t

500= 0t = 20 second

Example 32. Fig. shows a pirate ship 560 m from a fort defending the harbor entrance of an island. A defensecannon, located at sea level, fires balls at initial speed v0 = 82 m/s.

(a) At what angle 0 from the horizontal must a ball be fired to hit the ship ?(b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs?

x

y

2763

R = 560 m

Sol. (a) Because the cannon and the ship are at the same height, the horizontal displacement is the range.The horizontal range is

20vR sin 2

g ...(i)

which gives us

10 2

0

gR2 sin

v

10 2

9.8 5602 sin(82)

20 = sin1(0.816) ...(ii)

01 (46.7 ) 232

and = 90 0 = 90 23 = 67

The commandant of the fort can elevate the cannon to either of these two angles and (if only there were nointervening air!) hit the pirate ship.

(b) We have seen that maximum range corresponds to an elevation angle 0 of 45. Thus, from Eq. (i) with

KINEMATICS


0 = 45.2 20

0v (82)

R sin 2 sin(2 45 )g 9.8

R = 686 m 690 mAs the pirate ship sails away, the two elevation angles at which the ship can be hit draw together, eventuallymerging at 0 = 45 when the ship is 690 m away. Beyond that distance the ship is safe.

IMPORTANT FEATURES1. As we have seen in the above derivations that ax = 0, i.e., motion of the projectile in horizontal direction is

uniform. Hence, horizontal component of velocity u cos does not change during its motion.2. Motion in vertical direction is first retarded then accelerated in opposite direction. Because uy is upwards and

ay is downwards. Hence, vertical component of its velocity first decreases from O to A and then increases fromA to B. This can be shown as in fig.

u

B x

y

O

A

xu

yu

3. The co-ordinates and velocity components of the projectile at time t arex = sx = uxt = (u cos ) t

2 2y y y1 1y s u t a t u sin t gt2 2

vx = ux = u cos and vy = uy + ayt = u sin gt

Therefore, speed of projectile at time t is 2 2x yv v v and the angle made by its velocity vector with

positive x-axis isy1

x

vtan

v

4. Equation of trajectory of projectilex = (u cos )t

xt

u cos

Substituting this value of t in,

21y u sin t gt2

, we get2

2 2gxy x tan

2u cos

2

22

gxy x tan sec2u

2

22

gxy x tan 1 tan2u

These are the standard equations of trajectory of a projectile.The equation is quadratic in x. This is why the path of a projectile is a parabola. The above equation can also

KINEMATICS


be written in terms of range (R) of projectile as

xy x 1 tanR

5. Range R is given by

2R u cos u sing

Horizontal component Vertical component2Rof initial velocity of initial velocityg

6. There are two unique times at which the projectile is at the same height h(< H) and the sum of these two times

equals the time of flight T. Since, 21h u sin t gt2

is quadratic in time, so it has two unique roots t1 and

t2 (say) such that sum of roots (t1 + t2) is 2u sin

g

and product (t1t2) is 2hg . The time lapse (t1 t2) between

these two events is

(t1 t2)2 = (t1 + t2)

2 4t1 t22 2

1 24u sin 8ht t

g g

7. If K is the kinetic energy at the point of launch then kinetic energy at the highest point is

2 2 2x

1 1K mv mu cos2 2

K = K cos2 8. For complementary angles and 90 , if T and T90 are the times of flight and R is the range, then

9090

2R 2R 2RT Tg g g

e.g., 8911 892R2RT T

g g

Example 33. A particle is projected in the XY plane. 2 sec after projection the velocity of the particlemakes an angle 45 with the X-axis. 4 sec after projection, it moves horizontally. Find the velocity ofprojection.

Sol. After 4 sec the particle reach at maximum hight.At maximum height it move horizontally.

So that,T 42

=

2u sinT 8g

2usinT

g

v0

O x

y

80u sin 402

...(1)

u cos = v cos 45v sin45 = 45 sin 10 2

KINEMATICS


u sin 20 201u cos u cos

ucos = 20 ...(2)From eqn (1) and (2)

u2 sin2 + u2 cos2 = 402 + 202

u2 = 1600 + 400 = 2000u 20 100 10 20

20 5 m / s

PROJECTILE PROJECTED FROM SOME HEIGHTProjectile Projected in Horizontal Direction

Let a projectile is projected with velocity u.

Take observation point O at a height h from ground [Fig.].

v

P

h

yx

A

y(x, y)x

urO

a = g

vx

vy

y

Here, ux = u, uy = 0and ax = 0, ay = g

(a) Let at time t, the co-ordinates of position of projectile is (x, y), then

x = ut and 21y 0 gt2

Therefore, at time t position vector r x i y j

21 r (ut) i gt j2

22 2 2 21| r | x y (ut) gt

2

andytanx

(b) Let at time t the horizontal and vertical velocities of projectile be vx and vy, sovx = u and vy = 0 + (g)t = gt

x y v v i v j u i ( gt) j

and 2 2x yv v v 2 2v u ( gt)

andx

y

vtanv

(c) Let time taken by projectile from O to point A at ground is T, then

21h 0 ( g)T2

KINEMATICS


2hTg

The horizontal distance in time T2hPA uT ug

Therefore, the bomb dropped from an aeroplane moving with velocity u horizontally at height h, covers a

horizontal distance 2hug on the ground.

Projectile Projected Upwards at an Angle Let projectile is projected upward at angle with horizontal velocity u

[Fig.].

h

P D C

AB

u

O

u si

n

u cosu sin

u cos

u

x

a = gv = 0, v = u = u cos

y

y x x

ux = u cos and uy = u sin ax = 0 and ay = g

Now from 2nd equation of motion,

21h u sin T g T2

or gT2 (2u sin )T 2h = 0Solving this equation, we get horizontal distance covered in time T

PC = (u cos )T

Time taken in covering path 2u sinOAB

g

and horizontal distance covered in this time2u sin 2

OBg

In such case for range PC to become maximum, should be just less than 45.

NOTE : In such case on earths surface for maximum range,

2

usin =2u + 2gh .

If h = 0, then 1

=2 = 45.

Projectile Projected Downward at an Angle Let projectile is projected downward at an angle with horizontal velocity u

[Fig.].

KINEMATICS


P

h

yx

A

uO

a = gy

= u cosux

= u

sin

u y

ux = u cos and uy = u sin

ax = 0and ay = gFrom 2nd equation of motion,

21h = u sin T g T2

or gT2 + (2u sin )T 2h = 0To solve this euation, value of T can be evaluated. In this time the horizontal distance covered on the earth,

PA = (u cos )T

Example 34. A projectile is fired horizontal with a velocity of 98 m/s from thetop of a hill 490 m high. Find

(a) the time taken by the projectile to reach the ground(b) the distance of the point where the particle hits the ground from foot

B

x

A

u = 98 m/sO

vx

vy

y

of the hill and(c) the velocity with which the projectile hits the ground (g = 9.8 m/s2)Sol. Here, ux = 98 m/s, ax = 0, uy = 0 and ay = g(a) At A, sy = 490 m. So, applying

2y y y

1s u t a t2

21490 0 (9.8)t

2

t = 10 s

(b) 2x x x1BA s u t a t2

or BA = (98) (10) + (0) or BA = 980 m(c) vx = ux = 98 m/s

vy = uy + ayt = 0 + (9.8) (10) = 98 m/s

2 2 2 2x yv v v (98) (98) 98 2 m / s

andy

x

v 98tan 1

v 98

= 45Thus, the projectile hits the ground with a velocity 98 2 m / s at an angle of = 45 with horizontal.

Example 35. A train starts from rest and moves with a constant acceleration of 2.0 m/s2 for half a minute. Thebrakes are then applied and the train comes to rest in one minute. Find

KINEMATICS


(a) the total distance moved by the train,(b) the maximum speed attained by the train and(c) the position(s) of the train at half the maximum speed.

Sol. (a) S1 = The distance moved by the train in first half minute.

S1 = u1t1 + 12 a1t1

2

S1 = 0 t1 + 12 2 (30)

2 { t1 = 12 minute}

S1 = 12 2 900 S1 = 900 m

The distance moved by the train after brakes applied.

S2 = u2t2 + 12 a2t2

2

u2 = u1 + a1t1u2 = 0 + 2 30u2 = 60 m/sv2 = u2 a2t20 = 60 a2 60

22a 1 m / sec=

v22 = u2

2 + 2 a2S20 = 60 60 2 1 S2S2 = 1800 m

Total distance(s) = S1 + S2S = (900 + 1800) mS = 2700 m S 2.7 km=

(b) the maximum speed attained by the train is

2u 60 m / s=(c) the position of the train at half of minimum speed.

u22 = u1

2 + 2a1S(30)2 = 0 + 2 2 S900 S'

4= S' 225 m=

Example 36. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms1 by a constantacceleration of 5 ms2. It maintains a constant velocity of 20 ms2 for 5 seconds and then is brought to restagain by a constant acceleration of 2 ms2. Draw a velocity-time graph and find the distance covered by thecar.

Sol. v22 = u2 + 2aS

(20)2 = 0 + 2 5 S400S10

= S 40 m=

Car maintains a constant velocity of 20 m/s for 5 Second.S = v tS = 20 5 S '' 100 m=

KINEMATICS


The car comes to rest by a constant acceleration of 2 m/s2

(v)2 = v2 + 2 aS0 = (20)2 2 2 S

400S ''4

= S'' 100 m=

The total distance covered by the car.= S + S + S= 40 + 100 + 100= 240 m

Example 37. A ball rolls off the edge of a horizontal table top 4m high. If it strikes the floor at a point 5m horizontallyaway from the edge of the table, what was its speed at the instant it left the table ?

Sol. Using 21h gt2

we have,

2AB AC

1h gt2

or ABAC2h 2 4t 0.9 s

g 9.8

B

Av

C

5 m

4 m

Further, BC = vtAC or AC

BC 5.0v 5.55 m / st 0.9

Example 38. A ball is projected at an angle of 30 above with the horizontal from the top of a tower andstrikes the ground in 5 sec at an angle of 45 with the horizontal. Find the height of the tower and thespeed with which it was projected.

Sol. ux = v0 cos30uy = v0 sin30Vx = uX = v0 cos30vy = uy + ay tvy = v0 sin30 gt

0y

vv 10 52

30v0

y

xH

a =0, a = gx yBut

0

y

x 0

v 50v 2 tan 45v v cos30

00

vv cos30 502

01 3v 502 2

0100v3 1

=+

100 3 1

3 1 3 1

= 50 3 1

KINEMATICS


Also, 2y y1y H u t a t2

= - = +

20

1H v sin 30 5 10 52

H 125 2 3 m

Example 39. A ball is thrown horizontally from a cliff such that it strikes groundafter 5 sec. The line of sight from the point of projection to the point ofhitting makes an angle of 37 with the horizontal. What is the initial velocity

37

of projection.Sol. vx = v0cos37

vy = v0sin37ax = gsin37 = 6 m/s

2

ay = gcos37 = 8 m/s2

From O to A, displacement along y-axis is zero.

37

37x

v0

yO

Ay = uyt + ayt20 = uy 5 + 8 25 { t = 5 sec.}5uy = 8 25

y100u 20 m / s

5= =

0320 v5

0100

v m / s3

=

Example 40. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53 to horizontal.Its speed when it is at a height of 0.45 m from the point of projection is :(A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient

Sol. According to conservation principle of machenic energyUi + Ti = Uf + Tf

2 21 10 mu mgh mv2 2

+ = +

u2 = 2gh + v22v u 2gh= -2v 5 2 10 0.45

v 25 9 16 v 4 m / s

Example 41. In the figure shown, the two projectiles are fired simultaneously.What should be the initial speed of the left side projectile for the twoprojectile to hit in mid-air ?

60 4520m/s

u

10 mSol. When two projectiles are projected from same height, then for collision, vertical component of velocitiesof both projectiles should be same

KINEMATICS


u sin 60 = 20 sin45

20sin 45u

sin 60=

20 2u2 3

2u 20 m / s3

=

Example 42. The speed of a particle when it is at its greatest height is 2 / 5 times of its speed when it is atits half the maximum height. the angle of projection is _____ and the velocity vector angle at half themaximum height is ______.

Sol. uy = usinux = ucosvx = ucos ...(1)

u u cos

2 2y y yv u 2a S= +2 2 2y

Hv u sin 2g2

...(2)

2 2x y

2u cos v v5

2 2 2 22 Hu cos u cos u sin 2g5 2

2 222 u sinu cos u g

5 2g

22 sinu 15 2

2 sincos 15 2

22 2 sincos 1

5 2

or2

2 2 sincos5 5

,2

2 sin 21 sin5 5

or 2 25 5sin sin 2 or 3 4 sin2 = 0

3sin2

60

and2 2

y

x

v u sin gHtan

v u cos

{from eqn. (1) and (2)}

2 22 2 u sinu sin g

2gtan

u cos

22 sinsin

2tancos

2sin2tan

cos

tan tan 60tan

2 2

KINEMATICS


3tan2

13tan2

Example 43. A projectile is to be thrown horizontally from the top of a wall of height 1.7 m. Calculate theinitial velocity of projection if its hits perpendicularly an incline of angle 37 which starts from the groundat the bottom of the wall. The line of greatest slope of incline lies in the plane of motion of projectile.

Sol. ux = v0uy = 0ay = g = 10 m/s

2

vx = v1 cos53 = 0.6 v1vy = v1 sin53 = 0.8 v1

vx = ux = v0or 0.6 v1 = v0

0 0 01

v 10v 5vv0.6 6 3

= = = ...(1)

vy = uy + ayt0.8v1 = 10 t

05v0.8 10 t3

A

B

37

37 5353grad x

y

v0

v1

v1

h = 1.7 m

12

gt2C

O x

06 60v t t

0.8 8= = ...(2)

In BAC

ABtan 37AC

= 2

2

x 0

11.7 gt3 1.7 5t24 u t v t

- -= =

or 3v0t = 6.8 20t2

or2

0 00

8v 64 v3v 6.8 20

60 3600

or2

2 00

64v24v 6.8

60 180+ =

20

72 64 v 6.8180

or 206.8 180v 9

136

0v 3 m / s=

Example 44. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2 m/sec. A dear running with a speed V in front at a distance of 4 5m moving perpendicular to the directionof motion of the elephant. If hunter can throw his spear with a speed of 10 m/sec, relative to the elephant,then at what angle to its direction of motion must he thrown his spear horizontally for a successful hit.Find also the speed V of the dear.

Sol. 21h gt2

=

2h 2 4 8tg 10 10

KINEMATICS


Assume horizontal plane at xyplane.

rel v 10cos i 10sin j

rel v u 2 i

rel u v 2 i 10cos 2 i 10sin j

The deer is moving along yaxis.So, displacement of deer and displacement of spear along yaxis will be same in time t. v t = uyt v = 10 sin ...(1)Also, along xaxis :

xu t 4 5=

or 810cos 2 4 510 2m/s4 m

x

or 1010cos 2 4 5 8 4 5 10

2

10 cos = 8

8 4cos

10 5 = 37

From eqn (1)

v = 10 sin = 10 sin37 = 3105

= 6 m / s

Example 45. An object A is kept fixed at the point x = 3 m and y = 1.25 m on aplank P raised above the ground. At time t = 0 the plank starts moving alongthe + x direction with an acceleration 1.5 m/s2. At the same instant a stone isprojected from the origin with a velocity u as shown. A stationary person on

y1.25m

uO x3.0m

PA

the ground observes the stone hitting the object during its downward motion at an angle of 45 to the horizontal.All the motions are in x y plane. Find u and the time after which the stone hits the object. Take g = 10 m/s2.

Sol. ux = u cosuy = u sinax = 0ay = g

If the stone hitt the object after time t.45

u =ucosx

1.25

u cos

u si

n

u

So that virtical displacement of stone is 1.25 m2

y y1y u t a t2

= +

therefore 211.25 (u sin )t gt2

211.25 (u sin )t 10 t2

1.25 = (u sin) t 5 t2

(u sin)t = 1.25 + 5t2 ...(1)Hotizontal displacement of stone is

x = 3 + displacement of object A.Initial velocity of object is zero.

KINEMATICS


so displacement of object is2 2 21 11.25 at 1.5 t 0.75 t

2 2

x = 3 + 0.75 t2

(u cos)t = 3 + 0.75 t2 ...(2)Since velocity vector inclined at 45 with horizontals.

y

x

u u sin gttan(45 )

u u cos

u cos = (u sin gt)u cos = gt u sin ...(3)(u cos) t + (u sin) t = 10 t2

Add eqn (1) and (2)(u cos) t + (u sin) t = 4.25 + 5.75 t2 ...(4)

from eqn (3) and (4)10 t2 = 4.25 + 5 .27 t2

4.25 t2 = 4.25t2 = 1 t 1 sec.=

From eqn (1) and (2)uy = u sin = 6.25 m/suy = 6.25 m/sux = 3.75 m/s

2 2x yu u u= +

2 2u (6.25) (3.75)= + u 7.29 m / s=

IMPORTANT FEATURESProjectile motion is a two dimensional motion with constant acceleration (g). So, we can use

v u a t

21a u t a t2

etc. in projectile motion as well. Here, u u cos i u sin j

and a g j

Now, suppose we want to find velocity at time t.

v u a t

v u cos i u sin j gt j

x

y

u

Ogor v u cos i u sin gt j

Similarly, displacement at time t will be,

21S u t a t2

21 S u cos i u sin j t gt j2

KINEMATICS


21 S u cos i ut sin gt j2

PROJECTILE MOTION ON AN INCLINED PLANEOne an inclined plane projectile is projected into two cases, one upwards and the other downwards.

Up the PlaneA projectile is projected up the inclined plane from the point O with an initial velocity u at an angle withhorizontal. The angle of inclination of the plane with horizontal is [Fig.].

u

y

x

A

BOx y

Oxy

y

x

g sin(90 ) = g cosg co

s(90 )

= g sin

90

ux = u cos ( ) and ax = g sin uy = u sin ( ) and ay = g cos

(i) Time of flight : During motion from point O to A, the displacement along y-axis is zero. sy = 0 at t = T

2y y y1s u t a t2

or 210 u sin T g cos T2

2u sinT

g cos

NOTE : Substituting = 0, in the above expression, we get 2u sinT =

g .

Which is quite obvious because = 0 is the situation shown in Fig.

x

y

u

Og

(ii) Range : As shown in Fig, OA is the range of proectile.Horizontal component of initial velocity uH = u cos OB = uHT (as aH = 0)

2u cos 2u sin 2u sin cos=g cos g cos

KINEMATICS


2

2

2u sin cosOBR OAcos g cos

Using, 2 sin A cos B = sin (A + B) + sin (A B)Range can also be written as,

2

2uR sin 2 sin

g cos

This range will be maximum when

22

or 4 2

and 2

max 2uR 1 sin

g cos

we see that for = 0, range will be maximum for 4

or 45.

2

max 2uR 1 sin 0

g cos 0

2

maxu

Rg

Alternative method :For range, sx = R, t = T

2x x x

1s u t a t2

or 21R u cos T g sin T2

Substituting the value of 2u sinT

g cos

, in above equation for R.

2

2uR sin 2 sin

g cos

Down the PlaneA projectile is projected down the plane from the point O with an initial velocity u at an angle with horizontal[Fig.]. The angle of inclination of plane with horizontal is .

u x

A

g cos(

90 )

= g sin

90

y

O

g sin(90 ) = g cos

Therefore, ux = u cos ( + ), ax = g sin uy = u sin ( + ), ay = g cos

Proceeding in the similar manner, we get the following results :

2u sinTg cos

KINEMATICS


2

2uR sin 2 sin

g cos

IMPORTANT FEATURESIf two particles are projected at angles 1 and 2 respectively as shown in fig., then the relative motion of 1with respect to 2 is a straight line at an angle.

1

u1

y

x

u2y

x2

12y1

12x

utan

u with positive x-axis.

where u12x = u1x u2x = u1 cos 1 u2 cos 2u12y = u1y u2y = u1 sin 1 u2 sin 2

Example 46. A particle is projected with a velocity of 20 m/s at an angle of 30 to an inclined plane of inclination30 to the horizontal. The particle hits the inclined plane at an angle of 30, during its journey. Find the

(a) time of impact,(b) the height of the point of impact from the horizontal plane passing through the point of projection.Sol. The particle hits the plane at 30 (the angle of inclination of plane). It means particle hits the plane horizontally.

(a)T u sint2 g

20sin 30 30t 1.76 s9.8

3030

u

(b)2 2u sin

H2g

2 220 sin 60H 15.3 m2 9.8

Example 47. A particle is projected up an inclined plane with initial speed v = 20 m/s at an angle = 30 with plane.Find the component of its velocity perpendicular to plane when it strikes the plane.

Sol. Component of velocity perpendicular to plane remains the same (in opposite direction)i.e., u sin = 20 sin 30 = 10 m/s

Example 48. A particle is thrown horizontally with relative velocity 10 m/s from aninclined plane, which is also moving with acceleration 10 m/s2 vertically upward.Find the time after which it lands on the plane (g = 10 m/s2).

10m/s2

30

Sol. ux = 10 cos30 = 5 3 m/suy = 10 sin30 = 5 m/sarel =20 m/s

2

KINEMATICS


ax = arel sin30 = 10 m/s2

30x

y10 m/s2

10 cos3

030O

v0= 10m/s10sin30

Aay = arel cos30

2320 10 3 m / s2

when ball lands on inclined plane, y = 0

y = uyt + 12 ay t

2

210 5 t 10 3 t2

5 3 t 5=

1t sec.3

=

Example 49. A particle is projected from point P with velocity 5 2 m/s perpendicular tothe surface of a hollow right angle cone whose axis is vertical. It collides at Qnormally. Find the time of the flight of the particle.

P Q45

y

xSol. u0x = 5 2m / su0y = 0ax = g sin45ay = g cos45

At point Q, vx = 0 { the particle collied normally at point Q}

45

45 45

QP

y x

gThe time taken by particle to go from P to Q is t0.

vx = ux + ax t0 = 5 2 g sin45 t0

05 2 5 2 2

t 1 sec.gsin 45 10

0t 1sec.=

Example 50. A ball is projected on smoot inclined plane in direction perpendicular toline of greatest slope with velocity of 8m/s. Find its speed after 1 sec. 37

8 m/s

Sol. vy = uy + aytvy = 0 + 10 sin37 1

vy = 310 15

vy = 6 m/sgsin

37

xy

37vx = ux = 8 m/s

2 2x yv v v= +2 28 6= +

Ans. 10 m / s

Example 51. Two inclined planes OA and OB having inclination (with horizontal) 30 and 60 respectively,intersect each other at O as shown in fig.A particle is projected from point P with velocity 1u 10 3 ms-=

KINEMATICS


along a direction perpendicular to plane OA. If the particle strikesplane OB perpendicularly at Q, calculate

30 60

Au

Ph

O

Q

B

(a) time of flight,(b) velocity with which particle strikes the plane OB,(c) vertical height h of P from O,(d) maximum height from O attained by the particle.

Sol. (a) xu 10 3 m / s=uy = 0ax = g sin60

30 60

Au

Ph

O

Q

B

ay = g cos60At point Q

vx = 0The time taken by particle to go from P to Q is to

vx = ux + axt

00 10 3 gsin 60 t= -

010 3

tgsin 60

= 0t 2sec=

(b) vy = uy ayt0vy = 0 g cos60 2

y1v 10 2 10 m / sec2

(c)hsin 30x

=

1 h2 x

= x = 2h m30 60

Yu

O

X

h x

2y y

12h u t a t2

= +

212h 0 g sin 30 (2)2

1 12h 10 42 2

h 5 m=

(d) At maximum heightvy = 0

2 2y yv u 2gH= -

0 = (u cos30)2 2gH

30

60

u = u cos30yX

h

30Y

2 2u cos 30H2g

=

2

2 310 32

H2g

3100 3 4H2 10

KINEMATICS


25 920

45H 11.25 m4

= =

Maximum height from O attained by the particle= H + h= 11.25 + 5Ans. 16.25 m

Example 52. A large heavy box is sliding without friction down a smooth plane of inclination. From a point P on the bottom of a box, a particle is projected. inside the box. Theinitial speed of the particle with respect to box is u and the direction of projection makes P Qan angle with the bottom as shown in figure.

(a) Find PQ if particle lands on Q.(b) If horizontal displacement of particle with respect to ground is zero. Find the velocity of box.Sol. ux = u cos

uy = usinax = g sinay = g cos

When the particle hitt the inclined plane then displacement iny-direction is zero.

2y y

1y u t a t2

= +

210 (u sin )t g cos t2

yx

90

P u =uco

sx

a = g

cosy

a = gsinx

u =usiny

g

Q

21(usin )t g cos t2

2u sintg cos

...(1)

with respect to boxux = u cosax = 0

2x x

1x u t a t2

= +

x = u cost(2u sin )x (u cos )

g cos

(from eqn 1)2u sin 2xgcos

According to questionux = Uax = gsin

2x x

1x u t a t2

= +22u sin 2 2u sin 1 2u sinU g sin

g cos g cos 2 g cos

KINEMATICS


usin cosu cos Ucos

u sin cosu cos Ucos

u sin cosU u coscos

cos cos cos sinU ucos

u cos( )U

cos

THINKING PROBLEMS

1. Can you use the equations of kinematics to find the height attained by a body projected upwards with anyvelocity ?

2. Can the relative velocity of two bodies be greater than the absolute velocity of either ?

3. A boy sitting in a car moving with a constant velocity throws a ball straight up into the air. Will the ball fallbehind him, in front of him or into his hand? What would happen if the car accelerated forward or went rounda curve while the ball was in the air ?

4. A student argues that the mean velocity during an interval of time can also be expressed as f iv vv

2

and this should always be equal to f i

f

2 1

r rvt t

. Is he right .

5. Consider a collection of a large number of particles moving with the same speed v in random directions. Couldyou, by using simple logic, show that the magnitude of the relative velocity of a pair of particles averaged overall the pairs of the collection is greater than v ?

6. The barrel of a gun and a target lie along the same horizontal. If the target

is released and the gun is fired at the same time, the bullet will always

hit the target whatever be the distance between the gun and the target. Is this true or false ?

7. A black dot is made at the tip of an aerofoil of an aeroplane. What is the trajectory of the black dot as itappears to the pilot and to an observer on the ground?

8. A body is dropped fron the window of a train. Will the time of the free fall be equal if the train is stationary,moves with constant velocity, moves with constant acceleration?

9. We can order events in time, such as past, present and future. Hence there is a sense of time. So is time avector? If not, why not?

10. Average speed can mean the magnitude of the average velocity vector. Another meaning given to it is that theaverage speed is the total length of the path traversed divided by the elapsed time. Are these meaningsdifferent? if so, give an example.

11. Can a body have zero velocity and still be accelerating? can a body have a constant speed and still have avarying velocity? Can a body have a constant velocity and still have a varying speed?

12. Can an object have an eastward velocity while experiencing westward acceleration?

KINEMATICS


THINKING PROBLEMS SOLUTION

1. No, because the equations are applicable only so long as the acceleration is uniform. The acceleration due togravity is uniform only near the surface of the earth.

2. Yes, e.g., when two bodies move in opposite directions, the relative velocity of each is greater than the individualvelocity of either.

3. It will fall into his hand because of the physical independence of vectors or the fact that the forward velocity ofthe ball in the air is the same as that of the car. But if the car accelerated, the ball would fall behind him. If thecar went round a curve the ball would fall in front of him because it would move along the tangent to the curvedpath through a certain distance while the car moved along the curved path through the same distance.

4. No, he is not right. The correct definition of average velocity is the latter one. the first one can be used onlywhen there is uniform acceleration.

5. 2 2 2relv v v 2v cos 2v sin / 2. In other words between 0 and 60 the relative velocity is lessthan v, while between 60 and 180 it is greater than v. Hence the relative velocity averaged over all directions

relv v.6. True, provided the target is within the range of the bullet, because the downward acceleration of both is the

same.

7. A circular path relative to the pilot and a helical path relative to the observer fixed to the earth.

8. Same in all the three cases due to the physical independence of vectors.

9. Time is not a vector though it may have a sense because in order to be a vector it must add and subtract byvector algebra rules and it does not.

10. The meanings are different. The point can be made clear by considering the motion of a particle along acircle. Let a particle describe half a circle of radius R in time .

v

= average velocity vector R i R i 2R i

2R| v |

v = average speed

R

Obviously, v | v |

, that is, average speed as the modulus of the average velocity vector is not the sameas the average speed defined as the total length of the path traversed divided by time elapsed.

11. Yes, for example a particle in SHM at the extreme position has zero velocity but its acceleration is maximum.Yes, a particle in circular motion has constant speed but varying velocity. No, a body cannot have a constantvelocity while having a varying speed.

12. Yes, a body can have eastward velocity and westward acceleration, e.g., SHM.

KINEMATICS


ASSERTION & REASONTHE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS

A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just belowit of the statements, mark the correct answer as (A) If both A and R are true and R is the correct explanation of A.(B) If both A and R are true but R is not the correct explanation of A.(C) If A is true but R is false.(D) If both A and R are false.(E) If A is false but R is true.

1. Assertion (A) : Horizontal component of the velocity of angular projectile remains unchange during itsflight.

Reason (R) : At highest point the velocity reduces to zero.(A) (B) (C) (D) (E)

2. Assertion (A) : Suppose a particle starts moving in a straight line with initial velocity +u and an accelerationa, then velocity at displacement s comes out to be, 2 2 2v u as . If we draw a graphbetween 2v and s, it will be a straight line as shown in figure.

SS0

v2

Reason (R) : 2v versus s graph is a straight line passing through origin with positive intercept andnegative slope.(A) (B) (C) (D) (E)

3. Assertion (A) : A body can have acceleration even if its velocity is zero at a given instant of time.Reason (R) : A body is momentarily at rest when it reverses its direction of motion.

(A) (B) (C) (D) (E)

4. Assertion (A) : When a body is projected with an angle 45, its range is maximum.Reason (R) : For maximisation of range sin 2 should be equal to one.

(A) (B) (C) (D) (E)

5. Assertion (A) : Rocket in flight is not an illustration of projectile.Reason (R) : Rocket takes flight due to combustion of fuel and does not move under the gravity effect

alone.(A) (B) (C) (D) (E)

6. Assertion (A) : The slope of displacement-time graph of a body moving with high velocity is steeperthan the slope of displacement-time graph of a body with low velocity.

Reason (R) : Slope of displacement-time graph = Velocity of the body.(A) (B) (C) (D) (E)

7. Assertion (A) : Displacement of a body may be zero when distance travelled by it is not zero.Reason (R) : The displacement is the longest distance between initial and final position.

(A) (B) (C) (D) (E)

KINEMATICS


8. Assertion (A) : The relative velocity between any two bodies moving in opposite direction is equal to sumof the velocities of two bodies.

Reason (R) : Sometimes relative velocity between two bodies is equal to difference in velocities of thetwo.(A) (B) (C) (D) (E)

9. Assertion (A) : The displacement-time graph of a body moving with uniform acceleration is a straight line.Reason (R) : The displacement is proportional to time for uniformly accelerated motion.

(A) (B) (C) (D) (E)

10. Assertion (A) : A body falling freely may do so with constant velocity.Reason (R) : The body falls freely, when acceleration of a body is equal to acceleration due to gravity.

(A) (B) (C) (D) (E)

11. Assertion (A) : The speedometer of an automobile measure the average speed of the automobile.Reason (R) : Average velocity is equal to total displacement per total time taken.

(A) (B) (C) (D) (E)

KINEMATICS


Level # 11. A train starting from rest travels the first part of its journey with constant acceleration a, second part with

constant velocity v and third part with constant retardation a, being brought to rest. The average speed for

the whole journey is 8v7

. The train travels with constant velocity for .... of the total time

(A) 43 (B) 8

7(C)

65

(D) 79

2. A particle moving in a straight line with uniform acceleration is observed to be at a distance a from a fixedpoint initially. It is at distances b, c, d from the same point after n, 2n, 3n second. The acceleration of theparticle is

(A) 2nab2c

(B) 2n9abc

(C) 2n4ab2c

(D) 2nabc

3. If a, b and c be the distances travelled by the body during xth, yth and zth second from start, then which ofthe following relations is true ?(A) a(y z) + b(z x) + c(x y) = 0 (B) a(x y) + b(y z) + c(z x) = 0(C) a(z x) + b(x y) + c(y z) = 0 (D) ax + by + cz = 0

4. A body of mass 3 kg falls from the multi-storeyed building 100 m high and buries itself 2m deep in the sand.The time of penetration will be(A) 0.09 s (B) 0.9 s (C) 9 s (D) 10 s

5. Two cars A and B, each having a speed of 30 km/hr are heading towards each other along a straight path. Abird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km, heads directlytowards car B, on reaching B, the bird directly flies back to A and so forth, then the total distance the birdtravels till the cars meet is.(A) infinite (B) 30 km (C) 60 km (D) 120 km

6. Two cars A and B, each having a speed of 30 km/hr are heading towards each other along a straight path. Abird that

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