Resonance Kinematics

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MECHANICSMECHANICSMECHANICSMECHANICSMECHANICSMechanics is the branch of physics which deals with the cause andeffects of motion of a particle, rigid objects and deformable bodiesetc. Mechanics is classified under two streams namely Kinematicsand Dynamics.

Mechanics

Kinematics Dynamics (or Kinetics)The word kinematics means It is branch of mechanics which isscience of motion.branch of concerned about the causes (i.e.

themechanics which deals with force , torque) that cause motionstudy of motion without going of bodies.into the cause of motion, i.e.force, torque etc.

1.1.1.1.1. MOTION AND RESTMOTION AND RESTMOTION AND RESTMOTION AND RESTMOTION AND RESTMotion is a combined property of the object and the observer. Thereis no meaning of rest or motion without the observer. Nothing is inabsolute rest or in absolute motion.An object is said to be in motion with respect to a observer, if itsposition changes with respect to that observer. It may happen byboth ways either observer moves or object moves.

2.2.2.2.2. RECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTIONRectilinear motion is motion, along a straight line or in one dimension.It deals with the kinematics of a particle in one dimension.

2.12.12.12.12.1 PositionPositionPositionPositionPositionThe position of a particle refers to its location in the space at a certainmoment of time. It is concerned with the question where is theparticle at a particular moment of time ?

2.22.22.22.22.2 DisplacementDisplacementDisplacementDisplacementDisplacementThe change in the position of a moving object is known as displacement.It is the vector joining the initial position of the particle to its finalposition during an interval of time.2.32.32.32.32.3 DistanceDistanceDistanceDistanceDistanceThe length of the actual path travelled by a particle during a given timeinterval is called as distance. The distance travelled is a scalar quantitywhich is quite different from displacement. In general, the distancetravelled between two points may not be equal to the magnitude of thedisplacement between the same points.

E pur, si muove.(Translated: Even so, it doesmove.)When Galileo was declareda heretic by the CatholicChurch and chose torecant his beliefs in orderto spare his life, he is knownto have said:I must altogetherabandon the false opinionthat the Sun is the centerof the world andimmovable, and that theEarth is not the center ofthe world, and moves, andthat I must not hold,defend, or teach... the saiddoctrine.But afterwards, as he rosefrom his kneeling position,his body weary fromseventy years of age,legend has it that hemurmured, in Italian theabove statement.

RECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTIONRECTILINEAR MOTION


Example 1. Ram takes path 1 (straight line) to go from P to Q and Shyam takes path 2 (semicircle).(a) Find the distance travelled by Ram and Shyam?

P Q1

2

100 m

(b) Find the displacement of Ram and Shyam?Sol. (a) Distance travelled by Ram = 100 m

Distance travelled by Shyam = (50 m) = 50 m(b) Displacement of Ram = 100 m

Displacement of Shyam = 100 m

2.42.42.42.42.4 Average Velocity Average Velocity Average Velocity Average Velocity Average Velocity (in an interval) :The average velocity of a moving particle over a certain time interval is defined as the displacement divided bythe lapsed time.

Average Velocity = erval inttime

ntdisplaceme

for straight line motion, along xaxis, we have

vav

= v = = tx

=

if

ifttxx

The average velocity is a vector in the direction of displacement. For motion in a straight line, directionalaspect of a vector can be taken care of by +ve and -ve sign of the quantity.2.52.52.52.52.5 Average Speed Average Speed Average Speed Average Speed Average Speed (in an interval)(in an interval)(in an interval)(in an interval)(in an interval)Average speed is defined as the total path length travelled divided by the total time interval during which themotion has taken place. It helps in describing the motion along the actual path.

Average Speed = interval timetravelled distance

NOTE :(a) Average speed is always positive in contrast to average velocity which being a vector, can be positive

or negative.(b) If the motion of a particle is along a straight line and in same direction then,

average velocity = average speed.(c) Average speed is, in general, greater than the magnitude of average velocity.

The dimension of velocity and speed is [LT-1] and their SI unit is meters per second (m/s)Example 2. In the example 1, if Ram takes 4 seconds and Shyam takes 5 seconds to go from P to Q, find

(a) Average speed of Ram and Shyam?(b) Average velocity of Ram and Shyam?

Sol. (a) Average speed of Ram = 4100

m/s = 25 m/s

Average speed of Shyam = 550

m/s = 10 m/s

(b) Average velocity of Ram = 4100

m/s = 25 m/s

Average velocity of Shyam = 5100

m/s = 20 m/s

Example 3. A particle travels half of total distance with speed v1 and next half with speed v2 along a straight line.Find out the average speed of the particle?

Sol. Let total distance travelled by the particle be 2s.

Time taken to travel first half = 1v

s

Time taken to travel next half = 2v

s


Average speed = taken time Totalcovered distance Total

=

21 v

sv

ss2

+ =

21

21vv

vv2+

Q.1 A particle covers 43

of total distance with speed v1 and next 41

with v2. Find the average speed of the

particle?

Ans.21

213vvv4v

+

Q.2. A car is moving with speed 60 Km/h and a bird is moving with speed 90km/h along the same direction as shown in figure. Find the distance trav-elled by the bird till the time car reaches the tree?

240 m

Ans. 360 m

2.62.62.62.62.6 Instantaneous VelocityInstantaneous VelocityInstantaneous VelocityInstantaneous VelocityInstantaneous Velocity (at an instant) :The velocity at a particular instant of time is known as instantaneous velocity. The term velocity usuallymeans instantaneous velocity.

Vinst. =

txlim

0t = dtdx

In other words, the instantaneous velocity at a given moment (say , t) is the limiting value of the averagevelocity as we let t approach zero. The limit as t 0 is written in calculus notation as dx/dt and is calledthe derivative of x with respect to t.

2.72.72.72.72.7 Average acceleration (in an interval):Average acceleration (in an interval):Average acceleration (in an interval):Average acceleration (in an interval):Average acceleration (in an interval):The average acceleration for a finite time interval is defined as :

Average acceleration = ervalinttimevelocityinchange

Average acceleration is a vector quantity whose direction is same as that of the change in velocity.

avar

=

tv

r

=

tvv if

rr

Since for a straight line motion the velocities are along a line, therefore

aav

=

tv

=

if

ifttvv

(where one has to substitute vf and vi with proper signs in one dimensional motion)

2.82.82.82.82.8 Instantaneous Acceleration (at an instant):Instantaneous Acceleration (at an instant):Instantaneous Acceleration (at an instant):Instantaneous Acceleration (at an instant):Instantaneous Acceleration (at an instant):The instantaneous acceleration of a particle is its acceleration at a particular instant of time. It is defined asthe derivative (rate of change) of velocity with respect to time. We usually mean instantaneous accelerationwhen we say acceleration. For straight motion we define instantaneous acceleration as :

a = dtdv

=

tvlim

0t

and in general ar

= dtvdr

=

tvlim

0t

r

The dimension of acceleration is [LT-2] and its SI unit is m/s2.


Example 4. Position of a particle as a function of time is given as x = 5t2 + 4t + 3. Find the velocity andacceleration of the particle at t = 2 s?

Sol. Velocity; v = dtdx

= 10t + 4At t = 2 s v = 10(2) + 4 v = 24 m/s

Acceleration; a = 22

dtxd

= 10

Acceleration is constant, so at t = 2 sa = z10 m/s2

Q.3 The position of a particle moving on X-axis is given byx = At3 + Bt2 + Ct + D.

The numerical values of A, B, C, D are 1, 4, -2 and 5 respectively and SI units are used. Find (a) thedimensions of A, B, C and D, (b) the velocity of the particle at t = 4 s, (c) the acceleration of the particle att = 4s, (d) the average velocity during the interval t =0 to t = 4s, (e) the average acceleration during the intervalt = 0 to t = 4 s.

Ans. [(a) [A] = [LT -3], [B] = [LT -2], [C] = [LT -1] and [D] = [L] ; (b) 78 m/s ; (c) 32 m/s2 ; (d) 30 m/s ; (e) 20 m/s2]

3.3.3.3.3. MOMOMOMOMOTION WITH UNIFTION WITH UNIFTION WITH UNIFTION WITH UNIFTION WITH UNIFORM VELORM VELORM VELORM VELORM VELOCITOCITOCITOCITOCITYYYYYConsider a particle moving along xaxis with uniform velocity u starting from the point x = xi at t = 0.Equations of x, v, a are : x (t) = xi + ut ; v (t) = u ; a (t) = 0 x t graph is a straight line of slope u through xi. as velocity is constant, v t graph is a horizontal line. at graph coincides with time axis because a = 0 at all time instants.

x

O

xi

t u is negative

slope =

u

v

O

u

t

positive velocity

v

O

u

t

negative velocity

4.4.4.4.4. UNIFUNIFUNIFUNIFUNIFORMLORMLORMLORMLORMLY ACCELERAY ACCELERAY ACCELERAY ACCELERAY ACCELERATED MOTED MOTED MOTED MOTED MOTIONTIONTIONTIONTIONIf a particle is accelerated with constant acceleration in an interval of time, then the motion is termed asuniformly accelerated motion in that interval of time.For uniformly accelerated motion along a straight line (xaxis) during a time interval of t seconds, thefollowing important results can be used.

(a) v = u + at(b) s = ut + 1/2 at2

s = vt 1/2 at2xf = xi + ut + 1/2 at2(c) v2 = u2 + 2as

(d) s = 1/2 (u + v) t(e) s

n = u + a/2 (2n 1)

u = initial velocity (at the beginning of interval)a = accelerationv = final velocity (at the end of interval)s = displacement (xf xi)xf = final coordinate (position)xi = initial coordinate (position)s

n = displacement during the nth sec


Example 5. A particle moving rectilinearly with constant acceleration is having initial velocity of 10 m/s. Aftersome time, its velocity becomes 30 m/s. Find out velocity of the particle at the mid point of itspath?

Sol. Let the total distance be 2x. distance upto midpoint = xLet the velocity at the mid point be vand acceleration be a.From equations of motion

v2 = 102 + 2ax ____ (1)302 = v2 + 2ax ____ (2)

(2) - (1) givesv2 - 302 = 102 - v2 v2 = 500 v = 510 m/s

Example 6. A police inspector in a jeep is chasing a pickpocket an a straight road. The jeep is going at itsmaximum speed v (assumed uniform). The pickpocket rides on the motorcycle of a waitingfriend when the jeep is at a distance d away, and the motorcycle starts with a constant accelera-tion a. Show that the pick pocket will be caught if 2adv .

Sol. Suppose the pickpocket is caught at a time t after motorcycle starts. The distance travelled by the motor-cycle during this interval is

2at21

s = ____ (1)

During this interval the jeep travels a distancevtds =+ ____ (2)

By (1) and (2), vtdat21 2

=+

or,a

ad2vvt2

=

The pickpocket will be caught if t is real and positive.This will be possible if 2adv2 or, 2adv

Q.4 A car deccelerates from a speed of 20 m/s to rest in a distance of 100 m. What was its acceleration,assumed constant?Ans. [- 2 m/s2]

Q. 5 A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 50 m awayfrom the station at a speed of 25 m/s, what will be the speed of the back part of the train as itpasses the worker?Ans. [50 m/s]

5 .5 .5 .5 .5 . GRAPHS IN UNIFGRAPHS IN UNIFGRAPHS IN UNIFGRAPHS IN UNIFGRAPHS IN UNIFORMLORMLORMLORMLORMLY ACCELERAY ACCELERAY ACCELERAY ACCELERAY ACCELERATED MOTED MOTED MOTED MOTED MOTION TION TION TION TION (a 0) x is a quadratic polynomial in terms of t. Hence x t graph is a parabola.

xi

x

a > 0

t0

xi

x

a < 0

t0

x-t graph v is a linear polynomial in terms of t. Hence vt graph is a straight line of slope a.

v

ua is positive

slope

=a

t0

v

u

a is negative

slope =

a

t0


v-t graph at graph is a horizontal line because a is constant.

a

apositiveacceleration

t0a

a

negativeacceleration

0

a-t graph

6.6.6.6.6. REACTION TIMEREACTION TIMEREACTION TIMEREACTION TIMEREACTION TIMEWhen a situation demands our immediate action. It takes some time before we really respond. Reactiontime is the time a person takes to observe, think and act.

77777..... DIRECTIONS OF VECTORS IN STRAIGHT LINE MOTIONDIRECTIONS OF VECTORS IN STRAIGHT LINE MOTIONDIRECTIONS OF VECTORS IN STRAIGHT LINE MOTIONDIRECTIONS OF VECTORS IN STRAIGHT LINE MOTIONDIRECTIONS OF VECTORS IN STRAIGHT LINE MOTIONIn straight line motion, all the vectors (position, displacement, velocity & acceleration) will have only onecomponent (along the line of motion) and there will be only two possible directions for each vector. For example, if a particle is moving in a horizontal line (xaxis), the two directions are right and left.

Any vector directed towards right can be represented by a positive number and towards left can berepresented by a negative number.

For vertical or inclined motion, upward direction can betaken +ve and downward as ve

line of motion

lin

e of

m

otio

n

line

of

mot

ion

+

+ +

-

-

-

For objects moving vertically near the surface of the earth, the only force acting on the particle is itsweight (mg) i.e. the gravitational pull of the earth. Hence acceleration for this type of motion willalways be a = g i.e. a = 9.8 m/s2 (ve sign, because the force and acceleration are directeddownwards, If we select upward direction as positive).

NOTE :(a) If acceleration is in same direction as velocity, then speed of the particle increases.(b) If acceleration is in opposite direction to the velocity then speed decreases i.e. the particle slows

down. This situation is known as retardation.Example 7. Mr. Sharma brake his car with constant acceleration from a velocity of 25 m/s to 15 m/s over a

distance of 200m.(a) How much time elapses during this interval?(b) What is the acceleration?(c) If he has to continue braking with the same constant acceleration, how much longer would it takefor him to stop and how much additional distance would he cover?

Sol. (a) We select positive direction for our coordinate system to be the direction of the velocity and choosethe origin so that xi = 0 when the braking begins. Then the initial velocity is ux = +25 m/s at t = 0, andthe final velocity and position are v

x = +15 m/s and x = 200 m at time t.

Since the acceleration is constant, the average velocity in the interval can be found from the averageof the initial and final velocities.

vav, x

= 21 (u

x + v

x) = 2

1 (15 + 25) = 20 m/s.

The average velocity can also be expressed as vav, x

=

tx

. With x = 200 m and t = t 0,

we can solve for t:

t = xav,v

x = 20

200 = 10 s.


(b) We can now find the acceleration using vx = u

x + a

xt

ax = t

uv xx = 10

2551 = 1 m/s2.

The acceleration is negative, which means that the positive velocity is becoming smaller as brakesare applied (as expected).

(c) Now with known acceleration, we can find the total time for the car to go from velocity ux = 25 m/s to

vx = 0. Solving for t, we find

t = x

xx

a

uv = 1

250

= 25 s.

The total distance covered is

x = xi + uxt + 21

axt2 = 0 + (25)(25) + 2

1 (1)(25)2 = 625 312.5 = 312.5 m.Additional distance covered = 312.5 200 = 112.5 m.

Example 8. A particle is dropped from a tower. It is found that it travels 45 m in the last second of its journey. Findout the height of the tower?

Sol. Let the total time of journey be n seconds.

Using; )1n2(2a

usn +=

45 = 0 + 210 )12( n n = 5 sec

Height of tower;

h = 21

gt2 = 21

10 52 = 125 m

Example 9. A particle is dropped from height 100 m and another particle is projected vertically up with velocity50 m/s from the ground along the same line. Find out the position where two particle will meet?

Sol. Let the upward direction as positive.Let the particles meet at a distance y from the ground.

y=0m

y=100m u=0 m/s

u=50 m/s

A

B

For particle A,y0 = + 100 m

u = 0 m/s

a = 10 m/s2

y = 100 + 0(t) 21

10 t2 [y = y0 + ut + 21

at2] = 100 - 5t2 ---- (1)

For particle B,y0 = 0 m u = + 50 m/s a = 10 m/s2

y = 50(t) 21

10 t2

= 50t 5t2 ---- (2)According to the problem;

50t 5t2 = 100 5t2t = 2 sec

Putting t = 2 sec in eqn. (1),y = 100 20 = 80 m

Hence, the particles will meet at a height 80 m above the ground.


Q. 6 A particle is thrown vertically with velocity 20 m/s . Find (a) the distance travelled by the particle in first 3seconds, (b) displacement of the particle in 3 seconds.Ans. [25m, 15m]

Q.7 A stone is dropped from a balloon going up with a uniform velocity of 5 m/s. If the balloon was 50 m high whenthe stone was dropped, find its height when the stone hits the ground. Take g = 10 m/s2.Ans. [68.5 m]NOTE:- As the particle is detached from the balloon it is having the same velocity as that of balloon, but its

acceleration is only due to gravity and is equal to g.

8.8.8.8.8. GRAPHICGRAPHICGRAPHICGRAPHICGRAPHICAL INTERPRETAL INTERPRETAL INTERPRETAL INTERPRETAL INTERPRETAAAAATION OF SOME QUTION OF SOME QUTION OF SOME QUTION OF SOME QUTION OF SOME QUANTITIESANTITIESANTITIESANTITIESANTITIES8.18.18.18.18.1 Average VelocityAverage VelocityAverage VelocityAverage VelocityAverage VelocityIf a particle passes a point P (xi) at time t = ti and reaches Q (xf) at a later time instant t = tf , its average

velocity in the interval PQ is Vav

= tx

=

if

ifttxx

This expression suggests that the average velocity is equal to the slope ofthe line (chord) joining the points corresponding to P and Q on the xt graph.

x

Q

P

tO

xf

xi

ti tf8.28.28.28.28.2 Instantaneous VelocityInstantaneous VelocityInstantaneous VelocityInstantaneous VelocityInstantaneous VelocityConsider the motion of the particle between the two points P and Q on the xt graph shown. As the point Qis brought closer and closer to the point P, the time interval between PQ (t, t, t,......) get progressivelysmaller. The average velocity for each time interval is the slope of the appropriate dotted line (PQ, PQ,PQ......). As the point Q approaches P, the time interval approaches zero, but at the same time the slopeof the dotted line approaches that of the tangent to the curve at the point P. As t 0, V

av (=x/t) Vinst.

Geometrically, as t 0, chord PQ tangent at P.

Hence the instantaneous velocity at P is the slope of the tangent at P in the x tgraph. When the slope of the x t graph is positive, v is positive (as at the point Ain figure). At C, v is negative because the tangent has negative slope. Theinstantaneous velocity at point B (turning point) is zero as the slope is zero.

8.38.38.38.38.3 Instantaneous AccelerationInstantaneous AccelerationInstantaneous AccelerationInstantaneous AccelerationInstantaneous AccelerationThe derivative of velocity with respect to time is the slope

of the tangent in velocity time (vt) graph.v

ta0a=0

a=0

8 . 48 . 48 . 48 . 48 . 4 Displanncement from v - t graphDisplanncement from v - t graphDisplanncement from v - t graphDisplanncement from v - t graphDisplanncement from v - t graphDisplacement = x = area under vt graph.Since a negative velocity causes a negative displacement, areas below the timeaxis are taken negative. In similar way, can see that v = a t leads to the conclusionthat area under a t graph gives the change in velocity v during thatinterval.


Example 10. Describe the motion shown by the following velocity-time graphs.

(a) (b)

Sol. (a) During interval AB: velocity is +ve so the particle is moving in +ve direction, but it is slowing downas acceleration (slope of v-t curve) is negative. During interval BC: particle remains at rest asvelocity is zero. Acceleration is also zero. During interval CD: velocity is -ve so the particle ismoving in -ve direction and is speeding up as acceleration is also negative.

(b) During interval AB: particle is moving in +ve direction with constant velocity and acceleration iszero. During interval BC: particle is moving in +ve direction as velocity is +ve, but it slows downuntil it comes to rest as acceleration is negative. During interval CD: velocity is -ve so the particleis moving in -ve direction and is speeding up as acceleration is also negative.

Important Points to RememberImportant Points to RememberImportant Points to RememberImportant Points to RememberImportant Points to Remember For uniformly accelerated motion (a 0), xt graph is a parabola (opening upwards if a > 0 and opening

downwards if a < 0). The slope of tangent at any point of the parabola gives the velocity at that instant. For uniformly accelerated motion (a 0), vt graph is a straight line whose slope gives the acceleration of the

particle. In general, the slope of tangent in xt graph is velocity and the slope of tangent in vt graph is the acceleration. The area under at graph gives the change in velocity. The area between the vt graph gives the distance travelled by the particle, if we take all areas as positive. Area under vt graph gives displacement, if areas below the taxis are taken negative.

Example 11. For a particle moving along x-axis, velocity-time graph is as shownin figure. Find the distance travelled and displacement of the particle?

Sol. Distance travelled = Area under v-t graph (taking all areas as +ve.) Distance travelled = Area of trapezium + Area of triangle

= ( ) 86221

+ + 542

1

= 32 + 10 = 42 mDisplacement = Area under v-t graph (taking areas below time axis as -ive.) Displacement = Area of trapezium Area of triangle

= ( ) 86221

+

5421

= 32 10 = 22 mHence, distance travelled = 42 m and displacement = 22 m.

Q.8 For a particle moving along x-axis, following graphs are given. Find the distance travelled by the particle in 10s in each case?

Ans. [(a) 100m; (b) 50m]


9.9.9.9.9. Interpretation of some more GraphsInterpretation of some more GraphsInterpretation of some more GraphsInterpretation of some more GraphsInterpretation of some more Graphs

9.19.19.19.19.1 Position vs Time graphPosition vs Time graphPosition vs Time graphPosition vs Time graphPosition vs Time graph9.1.1 Zero VelocityAs position of particle is fix at all the time, so the body is at rest.

Slope; dtdx

= tan = tan 0 = 0

Velocity of particle is zero

9.1.29.1.29.1.29.1.29.1.2 Uniform VelocityUniform VelocityUniform VelocityUniform VelocityUniform Velocity

Here tan is constant tan = dtdx

dtdx

is constant.

velocity of particle is constant.9.1.39.1.39.1.39.1.39.1.3 Non uniform velocity Non uniform velocity Non uniform velocity Non uniform velocity Non uniform velocity (increasing with time)

In this case;As time is increasing, is also increasing.

dtdx

= tan is also increasing

Hence, velocity of particle is increasing.9.1.49.1.49.1.49.1.49.1.4 Non uniform velocity (decreasing with time)Non uniform velocity (decreasing with time)Non uniform velocity (decreasing with time)Non uniform velocity (decreasing with time)Non uniform velocity (decreasing with time)

In this case;As time increases, decreases.

dtdx

= tan also decreases.

Hence, velocity of particle is decreasing.

9.29.29.29.29.2 Velocity vs time graphVelocity vs time graphVelocity vs time graphVelocity vs time graphVelocity vs time graph9.2.19.2.19.2.19.2.19.2.1 Zero acceleration Zero acceleration Zero acceleration Zero acceleration Zero accelerationVelocity is constant.

tan = 0

dtdv

= 0

Hence, acceleration is zero.

9.2.29.2.29.2.29.2.29.2.2 Uniform accelerationUniform accelerationUniform accelerationUniform accelerationUniform accelerationtan is constant.

dtdv

= constant

Hence, it shows constant acceleration.

9.2.39.2.39.2.39.2.39.2.3 Uniform retardationUniform retardationUniform retardationUniform retardationUniform retardationSince > 90

tan is constant and negative.

dtdv

= negative constant

Hence, it shows constant retardation.


9.39.39.39.39.3 Acceleration vs time graphAcceleration vs time graphAcceleration vs time graphAcceleration vs time graphAcceleration vs time graph9.3.1 Constant acceleration9.3.1 Constant acceleration9.3.1 Constant acceleration9.3.1 Constant acceleration9.3.1 Constant acceleration

tan = 0

dtda

= 0

Hence, acceleration is constant.9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration9.3.2 Uniformly increasing acceleration

is constant.0 < < 90 tan > 0

dtda

= tan = constant > 0

Hence, acceleration is uniformly increasing with time.9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing acceleration9.3.3 Uniformly decreasing accelerationSince > 90

tan is constant and negative.

dtda

= negative constant

Hence, acceleration is uniformly decreasing with time

Example 12. The displacement vs time graph of a particle moving along a straight line isshown in the figure. Draw velocity vs time and acceleration vs time graph.

Sol: x = 4t2 v = dtdx

= 8t

Hence, velocity-time graph is a straight line having slope i.e. tan = 8.

a = dtdv

= 8

Hence, acceleration is constant throughout and is equal to 8.

Problem 10. The displacement vs time graph of a particle moving along a straight line is shown in the figure.Draw velocity vs time and acceleration vs time graph.


11.11.11.11.11. MOMOMOMOMOTION WITH NON-UNIFTION WITH NON-UNIFTION WITH NON-UNIFTION WITH NON-UNIFTION WITH NON-UNIFORM ACCELERAORM ACCELERAORM ACCELERAORM ACCELERAORM ACCELERATION TION TION TION TION (use of def(use of def(use of def(use of def(use of definitinitinitinitinite inte inte inte inte integrals)egrals)egrals)egrals)egrals)

x = fi

t

t

dt)t(v (displacement in time interval t = ti to tf)

The expression on the right hand side is called the definite integral of v(t) betweent = ti and t = tf. . Similarly change in velocity

v = vf vi = fi

t

t

dt)t(a

Table 2 : Some quantities defined as derivatives and integrals.

v(t) = dtdx

v = slope of xt graph

a (t) = dtdv

a = slope of vt graphs

F (t) = dtdp

F = slope of pt graph (p = linear momentum)

x = dx = fi

t

t

dt)t(v x = area under vt graph

v = dv = fi

t

t

dt)t(a v = area under at graph

p = dp = fi

t

t

dt)t(F p = area under Ft graph

W = dW = fi

x

x

dx)x(F W = area under Fx graph

12.12.12.12.12. SOLSOLSOLSOLSOLVING PRVING PRVING PRVING PRVING PROBLEMS WHICH INVOBLEMS WHICH INVOBLEMS WHICH INVOBLEMS WHICH INVOBLEMS WHICH INVOLOLOLOLOLVES NONUNIFVES NONUNIFVES NONUNIFVES NONUNIFVES NONUNIFORM ACCELERAORM ACCELERAORM ACCELERAORM ACCELERAORM ACCELERATIONTIONTIONTIONTION12.112.112.112.112.1 Acceleration depending on velocity v or time tAcceleration depending on velocity v or time tAcceleration depending on velocity v or time tAcceleration depending on velocity v or time tAcceleration depending on velocity v or time t

By definition of acceleration, we have a = dtdv

. If a is in terms of t, vv0

dv =

t

0

dt)t(a. If a is in

terms of v, =t

0

v

v

dt)v(adv

0

. On integrating, we get a relation between v and t, and then

using x

x0

dx =

t

0dt)t(v , x and t can also be related.

1 2 . 21 2 . 21 2 . 21 2 . 21 2 . 2 Acceleration depending on velocity v or position xAcceleration depending on velocity v or position xAcceleration depending on velocity v or position xAcceleration depending on velocity v or position xAcceleration depending on velocity v or position x

a = dtdv

a = dxdv

dtdx

a = dtdx

dxdv

a = v dxdv

This is another important expression for acceleration.

If a is in terms of x, v

v0

dvv =

x

x0

dx)x(a .


If a is in terms of v, = xx

v

v 00

dx)v(advv

On integrating, we get a relation between x and v. Using xx0

)x(vdx

= t

0

dt, we can relate x and t.

Example 13. An object starts from rest at t = 0 and accelerates at a rate given by a = 6t. What is(a) its velocity and(b) its displacement at any time t?

Sol. As acceleration is given as a function of time,

=t

t

v(t)

)v(t 00a(t)dtdv

Here t0 = 0 and v(t0) = 0

v(t) = t

0

6tdt= 0

t

2t6

2

= 6 (2t2

- 0) = 3t2

So, v(t) = 3t2

As = tt0

v(t)dtx

=t

0

2dt3tx= 0

t

3t3

3

=

03t33

= t3

Hence, velocity v(t) = 3t2 and displacement 3t=xQ.9. For a particle moving along x-axis, acceleration is given as 22va = . If the speed of the particle is v0 at

x = 0, find speed as a function of x.Ans. [ 2x0evv = ]

Q. 10. For a particle moving along x-axis, velocity is given as a function of time as v = 2t2 + sint. At t = 0, particle isat origin. Find the position as a function of time?

Ans. [ 1cos(t)t32

x3 += ]


S U M M A R YS U M M A R YS U M M A R YS U M M A R YS U M M A R YRectilinear MotionRectilinear MotionRectilinear MotionRectilinear MotionRectilinear Motion : Rectilinear motion is motion,along a straight line or in one dimension.Displacement : Displacement : Displacement : Displacement : Displacement : The vector joining the initial positionof the particle to its final position during an interval oftime.Distance : Distance : Distance : Distance : Distance : The length of the actual path travelled bya particle during a given time interval

Average Velocity Average Velocity Average Velocity Average Velocity Average Velocity = erval inttime

ntdisplaceme =

if

ifttxx

Average SpeedAverage SpeedAverage SpeedAverage SpeedAverage Speed = interval timetravelled distance

Instantaneous Velocity : Instantaneous Velocity : Instantaneous Velocity : Instantaneous Velocity : Instantaneous Velocity : Vinst. =

txlim

0t = dtdx

Average AccelerationAverage AccelerationAverage AccelerationAverage AccelerationAverage Acceleration

= ervalinttimevelocityinchange

=

if

ifttvv

Instantaneous Acceleration :Instantaneous Acceleration :Instantaneous Acceleration :Instantaneous Acceleration :Instantaneous Acceleration :

a = dtdv

=

tvlim

0t

Equations of MotionEquations of MotionEquations of MotionEquations of MotionEquations of Motion(a) v = u + at(b) s = ut + 1/2 at2

s = vt 1/2 at2xf = xi + ut + 1/2 at2(c) v2 = u2 + 2as

(d) s = 1/2 (u + v) t(e) s

n = u + a/2 (2n 1)

Important Points to RememberImportant Points to RememberImportant Points to RememberImportant Points to RememberImportant Points to Remember For uniformly accelerated motion (a 0), xt graph is

a parabola (opening upwards if a > 0 and openingdownwards if a < 0). The slope of tangent at any pointof the parabola gives the velocity at that instant.

For uniformly accelerated motion (a 0), vt graph isa straight line whose slope gives the acceleration ofthe particle.

In general, the slope of tangent in xt graph is velocityand the slope of tangent in vt graph is theacceleration.

The area under at graph gives the change in velocity. The area between the vt graph gives the distance

travelled by the particle, if we take all areas aspositive.

Area under vt graph gives displacement, if areasbelow the taxis are taken negative.

Maxima and MinimaMaxima and MinimaMaxima and MinimaMaxima and MinimaMaxima and MinimaConditions for maxima are:

dxdy

= 0 (b) 22

dxyd

< 0

Conditions for minima are:

dxdy

= 0(b) 22

dxyd

> 0

Motion with Non-Uniform AccelerationMotion with Non-Uniform AccelerationMotion with Non-Uniform AccelerationMotion with Non-Uniform AccelerationMotion with Non-Uniform Acceleration

x = fi

t

t

dt)t(v

v = vf vi = fi

t

t

dt)t(a

Solving Problems which Involves NonuniformSolving Problems which Involves NonuniformSolving Problems which Involves NonuniformSolving Problems which Involves NonuniformSolving Problems which Involves NonuniformAccelerationAccelerationAccelerationAccelerationAcceleration

If a is in terms of t, v

v0

dv =

t

0dt)t(a

If a is in terms of v, =t

0

v

v

dt)v(adv

0

If a is in terms of x, v

v0

dvv =

x

x0

dx)x(a .

If a is in terms of v, = xx

v

v 00

dx)v(advv


Exercise 1 OBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMS (RECTILINEAR MOTIN) (RECTILINEAR MOTIN) (RECTILINEAR MOTIN) (RECTILINEAR MOTIN) (RECTILINEAR MOTIN) BABABABABATTTTTCH - CCCH - CCCH - CCCH - CCCH - CC1. A motor car is going due north at a speed of 50 km/h. It makes a 90 left turn without changing the speed. The

change in the velocity of the car is about(A) 50 km/h towards west (B) 50 2 km/h towards south-west(C) 50 2 km/h towards north-west (D) zero

2. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant, Let xA and xB be themagnitude of displacements in the first 10 seconds and the next 10 seconds.(A) xA < xB (B) xA = xB (C) xA > xB(D) the information is insufficient to decide the relation of xA with xB.

3. A ball takes t seconds to fall from a height h1 and 2t seconds to fall from a height h2. Then h1/h2 is(A) 2 (B) 4 (C) 0.5 (D) 0.254. A body starts from rest and is uniformly acclerated for 30 s. The distance travelled in the first 10 s is x1, next 10 s is

x2 and the last 10 s is x3. Then x1 : x2 : x3 is the same as(A) 1 : 2 : 4 (B) 1 : 2 : 5 (C) 1 : 3 : 5 (D) 1 : 3 : 95. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is

(A) a upward (B) (g-a) upward (C) (g-a) downward (D) g downward6. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically

downward and the ball B is thrown vertically upward with the same speed. The ball A hits the ground with a speedA and the ball B hits the ground with a speed B. We have (A) A > B (B) A < B (C) A = B(D) the relation between A and B depends on height of the building above the ground.

7. Figure shows position-time graph of two cars A and B.(A) Car A is faster than car B.(B) Car B is faster than car A.(C) Both cars are moving with same velocity.

A

05

x(m)

B

(D) Both cars have positive acceleration.8. The displacementtime graph of a moving particle is shown below. The

instantaneous velocity of the particle is negative at the point :(A) C (B) D

t

x

C

D

E F

(C) E (D) F9. A particle starts from rest and moves along a straight line with constant acceleration. The variation of velocity v with

displacement S is :

(A) S

v

(B)

v

S

C)

v

S

(D)

v

S

10. The displacement time graphs of two particles A and B are straight lines making angles of respectively 300 and 600

with the time axis. If the velocity of A is vA and that of B is vB, then the value of BA

v

v is

(A) 1/2 (B) 1/ 3 (C) 3 (D) 1/311. The initial velocity of a particle is u (at t=0) and the acceleration f is given by (f = at). Which of the following

relations is valid?

(A) v = u + at2 (B) v = u + 2

at2 (C) v = u + at (D) v = u

12. A stone is dropped into a well in which the level of water is h below the top of the well. If v is velocity of sound,the time T after which the splash is heard is given by

(A) T = 2h/v (B) vh

gh2T += (C) v2

hgh2T += (D) v

h2g2

hT +=


13. A body is released from the top of a tower of height h metre. It takes T seconds to reach the ground. Whereis the ball at the time T/2 seconds ?(A) at h/4 metre from the ground (B) at h/2 metre from the ground(C) at 3h/4 metre from the ground (D) depend upon the mass of the ball

14. A stone is thrown vertically upward with an initial velocity u from the top of a tower, reaches the ground with avelocity 3u. The height of the tower is:

(A) g3u2 (B) g

4u2 (C) g6u2 (D) g

9u2

15. A particle starts from rest with uniform acceleration a. Its velocity after n seconds is v. The displacement of thebody in the last two seconds is :

(A) n

1)-2v(n (B) n

1)-v(n (C) n

1)v(n + (D) n

1)2v(2n +

16. Consider the motion of the tip of the minute hand of a clock. In one hour(A) the displacement is zero (B) the distance covered is zero(C) the average speed is zero (D) the average velocity is zero

17. A particle moves along the X-axis as x = u(t 2) + a(t 2)2(A) the initial velocity of the particle is u (B) the acceleration of the particle is a(C) the acceleration of the particle is 2a (D) at t =2s particle is at the origin.

18. Mark the correct statements for a particle going on a straight line:(A) If the velocity and acceleration have opposite sign, the object is slowing down.(B) If the position and velocity have opposite sign, the particle is moving towards the origin.(C) If the velocity is zero at an instant, the acceleration should also be zero at that instant.(D) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

19. The velocity of a particle is zero at t = 0(A) The acceleration at t = 0 must be zero(B) The acceleration at t = 0 may be zero.(C) If the acceleration is zero from t = 0 to t = 10 s, the speed is also zero in this interval.(D) If the speed is zero from t = 0 to t = 10 s the acceleration is also in the interval.

20. Mark the correct statements:(A) The magnitude of the velocity of a particle is equal to its speed.(B) The magnitude of average velocity in an interval is equal to its average speed in that interval.(C) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero(D) It is possible to have a situation in which the speed of the particle is never zero but the average speed in aninterval is zero.

21. The velocity-time plot for a particle moving on a straight line is shown in fig.(A) The particle has constant acceleration(B) The particle has never turned around.(C) The particle has zero displacement(D) The average speed in the interval 0 to 10s is the same as the 10 20 30

t(s)-10-20

100

v(m/s)

average speed in the interval 10s to 20s.SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS1. A particle covers each 1/3 of the total distance with speed v1, v2 and v3 respectively. Find the average speed of the

particle ?2. The position of a particle moving on x-axis is given by x = 4t3 + 3t2 + 6t + 4. Find

(a) The velocity and acceleration of particle at t = 5 s.(b) The average velocity and average acceleration during the interval t = 0 to t = 5 s, x = 4t3 + 3t2 + 6t + 4(a) The velocity and acceleration of particle at t = 5 s.(b) The average velocity and average acceleration during the interval t = 0 to t = 5 s.

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

4. A particle starts from rest with a constant acceleration. At a time t second, the speed is found to be 100 m/s and onesecond later the speed becomes 150 m/s. Find (a) the acceleration and (b) the distance travelled during the (t+1)thsecond.

5. For a particle moving along x-axis, following graphsare given. Find the distance travelled bythe particle in 10 s in each case.

t(s)

x(m)

10

10

(A) (B) t(s)

v(m/s)

100

10

2


6. For a particle moving along x-axis, velocity-time graph is as shown infigure. Find the distance travelled and displacement of theparticle? Also find the average velocity of the particle?

7. The acceleration of a cart started at t = 0, varies withtime as shown in figure . Find the distance travelledin 30 seconds and draw the position-time graph.

8. For a particle moving rectilinearly, acceleration as a function of speed is given as a = 8v2. Find the speed as afunction of x if the particle is having a speed of v0 at x = 0?

9. Under what conditions does the magnitude of the average velocity equal to the average speed.10. Can an object have increasing speed but its acceleration decreases? If yes, give an example; if not, explain why?11. Figure shows four paths along which obejcts move from a starting point to

a final point (particle is moving along the same straight line), all in thesame time. The paths pass over a grid of equally spaced straightlines. Rank the paths according to(a) the average velocity of the objects and(b) the average speed of the objects, greatest first.

12. A man walking with a speed '

v

' constant in magnitude and direction passes under a lantern hanging at aheight H above the ground. Find the velocity with which the edge of the shadow of the man's head moves overthe ground, if his height is '

h

'.

13. An elevator is descending with uniform acceleration. To measure the acceleration , a person in the elevator dropsa coin at the moment the elavator starts. The coin is 6 ft above the floor of the elevator at time it is dropped. Theperson observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of theelevator. [Take g = 32 ft/s2]

14. When a model rocket is launched, the propellant burnsfor a few seconds, accelerating the rocket upward. Afterburnout, the rocket moves upward for a while and thenbegins to fall. A parachute opens shortly after the rocketstarts down. The parachute slows the rocket to keep itfrom breaking when it lands. The figure here showsvelocity data from the flight of the model rocket. Use thedata to answer the following.(a) How fast was the rocket climbing when the engine stopped?(b) For how many seconds did the engine burn? (c) When did the rocket reach its highest point? What was its velocity then?(d) When did the parachute open up ? How fast was the rocket falling then?(e) How long did the rocket fall before the parachute opened?(f) When was the rocket's acceleration greatest?(g) When was the acceleration constant? What was its value then (to the nearest interger)?

1. A part icle of mass 102 kg is moving along the posit ive xaxis under the inf luence of a force

F(x) = 2x2K

where K = 102 N m2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Find

(i) its velocity when it reaches x = 0.50 m(ii) the time at which it reaches x = 0.25 m. [ JEE '98, 8 ]

2. In 1.0 sec. a particle goes from point A to point B moving in a semicircle ofradius 1.0 m. The magnitude of average velocity is: [JEE '99, 2](A) 3.14 m/sec (B) 2.0 m/sec(C) 1.0 m/sec (D) zero


3. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to aheight d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above theground as [ JEE '2000, 3 ]

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

4. A block is moving down a smooth inclined plane starting from rest at time t = 0. Let Sn be the distance travelled by

the block in the interval t = n 1 to t = n. The ratio 1n

n

SS

+

is [JEE Scr. 2004, 3]

(A) n2

1n2 (B) 1n21n2

+

(C) 1n21n2

+ (D) 1n2n2

5. A particle is initially at rest, It is subjected to a linear acceleration a , as shownin the figure. The maximum speed attained by the particle is[JEE Scr. 2004; 3](A) 605 m/s (B) 110 m/s (C) 55 m/s (D) 550 m/s

6. The velocity displacement graph of a particle moving along a straight line is shown. The most suitable acceleration-displacement graph will be [JEE Scr. 2005; 3]

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

EXERCISE # 1EXERCISE # 1EXERCISE # 1EXERCISE # 1EXERCISE # 1OBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMS1. B 2. D 3. D 4. C5. D 6. C 7. C 8. C9. B 10. D 11. B 12. B13. C 14. B 15. A 16. A,D17. C,D 18. A,B,D 19. B,C,D 20. A 21.A,D

SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS

1.313221

321vvvvvv

vv3v++

2. (a) vt = 5 s = 336 units, at = 5 s = 126 units ; (b) = 121units, = 66 units

3. (a) 2.7 km , (b) 60 m/s, (c) 225 m and 2.25 km4. (a) 50 m/s2; (b) 125 m 5. (A) 0; (B) 60m6. distance travelled = 10 m; displacement = 6 m; average

velocity = 1.2 m/s

7. 1000 ft. ,

straight line( )l j y j s[ kk

parabolic curve( )i j oy h; o

parabolic curve ( )i j oy h; o

8. v = v0 e8x

9. < vr

> = time

tdisplacmen =

timetravelled cetandis

When the distance travelled is equal to the displacementof the particle, i.e. particle moves along the straight linein the same direction without reversing the direction ofmotion.

10. Yes. Speed of object will increase if acceleration is inthe direction of velocity, yet its magnitude maydecrease.

a = dtdv

= slope of the curve.

acceleration (slope of the curve) is decreasing with timeyet the speed is increasing.

11. (a) >< 1vr

= >< 2vr

= >< 3vr

= >< 4vr

(b) v4, avg > v1, avg = v2, avg > v3, avg

12.

hHH

v 13. 20 ft/s2

14. (a) 190 ft/s (b) 2 s (c) 8 s, 0 ft/s(d) 10.8 s, 90 ft/s (e) 2.8 s(f) greatest acceleration happens 2 s after launch(g) constant acceleration between 2 and 10.8 s, 32 ft/s2.

EXERCISE # 2EXERCISE # 2EXERCISE # 2EXERCISE # 2EXERCISE # 2

1. (i) Vr

= 1 i m/s (ii) t = 4

33

+

2. B 3. A 4. B 5. C 6. B


RELARELARELARELARELATIVE MOTIVE MOTIVE MOTIVE MOTIVE MOTIONTIONTIONTIONTIONMotion is a combined property of the object under study and theobserver. Motion is always relative, there is no such term like absolutemotion or absolute rest. Motion is always defined with respect to anobserver or reference frame.

Reference frame :

Reference frame is an axis system from which motion is observed. Aclock is attached to measure time. Reference frame can be stationaryor moving. There are two types of reference frame:

(i) Inertial reference frame : A frame of reference in which Newtonsfirst law is valid is called as inertial reference frame.

(ii) Non-inertial reference frame : A frame of reference in whichNewtons first law is not valid is called as non-inertial reference frame.

Note : Earth is by definition a non-inertial reference frame because of its cen-tripetal acceleration towards sun. But, for small practical applicationsearth is assumed stationary hence, it behaves as an inertial referenceframe.

RELARELARELARELARELATIVE VELTIVE VELTIVE VELTIVE VELTIVE VELOCITOCITOCITOCITOCITYYYYYDefinition : Relative velocity of a particle (object) A with respect to Bis defined as the velocity with which A appears to move is B if consid-ered to be at rest. In other words, it is the velocity with which A appearsto move as seen by the B considering itself to be at rest.

Relative motion along straight line -

vA = dtdx A

, vB = dtdxB

xBA = xB xA

vBA = dtdxB

dtdxA

vBA = vB vA

vAA = vA vA = 0 (velocity of A with respect to A)Note : velocity of an object w.r.t. itself is always zero.

Impossibility is a

relative concept.

If my Theory of

relativity is proven

successful, Germany

will claim me as a

German and France

will declare that I

am a citizen of the

world. Should my

theory prove untrue,

France will say that

I am a German and

Germany will declare

that I am a Jew.

-Einstein

RELARELARELARELARELATIVE MOTIVE MOTIVE MOTIVE MOTIVE MOTIONTIONTIONTIONTION


Ex.1 An object A is moving with 5 m/s and B is moving with 20 m/s in the same direction. (Positive x-axis)(i) Find velocity of B with respect to A.

Sol. vB = 20 i m/s vA = 5 i m/s vB vA = 15 i m/s

(ii) Find velocity of A with respect to B

Sol. vB = 20 i m/s, vA = 15 i m/s vAB = vA vB = 15 i m/s

Note : vBA = vAB

Ex.2 Two objects A and B are moving towards each other with velocities 10 m/s and 12 m/s respectively a shown.

(i) Find out velocity of A with respect to B.Sol. vAB = vA vB = (10) (12) = 22 m/s towards right.

(ii) Find out velocity of B with respect to AvBA = vB vA = (12) (10) = 22 m/s towards left.

Velocity of Approach

It is the rate at which a separation between two moving particles decreases.

If separation decreases velocity of approach is positive,

Velocity of approach = 22 m/s

Velocity of approach = 15 m/s

If separation increases, velocity of approach is negative. It is mainly called velocity of separation.

Velocity of separation -

It is the rate with which separation between two moving object increases.Velocity of separation = 2 m/s

Velocity of separation = 15 m/s

Illustration :Two balls A and B are moving in the same direction with equal velocities, find out their relative velocity.

Velocity of A with respect to B ( )ABvr = 0


Illustration :A and B are thrown vertically upward with velocity, 5 m/s and 10 m/s respectively (g = 10 m/s2. Find separa-tion between them after one second

Sol. SA = ut 21

gt2 = 5t 21

10 t2

= 5 1 5 12 = 5 5 = 0

SB = ut 21

gt2 . = 10 1 21

10 12

= 10 5 = 5

SB SA = separation = 5m.

Alter :

By relative BAar

= Bar

Aar

= (10) (10) = 0Also BAv

r= Bv

r Av

r = 10 5 = 5 m/s

BAsr

(in 1 sec) = BAvr

t = 5 1 = 5 m

Distance between A and B after 1 sec = 5 m.

Illustration :A ball is thrown downwards with a speed of 20 m/s from top of a building 150 mhigh and simultaneously another ball is thrown vertically upwards with a speedof 30 m/s from the foot of the building. Find the time when both the balls willmeet. (g = 10 m/s2)

Sol. (I) S1 = 20 t + 5 t2 + S2 = 30 t 5 t2

____________________

150 = 50 t

t = 3 s.

(II) Relative acceleration of both is zero since both have acceleration in downward direction ABar

= Aar

Bar

= g g = 0

BAvr

= 30 (20)= 50

sBA = vBA t

t = BA

BAv

s = 50

150 = 3 s


Ex.5 Two cars C1 and C2 moving in the same direction on a straight road with velocities 12 m/s and 10 m/srespectively. When the separation between the two was 200 m C2 started accelerating to avoid collision.What is the minimum acceleration of car C2 so that they dont collide.

Sol. By relative

21CCar

= 1Car

2Car

= 0 a = (a)

21CCvr

= 1Cvr

2Cvr

= 12 10 = 2 m/s.

So by relativity we want the car to stop.

v2 u2 = 2as.

0 22 = 2 a 200 a = 1001

m/s2

= 0.1 m/s2 = 1 cm/s2.

Minimum acceleration needed by car C2 = 1 cm/s2

RELARELARELARELARELATIVE MOTIVE MOTIVE MOTIVE MOTIVE MOTION IN LIFTTION IN LIFTTION IN LIFTTION IN LIFTTION IN LIFTIllustration :

A lift is moving up with acceleration a. A person inside the lift throws theball upwards with a velocity u relative to hand.

(a) What is the time of flight of the ball?(b) What is the maximum height reached by the ball in the lift?

Sol. (a) BLar

= Bar

Lar

= (g + a) downwards

sr

= ur

t + 21

BLar

t2 0 = uT 21 (g + a)T2

T = )ag(u2

+

(b) v2 u2 = 2 as 0 u2 = 2(g + a) H

H = )ag(2u2

+

Projectile motion in a lift moving with acceleration a upwardsProjectile motion in a lift moving with acceleration a upwardsProjectile motion in a lift moving with acceleration a upwardsProjectile motion in a lift moving with acceleration a upwardsProjectile motion in a lift moving with acceleration a upwards(1) Initial velocity = u(2) Velocity at maximum height = u cos

(3) T = agsinu2+

(4) Maximum height (H) = )ag(2sinu 22

+

(5) Range = ag2sinu2

+


RELARELARELARELARELATIVE MOTIVE MOTIVE MOTIVE MOTIVE MOTION IN TTION IN TTION IN TTION IN TTION IN TWWWWWO DIMENSIONO DIMENSIONO DIMENSIONO DIMENSIONO DIMENSION

Arr

= position of A with respect to O

Brr

= position of B with respect to O

ABrr

= position of A with respect to B.

BAAB rrrrrr

=

dt)r(d AB

r = dt

)r(d Ar

dt)r(d B

r. BAAB vvv

rrr=

dt)v(d AB

r = dt

)v(d Ar

dt)v(d B

r BAAB aaa

rrr=

Note : These formulae are not applicable for light.

Illustration :Object A and B has velocities 10 m/s. A is moving along East while B is moving towards North from the samepoint as shown. Find velocity of A relative to B ( ABv

r )

Sol. ABvr

= Avr

Bvr

ABvr

= 102

Note : BA vvrr

= + cosvv2vv BA2B2A

Illustration :Two particles A and B are projected in air. A is thrown horizontally, B is thrownvertically up. What is the separation between them after 1 sec.

Sol. ABar

= Aar

Bar

= 0

ABvr

= 22 1010 + = 210

sAB = vABt = (10 2 ) t = 10 2 m

Consider the situation, shown in figure


Ex.23 (i) Find out velocity of B with respect to A

ABBA vvvrrr

= = 20 j 20 i

(ii) Find out velocity of A with respect to B

BAAB vvvrrr

= = 20 i 20 j

Ex.24

(1) Find out motion of tree, bird and old man as seen by boy.(2) Find out motion of tree, bird, boy as seen by old man(3) Find out motion of tree, boy and old man as seen by bird.

Sol. (1) With respect to boy :vtree = 16 m/s ()vbird = 12 m/s ()v

old man = 18 m/s ()(2) With respect to old man :

vBoy = 18 m/s ()vTree = 2 m/s ()vBird = 18 m/s () and 12 m/s ()

(3) With respect to Bird :vTree = 12 m/s () and 16 m/s ()v

old man = 18 m/s () and 12 m/s ().vBoy = 12 m/s ().

MOMOMOMOMOTION OF A TRAIN MOTION OF A TRAIN MOTION OF A TRAIN MOTION OF A TRAIN MOTION OF A TRAIN MOVING ON EQUVING ON EQUVING ON EQUVING ON EQUVING ON EQUAAAAATTTTTOR :OR :OR :OR :OR :If a train is moving at equator on the earths surface with a velocity vTE relative to earths surface and a pointon the surface of earth with velocity vE relative to its centre, then

ETTE vvvrrr

= or ETET vvvrrr

+=

So, if the train moves from west to east and if the train moves from east to west(the direction of motion of earth on its axis) (i.e. opposite to the motion of earth)


vT = vTE + vE vT = vTE vE

Relative Motion on a moving traing :If a boy is running with speed BTv

r on a train moving with velocity Tv

r relative to ground, the speed of the boy

relative to ground Bvr

will be given by:

BTvr

= Bvr

Tvr

or Bvr

= BTvr

+ Tvr

so, if the boy is running in the direction of trainvB = u + v

and if the boy is running on the train in a direction opposite to the motion of trainvB = u v

RELARELARELARELARELATIVE MOTIVE MOTIVE MOTIVE MOTIVE MOTION IN RIVER FLTION IN RIVER FLTION IN RIVER FLTION IN RIVER FLTION IN RIVER FLOOOOOW W W W W :

If a man can swim relative to water with velocity mRvr

and water is following relative to ground with velocity

Rvr

, velocity of man relative to ground mvr

will be given by :

mRvr

= mvr

Rvr

or mvr

= mRvr

+ Rvr

So, if the swimming is in the direction of flow of water,

vm = v

mR + vR

and if the swimming is opposite to the flow of water,

vm = v

mR vR

Illustration :A swimmer capable of swimming with velocity v relative to water jumps in a flowing river having velocity u. Theman swims a distance d down stream and returns back to the original position. Find out the time taken incomplete motion.

Sol. t = tdown + tup

= uv

d+

+ uv

d

= 22 uv

dv2

Ans.


CROSSING RIVERCROSSING RIVERCROSSING RIVERCROSSING RIVERCROSSING RIVERA boat or man in a river always moves in the direction of resultant velocity of velocity of boat (or man) andvelocity of river flow.

1. Shortest Time :

The person swims perpendicular to the river flow crossing a river : consider a river having flow velocity Rvr

andswimmer jump into the river from a point A, from one bank of the river, in a direction perpendicular to thedirection of river current. Due to the flow velocity of river the swimmer is drifted along the river by a distanceBC and the net velocity of the swimmer will be mv

r along the direction AC.

If we find the components of velocity of swimmer along and perpendicular to the flow these are.

Velocity along the river, vx = vR.

Velocity perpendicular to the river, vf = vmR vR ie

Practically it is not possible to reach at B if the river velocity (vR) is too high.Illustration :

A man can swim at the rate of 5 km/h in still water. A river 1 km wide flows at the rate of 3 km/h. The manwishes to swim across the river directly opposite to the starting point.

(a) Along what direction must the man swim?(b) What should be his resultant velocity?(c) How much time the would take to cross?

Sol. The velocity of man with respect to river vmR= 5 km/hr, this is greater than the river flow velocity, therefore, he

can cross the river directly (along the shortest path). The anlge of swim must be

= 2

+ sin1

mR

r

v

v = 90 + sin-1

mR

r

v

v

= 90 + sin1

53

= 90 + 37 = 127, with the river flow (upstream) Ans.


(b) Resultant velocity will be vm

= 2R

2mR vv

= 22 35

= 4 km/hr

along the direction perpendicular to the river flow.

(c) time taken to cross the

t = 2R

2mR vv

d

= hr/km4km1

= 41

h = 15 min

Ex. The velocity of about in still water is 5 km/h it crosses 1 km wide river in 15 minutes along the shortestpossible path. Determine the velocity of water in the river in km/h

Ans. 3km/h

Illustration :A man wishes to cross a river flowing with velocity u jumps at an anlge with the river flow. Find out the netvelocity of the man with respect to ground if he can swim with speed v. Also findHow far from the point directly opposite to the starting point does the boat reach the opposite bank. in whatdirection does the boat actually move. If the width of the river is d.

Sol. Velocity of man = vM = ++ cosuv2vu 22 Ans.

tan = +

cosvu

sinvAns.

(v sin) t = d t = sinvd

x = (u + v cos ) t = (u + v cos) sinvd

Ans.

Illustration :A boat moves relative to water with a velocity which is n times less than the river flow velocity. At what anlgeto the stream direction must the boat move to minimize drifting?

Sol. In this problem, one thing should be carefully meted thatthe velocity of boat is less than the river flow velocity. Insuch a case, boat cannot reach the point directly oppo-site to its starting point. i.e. drift can never be zero. Thus,to minimize the drift, boat starts at an angle from thenormal direction up stream as shown.

Now, again if we find the components of velocity of boat along and perpendicular to the flow, these are,velocity along the river, v

x = u v sin .

and velocity perpendicular to the river, vy = v cos .

time taken to cross the river is t = yv

d = cosv

d.

In this time, drift x = (vx)t

= (u v sin ) cosvd

or x = v

ud sec d tan

The drift x is minimum, when ddx

= 0,


or

v

ud (sec . tan ) d sec2 = 0

orv

u sin = 1

or sin = u

v=

n

1 (as v = n

u )

so, for minimum drift, the boat must move at an angle = sin1

u

v from normal direction or

an angle 2

+ sin1

u

v from stream direction.

RAIN PROBLEMSRAIN PROBLEMSRAIN PROBLEMSRAIN PROBLEMSRAIN PROBLEMS

If rain is falling vertically with a velocity Rvr

and on observer is moving horizontally with velocity mvr

, thevelocity of rain relative to observer will be :

Rmvr

= Rvr

mvr

or vRm = 2m

2R vv +

and direction = tan1

R

m

v

v with the vertical as shown in figure.

Illustration :Rain is falling vertically and a man is moving with velocity 6 m/s. Find the angle with which umbrella shouldbe hold by man to avoid getting wet.

Sol.

vr

rain = 10 j vr man = 6 i

Velocity of rain with respect to man = vr r m = 10 j 6 i

tan = 106

= tan1

53

Where is angle with vertical


Illustration :A man moving with 5m/s observes rain falling vertically at the rate of 10 m/s. Find the speed and direction ofthe rain with respect to ground.

vRM = 10 m/s, vM = 5 m/s

RMvr

= Ruvr

Mvr

Ruvr

= RMvr

Mvr

Rvr

= 55

tan = 21

, = tan1 21

.

Illustration :A man standing, observes rain falling with velocity of 20 m/s at an angle of 30 with the vertical.

(1) Find out velocity of man so that rain appears to fall vertically.(2) Find out velocity of man so that rain again appears to fall at 30 with the vertical.

Sol. (1) mvr

= v i (let)

Rvr

= 10 i 103 j

RMvr

= (10 v) i 103 j

(10 v) = 0 (for vertical fall, horizontal component must be zero)or v = 10 m/s Ans.

(2) Rvr

= 10 i 103 j

mvr

= vx i

RMvr

= (10 vx) i 103 j

Angle with the vertical = 30

tan 30 = 310v10 x

vx = 20 m/s


WIND AIRPLANEWIND AIRPLANEWIND AIRPLANEWIND AIRPLANEWIND AIRPLANEThis is very similar to boat river flow problems the only difference is that boat is replaced by also plane andriver is replaced by wind.

Thus,

velocity of aeroplane with respect to wind

waaw vvvrrr

=

or wawa vvvrrr

+=

where, avr

= absolute velocity of aeroplane

and, wvr

= velocity of wind.

Illustration :

An aeroplane flies along a straight path A to B and returns back again. The distance between A and B is land the aeroplane maintains the constant speed v. There is a steady wind with a speed u at an angle withline AB. Determine the expression for the total time of the trip.

Sol. A to B :

Velocity of plane along AB = v cos ucos ,and for no-drift from line

AB : v sin = usin sin = v

sinu

time taken from A to B : tAB = cosucosvl

B to A :velocity of plane along BA = vcos + u cos

and for no drift from line AB : vsin = usin

sin = v

sinu

time taken from B to A : tBA = + cosucosvl

total time taken = tAB + tBA

= + cosucosvl

+ + cosucosvl

= +

2222 cosucosv

cosv2 l= 22

2

22

uv

v

sinu1v2

l

.

Ex. Find the time an aeroplane having velocity v, take to fly around a square with side a and the wind blowing ata velocity u, in the two cases,

(a) if the direction of wind is along one side of the square,(b) If the direction of wind is along one of the diagonals of the square

Ans. (a) 22 uva2

+ 22 uvv (b) a22

22

22

uv

uv2.


Condition to collide or to reach at the same pointCondition to collide or to reach at the same pointCondition to collide or to reach at the same pointCondition to collide or to reach at the same pointCondition to collide or to reach at the same pointWhen the relative velocity of one particle w.r.t. to other particle is directed towards each other then they willcollide. (If there is a zero relative acceleration).

Illustration :There are particles A, B and C are situated at the vertices of an equilateral triangle ABC of side a at t = 0.Each of the particles moves with constant speed v. A always has its velocity along AB, B along BC and Calong CA. At what time will the particle meet each other?

Sol. The motion of the particles is roughly sketched in figure. By symmetry they will meet at the centroid O of thetriangle. At any instant the particles will from an equilateral triangle ABC with the same

Centroid O. All the particles will meet at the centre. Concentrate on the motion of any one particle, say B.At any instant its velocity makes angle 30 with BO.The component of this velocity along BO is v cos 30. This component is the rate of decrease of the distanceBO. Initially.

BO =30cos

2/a = 3

a = displacement of each particle.

Therefore, the time taken for BO to become zero

= 30cosv

3/d = 3v3

d2

= v3d2

.

Alternative : Velocity of B is v along BC. The velocity of C is along CA. Itscomponent along BC is v cos 60 = v/2. Thus, the separation BC de-creases at the rate of approach velocity.

approach velocity = v + 2v

= 2v3

Since, the rate of approach is constant, the time taken in reducing the separation BC from a to zero is

t = v3a2

2v3

a=

Q. Six particles situated at the corners of a regular hexagon of side a move at a constant speed v. Each particlemaintains a direction towards the particle at the next corner. Calculate the time the particles will take to meeteach other.

Ans. 2 a/v.Q. A moves with constant velocity u along then x axis. B always has veloc-

ity towards A. After how much time will B meet A if B moves with constantspeed v. What distance will be travelled by A and B.

Ans. distance travelled by A = 222

uv

v

l,

distance travelled by B = 22 uvuv

l


Illustration :Two cars A and B are moving west to east and south to north respectively along crossroads. A moves with aspeed of 72 kmh1 and is 500 m away from point of intersection of cross roads and B moves with a speed of54 kmh1 and is 400 m away from point of intersection of cross roads. Find the shortest distance betweenthem ?

Sol.Method I (Using the concept of relative velocity)

In this method we watch the velocity of A w.r.t. B. To do this we plot the resultant velocity Vr. Since theaccelerations of both the bodies is zero, so the relative acceleration between them is also zero. Hence therelative velocity will remain constant. So the path of A with respect to B will be straight line and along thedirection of relative velocity of A with respect to B. The shortest distance between A & B is when A is at pointF (i.e. when we drop a perpendicular from B on the line of motion of A with respect to B).From figure

tan = A

BVV

= 2015

= 43

........................(i)This is the angle made by the resultant velocity vector with the x-axis.Also we know that from figure

OE = 500x

= 43

..............................(ii)From equation (i) & (ii) we getx = 375 m

EB = OB OE = 400 375 = 25 mBut the shortest distance is BF.

From magnified figure we see that BF = EB cos = 25 54

BF = 20 mMethod II (Using the concept of maxima minima)

A & B be are the initial positions and A,B be the final positions after time t.B is moving with a speed of 15 m/sec so it will travel a distance of BB = 15t during time t.A is moving with a speed of 20 m/sec so it will travel a distance of AA = 20t during time t.So

OA =500 20 tOB = 400 15 t

AB2 = OA2 + OB2 = (500 20t)2 + (400 15t)2 ..................(i)For AB to be minimum AB2 should also be minimum


dt)'B'A(d 2 = dt

)t20500()t15400(d 22 += 0

= 2(400 15t) (15) + 2(500 20t) (20) = 0= 1200 + 45t = 2000 80 t

125 t = 3200

t = 5128

s. Hence A and B will be closest after 5128

s.

Now 22

dt'B'Ad comes out to be positive hence it is a minima.

On substituting the value of t in equation (i) we get

AB2 = 2

512815400

+

2

5128

.20500

= 22 )12(16 + = 20 m

Minimum distance AB = 20 m.Method III (Using the concept of relative velocity of approach) After time t let us plot the components of velocity of A & B in thedirection along AB. When the distance between the two is mini-mum, the relative velocity of approach is zero. VA cosf + VB sinf = 0(where f is the angle made by the line AB with the x-axis)

20 cosf = 15 sinf

tanf = 1520

= 34

Here do not confuse this angle with the angle in method (I)because that is the angle made by the resultant with x-axis.

Here f is the angle made with x-axis when velocity of approach in zero,

t20500t15400

= 34

t = 5128

So, OB = 16 m and OA = 12m

AB = 22 )12(16 + = 20 mQus. Two ships are 10 km apart on a line running south to north. The one farther north is steaming west at

20 km h1. The other is steaming north at 20 km h1. What is their distance of closest approach ? How longdo they take to reach it ?

Ans. 25 km/h ; 1/4 h = 15 min consider the situation shown in figure for the two particle A and B.

Qus. (1) Will the two particle will collide(2) Find out shortest distance between two particles

Ans. (1) The particles will not collide(2) 54 m.

Note : Muzzle Velocity is the velocity of bullet with respect to the gun i.e. it is Relative Velocity.

RESONANCE Page # 35TM

Exercise 1 OBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMS1. A stone is thrown upwards with a velocity 50 ms1. Another stone is simultaneously thrown downwards from

the same location with a velocity 50 ms1. When the first stone is at the highest point, the relative velocity ofthe second stone w.r.t. the first stone is :(A) Zero (B) 50 ms1 (C) 100 ms1 (D) 150 ms1

2. A thief is running away on a straight road in a jeep moving with a speed of 9 m s1. A police man chases himon a motor cycle moving at a speed of 10 m s1. If the instantaneous separation of the jeep from themotorcycle is 100m, how long will it take for the police man to catch the thief?(A) 1s (B) 19s (C) 90s (D) 100s

3. Two cars are moving in the same direction with a speed of 30 km h1. They are separated from each other by5 km. Third car moving in the opposite direction meets the two cars after an interval of 4 minutes. What is thespeed of the third car?(A) 30 km h1 (B) 35 km h1 (C) 40 km h1 (D) 45 km h1

4. Shown in the figure are the displacement time graph for two children going home from the school. Which ofthe following statements about their relative motion is true?Their relative velocity :(A) first increases and then decreases(B) first decreases and then increases(C) is zero(D) is non zero constant.

5. A person standing on the escalator takes time t1 to reach the top of a tower when the escalator is moving. Hetakes time t2 to reach the top of the tower when the escalator is standing. How long will he take if he walksup a moving escalator?(A) t2 t1 (B) t1 + t2 (C) t1 t2 /(t1 t2) (D) t1 t2/(t1 + t2)

6. Shown in the figure are the velocity time graphs of the two particles P1 and P2. Which of the followingstatements about their relative motion is true?Their relative velocity :(A) is zero(B) is non-zero but constant(C) continuously decreases(D*) continuously increases

7. Two particles P1 and P2 are moving with velocities v1 and v2 respectively. Which of the statement about theirrelative velocity v

r 12 is true?(A) vr 12 > (v1 + v2) (B) vr 12 cannot be greater than v1 v2(C) vr 12 cannot be greater than v1 + v2 (D) vr 12 < (v1 + v2)

8. Two identical trains take 3 sec to pass one another when going in the opposite direction but only2.5 sec if the speed of one is increased by 50 %. The time one would take to pass the other when goingin the same direction at their original speed is :(A) 10 sec (B) 12 sec (C) 15 sec (D) 18 sec

9. Two billiard balls are rolling on a flat table. One has velocity components vx = 1m/s, vy = 3 m/s and the other

has components vx = 2m/s and vy = 2 m/s. If both the balls start moving from the same point, the angle

between their path is -(A) 60 (B) 45 (C) 22.5 (D*) 15

10. A battalion of soldiers is ordered to swim across a river 500 ft wide . At what minimum rate should theyswim perpendicular to river flow in order to avoid being washed away by the waterfall 300 ft downstream.The speed of current being 3 m.p.h. :(A) 6 m.p.h. (B*) 5 m.p.h. (C) 4 m.p.h. (D) 2 m.p.h.

11. A boat, which has a speed of 5 km/hr in still water, crosses a river of width 1 km along the shortest possiblepath in 15 minutes. The velocity of the river water in km/hr is -(A) 1 (B*) 3 (C) 4 (D) 41


12. A bucket is placed in the open where the rain is falling vertically. If a wind begins to blow at double the velocityof the rain, how will be rate of filling of the bucket change?(A) Remain unchanged (B) Doubled (C) Halved (D) Become four times

13. A car with a vertical wind shield moves along in a rain storm at the speed of 40 km/hr. The rain drops fallvertically with a terminal speed of 20 m/s. The angle with the vertical at which the rain drop strike the windshield is -(A) tan1 (5/9) (B) tan1(9/5) (C) tan1 (3/2) (D) tan1(3)

SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS14. Men are running along a road at 15 km/h behind one another at equal intervals of 20 m. Cyclist are

riding in the same direction at 25 km/h at equal intervals of 30 m. At what speed an observer travelalong the road in opposite direction so that whenever he meets a runner he also meets a cyclist?s

15. Two perpendicular rail tracks have two trains A & B respectively. Train A moves north with a speed of 54 kmh1 and train B moves west with a speed of 72 km h1 . Assume that both trains starts from same point.Calculate the(a) rate of separation of the two trains(b) relative velocity of ground with respect to B(c) relative velocity of A with respect to B.

16. A man is swimming in a lake in a direction of 30 East of North with a speed of 5 km/hr and a cyclist isgoing on a road along the lake shore towards East at a speed of 10 km/hr. In what direction and withwhat speed would the man appear to swim to the cyclist.

17. A motor boat has 2 throttle position on its engine. The high speed position propels the boat at 10 kmhr1 in still water and the low position gives half the higher speed. The boat travels from its dockdownstream on a river with the throttle at low position and returns to its dock with throttle at highposition. The return trip took 15 % longer time than it did for the downstream trip. Find the velocity ofthe water current in the river.

18. () A man can swim with a speed of 4 km h1 in still water. How long does he take to cross a river 1 km wideif the river flows steadily at 3 km h1 and he makes his strokes normal to the river current ? How far down theriver does he go when he reaches the other bank ?() If he keeps himself always at an angle of 120C with the river flow while swimming.(a) Find the time he takes to cross the river. (b) At what point on the opposite bank will he arrive ?

19. A river is flowing from west to east at a speed of 5 m/min. A man on the south bank of the river, capable ofswimming at 10 m/min in still water, wants to swim across the river in shortest distance. In what directionshould he swim ?

20. An airplane is flying with velocity 250 km/hour in north-east direction. Wind is blowing at 25 km/hr fromnorth to south. What is the resultant displacement of airplane in 2 hours ?

21. When a train has a speed of 10 m s1 eastward, raindrops that are falling vertically with respect to the earthmake traces that are inclined 30 to the vertical on the windows of the train.(a) What is the horizontal component of a drop's velocity with respect to the earth ? With respect to the train ?(b) What is the velocity of the raindrop with respect to the earth ? With respect to the train ?

22. To a man walking at 7 km/h due west, the wind appears to blow from the north-west, but when he walks at 3km/h due west, the wind appears to blow from the north. What is the actual direction of the wind and what isits velocity ?

23. When a motorist is driving with velocity j8i6 + , the wind appears to come from the direction i . Whenhe doubles his velocity the wind appears to come from the direction ji + . Then the true velocity of thewind expressed in the form of jbia + is ______.

24. '

n

' numbers of particles are located at the vertices of a regular polygon of '

n

' sides having the edgelength '

a

'. They all start moving simultaneously with equal constant speed '

v

' heading towards eachother all the time. How long will the particles take to collide?

25. Two ships are 10 km apart on a line running south to north. The one further north is streaming west at40 km/hr. The other is streaming north at 40 km/hr. What is their distance of closest approach and howlong do they take to reach it?


26. A ship is sailing towards north at a speed of 2 m/s. The current is taking it towards East at the rateof 1 m/s and a sailor is climbing a vertical pole on the ship at the rate of 1 m/s. Find the velocity of thesailor in space.

27. A motorboat is observed to travel 10 km hr1 relative to the earth in the direction 37 north of east. If thevelocity of the boat due to the wind only is 2 km hr1 westward and that due to the current only is 4 km hr1southward, what is the magnitude and direction of the velocity of the boat due to its own power ?

28. A person P sitting on a wooden block (which does not move relative to water) in a flowing river sees twoswimmers A and B. A and B both have constant speed v

m relative to water. P observes that A starts from

one point of the river bank and appears to move perpendicular to the river flow. P also observes that Bstarts from some point on the other bank at the same time and moves downstream. The width of theriver is '

d

' and it flows with velocity vr. If A and B both reach a point at the same time, than find the initial

separation between A and B.29. A motorboat going down stream overcome a float at a point M. 60 minutes later it turned back and after

some time passed the float at a distance of 6 km from the point M. Find the velocity of the streamassuming a constant velocity for the motorboat in still water.

30. 2 swimmers start from point A on one bank of a river to reach point B on the other bank, lying directlyopposite to point A. One of them crosses the river along the straight line AB, while the other swims atright angles to the stream and then walks the distance which he has been carried away by the streamto get to point B. What was the velocity (assumed uniform) of his walking if both the swimmersreached point B simultaneously. Velocity of each swimmer in still water is 2.5 km hr1 and the streamvelocity is 2 km hr1.

31. An airplane pilot sets a compass course due west and maintains an air speed of 240 km. hr1. After flying for

21

hr, he finds himself over a town that is 150 km west and 40 km south of his starting point.(a) Find the wind velocity, in magnitude and direction.(b) If the wind velocity were 120 km. hr1 due south, in what direction should the pilot set his course in

order to travel due west ? Take the same air speed of 240 km. hr1.32. Two straight AOB and COD meet each other right angles. A person walking at a speed of 5 km/hr

along AOB is at the crossing O at noon. Another person walking at the same speed along COD reachesthe crossing O at 1:30 PM. Find at what time the distance between them is least and what is its value?

1. An aeroplane is flying vertically upwards with a uniform speed of 500 m/s. When it is at a height of1000 m above the ground a shot is fired at it with a speed of 700 m/s from a point directly below it.What should be the acceleration of the aeroplane so that it may escape from being hit?[REE 94, 6 ]

2. The width of a river is 25 m and in it water is flowing with a velocity of 4 m/min. A boatman is standingon the bank of the river. He wants to sail the boat to a point at the other bank which is directly oppositeto him. In what time will he cross the river, if he can sail the boat at 8 m/min, relative to the water.

[ REE '95, 6 ]3. On a frictionless horizontal surface, assumed to be the x

y plane a small trolley A is moving along astraight line parallel to the y

axis as shown in the figure with a constant velocity of ( 3 1) m/s.

Ata particular instant, when the line OA makes an angle of 45 with the x

axis, a ball is thrown along thesurface from the origin O. Its velocity makes an angle with the x

axis when it hits the trolley.(a) The motion of the ball is observed from the frame of the trolley.

Calculate the angle made by the velocity of the ball withthe x

axis in this frame.(b) Find the speed of the ball with respect to the surface, if = 4

/3.[ JEE 2002, 2 + 3 ]


EXERCISE # 1EXERCISE # 1EXERCISE # 1EXERCISE # 1EXERCISE # 1OBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMSOBJECTIVE PROBLEMS1. C 2. D 3. D 4. D5. D 6. D 7. C 8. C9. D 10. B 11. B 12. A

13. A

SUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMSSUBJECTIVE PROBLEMS

14. 5 km/h15. (a) 25 m/s or 90 km/hr (b) 20 m/s or 72 km/hr due

east (c) 25 m/s or 90 km/hr at 37N of E16. 30 N of W at 35 km/hr. 17. 3 km/hr.

18. () 0.75 km

() (a) 321

h (b) 321

km.

19. At an angle 30 west of north.

20. km550

21. (a) 0, 10 m/s West (b) 10 3 m/s, 20 m/s

22. Coming from 5 km/hr, 53 N of E

23. ( )j8i4 + 24.

n

2cos1v

a

25.2

10, 8

1 hr

26. 2m/s in a direction making an angle of 60 with E,45 with N and 60 with the vertical

27. 210 km/hr, 45 N of E

28. 2 d 29. 3 km/hr

30. 3 km/hr towards B

31. (a) 100 km/hr, 37 W of S (b) 30 N of W32. 12 : 45 PM

EXERCISE # 2EXERCISE # 2EXERCISE # 2EXERCISE # 2EXERCISE # 21. a > 10 m/s2 2. 3.6 minute

3. (a) = 45 ; (b) 2 m/s


I mentally conceive of somemovable projected on a hori-zontal plane all impedimentsbeing put aside. Now it isevident ... that the equablemotion on this plane wouldbe perpetual if the plane wereof infinite extent; but if weassume it to be ended, and[situated] on high, the mov-able ... , driven to the end ofthe plane and going on fur-ther, adds on to its previousequable and indelible motionthat downward tendencywhich it has from its ownheaviness. Thus there emergesa certain motion, compoundedfrom equable horizontal andfrom naturally accelerateddownward [motion], which Icall projection.

- Galileo(Two New Sciences 1638)

PRPRPRPRPROJECTILE MOOJECTILE MOOJECTILE MOOJECTILE MOOJECTILE MOTIONTIONTIONTIONTION

1 .1 .1 .1 .1 . BASIC CONCEPT :BASIC CONCEPT :BASIC CONCEPT :BASIC CONCEPT :BASIC CONCEPT :1.1 PROJECTILE

Any object that is given an initial velocity obliquely and that subse-quently follows a path determined by the gravitational force acting onit, is called a Projectile. A projectile may be a football, a cricket ball,or any other object.1.2 TRAJECTORY

The path followed by a particle (here projectile) during its motion iscalled its Trajectory.NOTE : 1. We shall consider only trajectories that are of sufficientlyshort range so that the gravitational force can be considered constantin both magnitude and direction.

2. All effects of air resi

Resonance Kinematics

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