kinematics

Kinematics

2

Chapter 3 : Kinematics3.1 Motion in One Dimension 04

3.2 Motion in Two & Three Dimension 26

3.3 Projectile Motion 31

3.4 Relative Motion in One and Two Dimension 42

3.5 Circular Motion 47

Solved Examples 52

Exercises 60

Previous Years’ IITJEE Questions 67

Answer Key 69

2010, NH Elite

Every effort has been made to avoid error or omissions in this publication. In spite of this errors may creep in. It isnotified that author or seller will not be responsible for any damage or loss of action to anyone of any kind in anymanner, therefrom. No part of this book may be reproduced or copied in any form or by any means (graphic,electronic or mechanical, including photocopying, recording, taping or information retrieval system) or reproducedon any disc, tape or perforated media or other information storage device, etc., without the written permission ofNH Elite Education Services Limited. Breach of this condition is liable for legal action.

All rights reserved.

All disputes are subject to Delhi Jurisdiction only.

Corporate Office: 30, Hauz Khas Village, 3rd FloorPower House Complex, New Delhi-110016

Phone: 011- 44160000

Interesting Fact

“Our nature

consist in

motion;

complete rest

is death.”Blaise Pascal

PhysicistGalileo Galilei

Galileo Galilei (15 February 1564– 8 January 1642)was an Italian physicist, mathematician, astronomer andphilosopher who played a major role in the ScientificRevolution. Galileo's theoretical and experimental workon the motions of bodies, was a precursor of the classicalmechanics developed by Sir Isaac Newton. He alsoconcluded that objects retain their velocity unless a force—often friction—acts upon them. According to StephenHawking, Galileo probably bears more of the responsibilityfor the birth of modern science than anybody else, andAlbert Einstein called him the father of modern science.

Galileo Galilei

Galileo had dropped balls of the samematerial, but different masses, from theLeaning Tower of Pisa to demonstrate thattheir time of descent was independent of theirmass.

4 |PHYSICS IIT STUDY CIRCLE

KINEMATICS IITJEE COURSE

Introduction

Our aim is to study the motion of a rigid body. A rigid body can also rotate as it moves. However, we shall restrict ourselves totranslatory motion for the time being , in which every portion of the body moves in the same direction at the same rate. As allparticles in the body move in the same way, we may as well consider the motion of a single particle in the body. Moreover, thecause of the motion will not be dealt with until we come to Newton’s laws of motion. In this lesson we study the motion itself.

Section - 3.1 Motion in One Dimension

To describe the motion of a particle, we introduce four important quantities, namely position, displacement, velocity andacceleration. In the general motion of a particle in two or three dimensions, these quantities are vectors, which have directionsas well as magnitude. In this section we confine ourselves to motion in one dimension. For such restricted motion, the particleis confined to move in a straight line and there are only two directions, distinguished by designating them as positive andnegative.

3.1.1 Displacement, Velocity and Speed

To understand the concept of displacement, let us set up a coordinate system by choosing some reference point on a line for theorigin O. Any other point on the line can be assigned a number x which indicates how far the point is from the origin. The valueof x will depend on the unit chosen as the measure of distance. If the point is towards right of origin, x is positive and if it istowards left of origin then x is negative.

Suppose that the particle is at x1 at time t

1 and at x

2 at time t

2 then the displacement of the particle is x

2 – x

1. The change in position

(displacement) is denoted by ∆x.

∆x = x2 – x

1

The average velocity of the particle is defined to be the ratio of displacement ∆xand time interval ∆t = t

2 – t

1:

2 1

2 1

.avg

x xxV

t t t

−∆= =∆ − x1

t t = 1 t t = 2

x = 0 x2 x-axis

The displacement and average velocity may be either positive or negative, dependingon whether x

2 is greater or less than x

1. A positive value would indicate the motion

towards the right and the negative value indicates the motion towards left. If weplot position (x) of the particle as a function of time (t), then in the graph

Vavg

= 2 1

2 1

x x

t t

−− =

BC

AC = tanθ

θA

x1

x2

x

∆t t t = – 2 1 C

B

∆x x x = – 2 1

t1 t2= Slope of line joining points A and B on x – t graph.

The distance travelled by the particle is not necessarily equal to the displacement. If the particle moves only in one direction thenthe displacement is equal to distance travelled, otherwise distance travelled is more than displacement.

x x = 1

t t = 1 t t = 2

x = 0 x x = 2

x-axisx x = 1x = 0 x x = 2 x x = 3

x-axis

A

C

B

Distance travelled = Displacement Distance travelled = AB + BC= (x

3 – x

1) + (x

3 – x

2)

Displacement = x2 – x

1

IITJEE COURSE KINEMATICS

IIT STUDY CIRCLE PHYSICS| 5

In general Distance travelled ≥ Displacement

The average speed of the particle is defined to be ratio of distance travelled and time interval.

Average speed = total distance

time taken

Note that while velocity is a vector quantity and has direction, speed has no direction. The SI units are meters per second.

Illustration 1:

If a particle is at x1 = 10 m at t

1 = 2s and at x

2 = 6 m at t

2 = 4s find its displacement and average velocity for this time interval.

Solution :

By definition, displacement is

∆x = x2 – x

1 = 6 – 10 = – 4 m

and the average velocity is

Vavg

= x

t

∆∆ =

4

4 2

−−

= – 2 m/s

Illustration 2:

A man runs 200 m in first 25 seconds and then turns back and runs 100m in next 15 seconds towards the starting point. Findout his

(a) average velocity (b) average speed.

Solution :

(a) Let us take origin at the starting point.

Then x1 = 0, x

2 = 200 – 100 = 100 m

∆x = x1 – x

1 = 100 – 0 = 100 m

∆t = 25 + 15 = 40 sec

(0, 0) (100, 0) (200, 0)

t = 0 t = 40 sec t = 25 sec

Vavg

= 100

40 = 2.5 m/s

(b) Distance travelled = 200 + 100 = 300 m

time taken = 25 + 15 = 40 sec

Average speed = 300

40 = 7.5 m/s

Instantaneous Velocity (Velocity at an instant)

The velocity at a particular instant of time is known as instantaneous velocity.

Avg. velocity for time interval ∆t is Vavg

= x

t

∆∆ = slope of line AB.

∆x

t1 t2''' t2'' t2' t2 t

x1

x2'''

x2''

x2'

x2

A

B

B2

B1

B3

Bn

x

∆t

As we reduce the interval ∆t, point B moves nearer and nearer to point A. (B→ B

1 → B

2 → B

3 -------- → B

n) and (t

2 → t

2' → t

2'' → t

2''')

As ∆t → 0, (∆t tends to zero), point B approaches point A and we can findinstantaneous velocity at point A.

THEORY



∴ Vinstantaneous

= 0

limt

x

t∆ →

∆∆ =

dx

dt

= Slope of tangent at point A.' ∴ Slope of x – t graph at any point gives velocity at that point

Magnitude of instantaneous velocity = Instantaneous speed

3.1.2 Average and Instantaneous Acceleration

When the velocity of the body changes continuously as the motion proceeds, the body is said to have accelerated motion.Suppose a particle moving along x-axis have velocity v

1 at point P

1 and velocity v

2 at point P

2. Figure below shows the graph of

instantaneous velocity v plotted as a function of time t.

t1 t2

P1

P2

v1

v2

t

v

Slope of tangent at P = Instantaneous acc at P1 1n

Slope of tangent at P = Instantaneous acc at P

Slope of line P P = Avg. acc during interval [ , ]2 2

1 2 1 2

n

n t t

The average velocity of the particle as it moves from P1 to P

2 is defined as the ratio of change in velocity to the elapsed time.

aavg

= 2 1

2 1

v v v

t t t

− ∆=− ∆

In the figure, the average acceleration is represented by slope of the line joining points P1 and P

2.

The instantaneous acceleration of a body, that is acceleration at some time t is defined in the same way as instantaneous velocity.Let the second point P

2 in figure be taken closer and closer to point P

1 and let average acceleration be calculated over shorter and

shorter time intervals. The instantaneous acceleration at the first point is defined as the limiting value of the average accelerationwhen the second point is taken closer and closer to the first point.

ainstantaneous

= 0

limt

v

t∆ →

∆∆ =

dv

dt

In the figure shown, the instantaneous acceleration at any point is equal to the slope of the tangent in the v – t graph at that point.

Illustration 3:

The position of a particle is given by x = (3t – 5t2 + t3) m where t is in seconds. Find the velocity and acceleration as afunction of time.

Solution :

To find the velocity of the particle we compute the derivative of x with respect to t.

v = dx

dt =

d

dt(3t – 5t2 + t3) = (3 – 10t + 3t2) m/sec

To find acceleration, we differentiate v with respect to time t.

a = dv

dt = – 10 + 6t = (6t – 10) m/s2



In-Chapter Exercise - 1

1. The velocity of a car travelling on a straight road is given by the equation v = 6 + 8t – t2 where v is in meters per second andt in seconds. The instantaneous acceleration when t = 4.5 s is :

(a) 0.1 m/s2 (b) 1 m/s2 (c) – 1 m/s2 (d) – 0.1 m/s2

2. The displacement x of a body varies with time t as x = – 2

3t2 + 16t + 2. The body will come to rest after time t = ..... .

3. An athlete takes 4 sec to reach his maximum speed of 36 km/hr. What is the magnitude of this avg. acceleration.

4. When a person leaves his home for shopping by his bike, the milometer reads 1032 miles. When he returns home after 1.5hours the reading is 1122 miles.

(a) What is the average speed of the car during this period? (b) What is the average velocity?

5. The position of a particle is given by x = 6t – 3t2 where t is expressed in seconds and x in meter.

(a) The acceleration of the particle is

(i) 6 m/sec2 (ii) – 6 m/sec2 (iii) – 3 m/sec2 (iv) None

(b) The maximum value of position co-ordinate of particle on positive x-axis is

(i) 3 m (ii) 1 m (iii) 2 m (iv) 4 m

(c) The total distance travelled by the particle between t = 0 to t = 2 sec is

(i) 0 (ii) 3 m (iii) 4 m (iv) 6 m

3.1.3 Motion with Constant Acceleration

If a particle is accelerated with constant acceleration for a time interval, then following important results can be used.

v = u + at Here u = initial velocity

s = ut + 1

2at2 v = final velocity

= vt – 1

2at2 a = acceleration

xf = x

i + ut +

1

2at2 x

f = final position coordinate

v2 = u2 + 2as xi = initial position coordinate

sn = u +

1

2a(2n – 1) s

n = Displacement during nth second

If acceleration is in same direction as velocity, then speed of the particle increases and if in opposite direction then speed decreases.

Illustration 4:

A person travels half of the total distance with speed 20m/sec and next half with speed 30m/sec along a straight line. Findout the average speed of the particle?

Solution :

Let the total distance travelled by the particle be 2D.

Time taken for 1st half = D

20

Time taken for 2nd half = D

30

Average speed = Total Distance Covered

Time take =

2DD D20 30

+ = 24 m/sec

THEORY



Illustration 5:

Position of a particle moving along x-axis is a function of time as x = 3t2 + 9t + 6 meters. Find

(a) Average velocity for time interval t : [2, 6] sec. (b) Instantaneous velocity at t = 3 sec.

(c) Average acceleration for time interval t : [0, 6] sec. (d) Instantaneous acceleration at t = 4 sec.

Solution :

(a) Vavg.

= Total Displacement

Time interval=

( ) ( )6 2

6 2

x t x t= − =−

= ( ) ( )2 23 6 9 6 6 3 2 9 2 6

4

× + × + − × + × + = 33 m/sec.

(b) Instantaneous velocity = dx

dt =

d

dt(3t2 + 9t + 6) = (6t + 9) m/sec

∴ V at t = 3 sec = 6 × 3 + 9 = 27 m/sec.

(c) aavg.

= Change in Velocity

Time interval=

V( 6) V( 0)

6

t t= − =

= (6 6 9) (6 0 9)

6

× + − × + = 6 m/sec2

(d) Instantaneous acceleration = dv

dt =

d

dt(6t + 9) = 6 m/sec2.

Illustration 6:

A car starts from rest at a constant acceleration of 8 m/s2. Find

(a) How fast is it going at t = 10 sec (b) How far has it gone till t = 10 sec

(c) Its average velocity for the interval t : [0, 8] sec.

Solution :

Using Equation of motion (Q acceleration = const.)

(a) v = u + at

= 0 + 8 × 10 = 80 m/sec∴ v(t = 10 sec) = 80 m/sec

(b) s = ut + 1

2at2

= 0 × 10 + 1

2 × 8 × 102 = 400 m

(c) Vavg

= Displacement

Time Interval=

( 8sec) ( 0)

8sec

x t x t= − =

=

210 8 8 8 0

28

× + × × − = 32 m/sec

Illustration 7:

A particle moves in a straight line with constant acceleration. If it covers 10 m in the first second and 20 m in the nextsecond, find its initial velocity and acceleration.

Solution :

Let v0 be the initial velocity and a be the acceleration



Applying s = ut + 1

2at2 for 1st sec we get

10 = v0 × 1 +

1

2a × 12 ⇒ 10 = v

0 +

9

2––––––– (1)

Applying s = ut + 1

2at2 for 1st two seconds we get

30 = v0 × 2 +

1

2a × 22 ⇒ 30 = 2v

0 + 2a ––––––– (2)

Solving we get v0 = 5 m/sec and a = 10 m/s2.

Illustration 8:

A particle with an initial velocity of 10 m/sec moves along a straight line with a constant acceleration. When the velocityof the particle is 50 m/sec, the direction of acceleration is reversed. Find the velocity of the particle when it reaches thestarting points again.

Solution :

Let the magnitude of acceleration be a0.

For AB : v2 = u2 + 2as

(50)2 = (10)2 + 2a0 × (AB)

∴ AB = 0

1200

a

AA B C

vi = 10 m/sec vf = 50 m/sec v = 0

a0 a0a0

For BC : v2 = u2 + 2as

02 = (50)2 + 2(– a0)BC

∴ BC = 0

1250

a

Now, CA = AB + BC = 0

2450

a

∴ vf2 = 02 + 2 × a

0 × CA = 2 × a

0 ×

0

2450

a = 4900

∴ vf = 70 m/sec. (In opposite direction to initial velocity)

Illustration 9:

A bus starts from a bus stop with a constant acceleration of 0.4 m/s2. A passenger arrives at the bus stop 6 sec after thebus left. Then he immediately starts running towards the bus with a constant speed 5 m/sec. At what distance from thestarting point will he catch the bus.

Solution : 5 m/sec

t = 6 sec t = T

0.4 m/s2

Let us assume that the bus starts at t = 0 from x = 0 and the man catches the bus at t = T.

Bus travelled for time interval = T and man ran for time interval (T – 6). Finally both of them reached at x = xf .

∴ For Bus xf = 0 × T +

1

2 × 0.4 T2 –––––––– (1)

For Man xf = 5(T – 6) –––––––– (2)

THEORY



Equating (1) and (2) we get 0.2 T2 = 5(T – 6)

On solving we get T = 10 sec and T = 15 sec

So, the man will catch the bus at t = 10 sec.

Distance travelled = 5(10 – 6) = 20 m.

Illustration 10:

A car moving rectilinearly with constant acceleration is having initial velocity v0. After some time its velocity becomes

7 v0. Find out velocity of the particle at the midpoint of its path.

Solution :

From the figure

D D

v0 v = ?? v07

( )2

07v = v02 + 2 × a

0 × 2D

or3

2v

02 = a

0D –––––––– (1)

Let velocity at midpoints be v. Then

v2 = v02 + 2 × a

0 × D –––––––– (2)

Putting (1) in (2) we get v2 = v02 + 2

20

3

2v

= 4v02

∴ v at midpoint = 2v0

3.1.4 Motion Under Gravity

When a particle moves under the influence of gravitational pull of Earth alone then its acceleration is g = 9.8 m/s2 (approx = 10m/s2) downwards.

Illustration 11:

A stone is thrown vertical upwards from ground level with u = 20 m/s.

(a) Find the maximum height attained by the stone

(b) time interval t after which it returns to the point of projection

(c) The velocity with which it strikes the ground (Take g = 10 m/s2)

Solution :

20 m/sec

Hmax

v = 0Let us choose our origin O at the point of projection with +ve x-axis pointingin the vertical upwards direction.

Note that in this coordinate system, acceleration due to gravity is negativebecause it points in the downward direction.

Thus a = – 10 m/s2, u = 20 m/s

(a) At the highest point, velocity of the particle will become zero. Let h bethe maximum height. Thus S = h.

using the relation,



v2 – u2 = 2a S

we get0 – 202 = 2 ×(– 10) × h ⇒ h = 400

20 = 20 m

(b) Let time of flight be T. As the particle returns to the same position i.e. S = 0 after time interval t = T from the time ofprojection,

using S = ut + 1

2at2 we have

210 20 10

2t t= − × ⇒ t =

40

10 = 4 sec

(c) To find the velocity when it returns,

we use v = u + at

⇒ v = 20 – 10 × 4

= – 20 m/s

Here, minus sign indicates that particle moves in the downward direction.

Note : It returns with same speed with which it was thrown.

In general, if a particle is thrown upwards with velocity v0, then

Time of flight = 02v

g

Max Height reached = 2

0

2

v

gSpeed just before hitting ground = v

0 (downwards)

Illustration 12:

If another stone was thrown from the same point after one second in the last example, with initial velocity of 25 m/s , find

(a) The coordinates of the point where they collide

(b) The velocities of the two stones when they collide. (Take g = 10 m/s2)

Solution :

(a) Let x1 and x

2 to be the coordinates of first and second stone after time t.

S1 = x

1 – 0 = 20t –

1

210t2 —————— (1)

S2 = x

2 – 0 = 25(t – 1) –

1

210(t – 1)2 —————— (2)

Please note that if the first stone is in flight for t second then second stone will be in flight for (t – 1) seconds as it wasthrown after 1 sec. When they collide, they are at the same place, hence

x1 = x

2

⇒ 20t – 5t2 = 25(t – 1) – 5(t – 1)2

⇒ t = 2 sec

Substituting t = 2 in eqn. (1) or eqn (2) , we find

h = x1 = x

2 = 20 m

(b) The velocities of the stones after time t are

v1 = 20 – 10t —————— (3)

v2 = 25 – 10(t – 1) —————— (4)

Substituting t = 2 in eqns (3) and (4), we find

v1 = 0 m/sec (i.e. stone is at its highest point.)v

2 = 15 m/sec

THEORY



Illustration 13:

A body is thrown down from the top of a tower of height h with velocity 10 m/s. Simultaneously, another body is projectedupward from bottom. They meet at a height 2h/3 from the ground level. If h = 60 m, find the initial velocity of the lowerbody.

Solution :

h

v = 0

B v0

A

A

B2

3

h

Let us choose the origin at the foot of the tower and y-axis pointing upwards. Let usrefer upper and lower ball as A and B respectively

For Ball A : yinitial

= h, yfinal

= 2

3

h

uinitial

= – 10 m/sec, a = – 10 m/s2

Applying yf = y

i + ut +

1

2at2 we have

2

3

h = h – 10t –

1

2 × 10t2 ————— (1)

For Ball B : yinitial

= 0, yfinal

= 2

3

h, u

initial = v

0, a = – 10m/s2

Applying y = yi + ut +

1

2at2 we have

2

3

h = 0 + v

0t –

1

2 × 10t2 ————— (2)

Subtrating (1) from (2) we have 0 = – h + (v0 + 10)t ⇒ t =

0 10

h

v +

Putting this value in eq. (1) we have

0 =

2

0 0

10

3 ( 10) 10

h h h

v v

×− − + + 2

0 0

2 410 10

h h

v v

+ − + +

= 0

On solving we have v0 = 15 5+ 5 = 38.5 m/sec (∴ h = 60 m)

Illustration 14:

A rocket is fired vertically and ascends with constant vertical acceleration of 19.6 m/s2 for 30 seconds. Its fuel is then allused up and it continues as a free particle.

(a) What is the maximum altitude reached? (b) What is total time after which it strikes the ground again?

Solution :

Let us choose the vertical upward to be the positive direction. Let A be the point at which fuel is exhausted and let pointB represent the maximum altitude.

From O → AA

O

Aa = 19.6 m/s2

a = 9.8 m/s2

v = 0Bu = 0, a = 19.6 m/s2, t = 30s

OA = ut + 1

2at2 = 0 + 9.8 × 900 = 8820 m

VA = u + at = 0 + 19.6 × 30 = 588 m/s

Form A → B

u = VA = 588 m/s, a = – 9.8 m/s2

v = VB = 0

using v2 – u2 = 2as we get

0 – (588)2 = 2 × – 9.8 × AB

⇒ AB = 17640 m

Hence,



maximum altitude = OA + AB = 26.46 km

To find time, let us consider the path A → B → O

a = – 9.8 m/s2, s = – OA = – 8820, u = 588 m/s

s = ut + 1

2at2

⇒ – 8820 = 588 × t – 4.9t2

⇒ t2 – 120t – 1800= 0

t = 120 14400 7200

2

+ +

= 133.5 sec

So, the total time = 30 + 133.5 = 163.5 sec

Illustration 15:

A balloon is rising with constant acceleration 2m/sec2. Two stones are released from the balloon at the interval of 2 sec.Find out the distance between the two stones 1 sec after the release of second stone. (g = 10 m/s2)

Solution :

Acceleration of balloon = 2 m/sec2. Let v0 be the velocity of the balloon at the time of release of 1st stone and y

0 be its

attitude w.r.t ground.

After 2 seconds, balloon’s new velocity is

v = v0 + 2 × 2 = (v

0 + 4) m/sec upwards v v = + 40

v v = 0

a g =

a g =

2nd stone released

1st stone released

v0 + 4

v0

y0

Its new altitude is y = y0 + v

0 × 2 +

1

2 × 2 × 22

= y0 + 2v

0 + 4

Altitude of 1st stone 3 seconds after its release

y1= y

0 + v

0 × 3 +

1

2(– 10) × 32

= y0 + 3v

0 – 45

Altitude of 2nd stone 1 second after its release

y2= (y

0 + 2v

0 + 4) + (v

0 + 4) × 1 +

1

2(– 10) × 12

= y0 + 3v

0 + 3

Distance between the stones = |y2 – y

1|

= |(y0 + 3v

0 + 3) – (y

0 + 3v

0 – 45)|

= 48 m

Note : As the particle is detached from the balloon it is having the same velocity as that of balloon, but itsacceleration is only due to gravity and is equal to g.


1. A ball is dropped from a height. If it takes 0.200s to cross the last 6.00 m before hitting the ground, the height from whichit was dropped is ................ (Take g = 10 m/s2)

2. A parachutist drops freely from an aeroplane for 10 s before the parachute opens out. Then he descends with a netretardation of 2.5 m/s2. If he bails out the plane at a height of 2495 m and g = 10 m/s2, his velocity on reaching the groundwill be v = .................. .

THEORY



3. A particle has an initial velocity of 9 m/s due east and a constant acceleration of 2 m/s2 due west. The distance covered bythe particle in the fifth second of its motion is .............. .

4. In a car race, car A takes a time of t sec. Less than car B at the finish and presses the finishing point with a velocity v morethan the car B. Assuming that the cars start from rest and travel with constant accelerations a

1 and a

2 respectively. Which

of the following relation is correct :

(a) v = 1 2a a t (b) v = 2 1 2a a t (c) v = 1 2

2

a at (d) None of these

5. A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time to the driver. If he is drivinga car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s2, the distance travelled by the car after he sees theneed to put the brakes on is :

(a) 20 m (b) 22 m (c) 24 m (d) 28 m

6. A ball is dropped from the top of a building. The ball takes 0.5 s to fall past the 3 m length of a window some distance fromthe top of the building. If the speed of the ball at the top and at the bottom of the window are v

T and v

B respectively, then

(g = 9.8 m/sec2)

(a) vT + v

B = 12 ms– 1 (b) v

T – v

B = 4.9 ms– 1 (c) v

Bv

T = 1 ms– 1 (d)

B

T

v

v = 1 ms– 1

7. A particle moving along a straight line with constant acceleration is having initial and final velocity as 5 m/s and 15 m/srespectively in time interval of 5 s. Find the distance travelled by the particle and the acceleration of the particle. If theparticle continues with same acceleration, find the distance covered by the particle in the 8th second of its motion.

8. An object is thrown vertically upward. It has a speed of 10 m/s when it has reached one half its maximum height.

(a) How high does it rise?

(b) What is its velocity and acceleration 1 sec after it is thrown?

(c) What is its velocity and acceleration 3 s after it is thrown?

(d) What is the average velocity during the first half second?

9. A lift starts from the top of a mine shaft and descends with a constant speed of 10 m/s. 4 s later, a boy throws a stonevertically upwards from the top of the shaft with a speed of 30 m/s. Find when and where stone hits the lift.

(Take: g = 10 m/s2)

10. (a) Can an object reverse the direction of its motion even though it has const. acceleration?

(b) Can body have (i) Zero instantaneous velocity and yet be accelerating(ii) Zero avg. speed but non-zero velocity

(iii) Negative acceleration and yet be speeding up

3.1.5 Graphs

For better understanding and visualization of motion of a particle we use various graphs like

1. x – t graph (position – time graph)

2. v – t graph (velocity – time graph)

3. a – t graph (acceleration – time graph)

Before proceeding for graphs let us first study various terms that will be used frequently while dealing with graphs.

1. Slope of a Line : If a line makes angle θ with the positive direction of x-axis, we define its slope = tanθ .

(a) If θ is acute, slope is positive.

(b) If θ is obtuse, slope is negative.

(c) If θ is zero or line is horizontal then slope is zero.

xLine with positive slope

x

Line with negative slope



2. Slope at a point on the curve : Slope at a point on the curve is slope of the tangent on the curve at that point.

x

Curve (1)

y

( , )x y0 0

x

Curve (2)

y

( , )x y0 0

Slope at (x0, y

0) on the curve (1) is tanθ i.e., Slope at (x

0, y

0) on the curve (2) is tanθ i.e.,

slope of tangent at (x0, y

0) (+ve in this case) slope of tangent at (x

0, y

0) (–ve in this case)

Parabola

Various types of parabola :

1. y = ax2 (a > 0) 2. y = ax2 (a < 0) 3. y = ax2 + b (a > 0, b > 0)

Mouth opening upwards

y

x

Mouth opening downwards

y

x

Mouth opening upwards

y

x(0, )b

4. y = ax2 + b (a < 0, b > 0) 5. y = ax2 + bx + c (a > 0)

Mouth opening downwards

y

x

(0, )b

Mouth opening upwardsy

x

Graph for Various Cases

1. Motion with uniform velocity : Consider a particle moving along x-axis with uniform velocity u from the point x = xi at t

= 0. Then x(t) = xi + ut ; v(t) = u ; a(t) = 0.

x – t graph : Straight line with slope = u passing through xi at t = 0

v – t graph : Horizontal line with slope = 0

a – t graph : a = 0 (Q velocity is constant)

Slope= tan= = velocity

uxi

u is positive t

x

Slope = tan= = velocity

u

xi

u is negativet

x

THEORY



u

t

v

u > 0

u

t

v

u < 0

t

a

a = 0

Uniformly Accelerated Motion

Case 1: (acceleration > 0)

xi

t

x

Initial velocityv i = 0

xi

t

x

Initial velocityv >i 0 xi

t

x

Initial velocityv <t

i 0 at = 0

Slope of x – t graph = 0 at t = 0 Slope of x – t graph is vi > 0 at t = 0 Slope of x – t graph is v

i < 0 at t = 0

vi t

Slope = a0

v

vi

t

v

Slope = a0

vi

t

v

Slope = a0

t

a0

a

a0

t

a

a0

a

t

⇒ For constant acceleration (a0 > 0), x – t graph is a parabola with mouth opening upwards.

⇒ v – t graph is a straight line with slope equal to acceleration.

Case 2: (acceleration < 0)



xi

t

x

Initial velocityv i = 0 xi

t

x Initial velocityv >i 0

xi

t

x

Initial velocityv <i 0

Slope of x – t graph at t = 0 is vi = 0 Slope of x – t graph at t = 0 is v

i > 0 Slope of x – t graph at t = 0 is v

i < 0

vi

t

v

Slope = a0

vi

t

v

Slope = a0

vi

t

v

Slope = a0

a0

t

a

negative acc.a0

t

a

negative acc.a0

t

a

negative acc.

⇒ For constant acceleration (a0 < 0), x – t graph is a parabola with mouth opening downwards.

⇒ v – t graph is a straight line with negative slope equal to acceleration (a0).

Key points

• The slope of tangent at any point on x – t graph gives velocity at that point and the slope of tangent in v –t graph is the acceleration.

• Area under a – t graph gives change in velocity.

• Area under v – t graph and the time axis gives the distance travelled by the particle if we take all areas aspositive and displacement if areas below t-axis are taken negative.

Illustration 16: a (m/s )2

3 6

2

– 2

t(sec)0

Acceleration - time graph of a particle moving along x-axis isshown in the figure. Draw its x – t graph and v – t graph. Initiallyparticle is at x = 4 and moving with speed 8 m/sec towardspositive x-axis.

THEORY



Solution :

For t : [0, 3], a = – 2 m/s2.

∴ v – t graph is a straight line with negative slope and x – t graph is a parabola

(mouth opening downwards).

x – t graph is a parabola (mouth opening downwards).

v(3) = v(0) + a × t = 8 + (– 2) × 3 = 2 m/sec

x(3) = x(0) + v(0) × t + 1

2 × a × t2 = 4 + 8 × 3 +

1

2 × – 2 × 32 = 19 m.

For t : [3, 6], a = 2 m/s2.

∴ v – t graph is a straight line with positive slope and x – t graph is a parabola (month opening upwards)

v(6) = v(3) + a × t = 2 + 2 × 3 = 8 m/sec

x(6) = x(3) + v(3) × t + 1

2 × a × t2 = 19 + 2 × 3 +

1

2× 2 × 32 = 34 m.

Also slope of x(t) graph at t = 0 is 8 m/sec and at t = 3 it is 2 m/sec.

4

10

19

30

34

03 6

x m( )

t(sec)

2

4

6

8

03 6

v(m/sec)

t(sec)

Illustration 17:

For the given velocity-time graph, select the correct alternatives.

t0

v

t0 2t0 3t0

(a) Particle turns back exactly once during its motion.

(b) Acceleration is always –ve.

(c) Average acceleration for [0, t0] is +ve.

(d) Average velocity for [0, t0] is +ve.

(e) Average speeds for [0, t0] and [t

0, 2t

0] are equal.

Solution :

(a) Correct : For t < t0, v > 0, at t = t

0, v = 0 and for t > t

0, v < 0

∴ Particle’s velocity changes sign at t = t0.

Before t = t0, particle is moving along the +ve x-direction and after t = t

0 it is moving along –ve x-direction.

∴ Particle turns back at t = t0.

(b) Correct : Velocity is always decreasing, therefore, a < 0.

Alternatively : Slope of v – t graph is always –ve, therefore acceleration is always –ve.

(c) Incorrect : Slope of ‘v – t’ is constant

∴ a = constant < a > = a

∴ for any time interval acceleration is –ve.



(d) Correct : For [0, t0] : Displacement is +ve.

⇒ average velocity is +ve.

(e) Correct : For [0, t0] and [t

0, 2t

0] : Magnitudes of areas (i.e. distances) are equal, therefore average speeds are equal.

(lengths of intervals are also equal).

Illustration 18:

A man throws a ball upwards with speed 20 m/sec from the top of atower of height 105 m. Taking base of the tower as origin and upwarddirection as positive draw v – t graph and position-time graph ofthe ball for next 12 sec.

(Take g = 10 m/s2 and assume elastic collision with ground)

TOWER

105 m g = 10 m/s2

20 m/sec

Solution :

For upward motion :

v = u + a × t

0 = 20 – 10 × t ⇒ at t = 2 sec v = 0

ymax

= y0 + ut +

1

2at2 = 105 + 20 × 2 +

1

2(– 10) × 22 = 125 m.

For downward motion :

125 = 1

2 × 10 × t2 ⇒ t = 5 sec

∴ Ball will reach ground at t = 7 sec.

125

02 4 6 7 8 10 12

y

t

105

v = u + at = 0 + 10 × 5 = 50 m/sec

0

– 10

1020304050

– 20– 30– 40– 50

1 2 3 4 5 6 8 9 10 11 12

v(m/sec)

t(sec)

7

∴ Velocity of the ball just before hitting ground is 50 m/secdownwards. or (v = – 50 m/sec)

After hitting ground ball will rebound with same speed upwards orv = 50 m/sec. It will reach the top most point after 5 more sec or at t= 12 sec.

Illustration 19:

Plot x – t and a – t graphs for given v – t graph. It is given that initially particle is at x = 10.

Solution :

For t : [0, 2]

01 2 3 4 5

t

1

2

(sec)

v(m/sec)

v – t graph is straight line with slope = 2

2 = 1 m/s2.

(constant acceleration)

x – t graph is a parabola (mouth opening upwards)

THEORY



x(2) = x(0) + ut + 1

2at2 = 10 + 0 × 2 +

1

2 × 1 × 22 = 12m.

For t : [2, 5]

Velocity remains constant = 2 m/sec ∴ acceleration = 0

x – t graph is a straight line.

x(5) = x(2) + ut + 1

2at2 = 12 + 2 × 3 +

1

2 × 0 × 32 = 18 m.

01 2 3 4 5

a(m/s )2

t

1

2

(sec)

01 2 3 4 5

x

t

5

1012

15

1820

(sec)

Illustration 20:

A car starts moving rectilinearly, first with acceleration a = 5 m/s2 (initial velocity = 0), then uniformly, and finally,decelerating at the same rate ‘a’ comes to a stop. The total time of motion is equal to 25 sec. The average velocity duringthat time is equal to <v> = 72 km/hr. How long does the car move uniformly?

Solution :

Let the particle accelerate for t0 sec with acceleration = 5 m/s2.

Its velocity after t0 sec is v = 0 + 5t

0 = 5t

0.

Then it will move with constant velocity for some time and finally retard for t' sec to come to rest.

While decelerating, 0 = v – 5t' ⇒ 5t0 – 5t' = 0

⇒ t' = t0.

∴ It will retard for t0 sec and hence will move with constant speed for (25 – 2t

0) sec.

Average velocity = 72 × 5

18 = 20 m/sec v

5t0

tt0 25 – 2t0 t0

20 m/sec = Displacement

Total time =

Displacement

25 sec

∴ Displacement = 500 m.

Now Displacement = Area under v – t graph

∴ 500 = 2 0 0

15

2t t × ×

+ 5t

0 × (25 – 2t

0)

t02 – 25t

0 + 100 = 0

⇒ t0 = 5 sec (t

0 = 20 sec is not possible)

∴ Particle moves uniformly for 25 – 2t0 = 15 sec.




1. The figure below shows displacement-time graph of a particle. Find the time interval (starting from t = 0) during motion suchthat the average velocity of the particle during that period is zero.

05 10 15 20 25

x(m)

t

10

20

30

(sec)

2. The displacement of a particle as a function of time is shown in figure.The figure indicates.

20

10

010 20 30 40

Dis

plac

emen

t(a) the particle starts with a certain velocity, but the motion is retarded andfinally the particle stops

(b) the velocity of the particle is constant throughout

(c) the particle starts with a zero velocity, the motion is accelerated.

(d) the acceleration of the particle is constant throughout.

3. For the given position-time (x – t) graph find the interval in which

O t1 t2 t3 t4 t5 t6

t

x

(a) velocity is zero

(b) both velocity and acceleration are negative

(c) velocity is positive but acceleration is negative

(d) velocity is negative but acceleration is positive.

4. The following shows the velocity-time graph for a moving object.The maximum acceleration will be :

(a) 1 m/sec2

v(m

/sec

)

t in sec.

40

60

20

010 20 30 40 50 60 70

D

C

BA(b) 2 m/sec2

(c) 3 m/sec2

(d) 4 m/sec2

5. The following figure shows the linear motion velocity-time graphof a body. The body will be displaced in 5 seconds by:

(a) 2 m

v(m

/sec

) B

2

1

0

– 1

– 2

1 2 3 4 5

D

t in sec.

C

A

E

(b) 3 m

(c) 4 m

(d) 5 m

THEORY



6. Acceleration-time graph of a body is shown. The corresponding velocity-time graph of the same body is :

a

t

v

t

v

t

v

t

v

t

(a) (b) (c) (d)

7. A parachutist jumps out of an aircraft, falls freely for some time and then opens his parachute. Identify the graph whichcorrectly represents his acceleration (a) versus time (t) graph :

t

a

0t

a

0t

a

0t

a

0

(a) (b) (c) (d)

3.1.6 Motion with Non-Uniform Acceleration

Use of definite integral

To find displacement when the particle is moving with a variable velocity in an interval ∆t = t2 – t

1, we can divide this interval into

n subintervals.

i.e., ∆t = t2 – t

1 = ∆t

1 + ∆t

2 + ∆t

3 + ................ + ∆t

n.

As the number of intervals is increased (i.e., n → ∝), each interval becomes infinitely small (∆ti → 0, i = 1, 2, 3 .......) and we can treat

velocity as constant during one such interval.

∴ ∆x = (v1 ∆t

1 + ∆v

2 ∆t

2 + .............. + v

n ∆t

n) as n → ∝ and ∆t

i → 0

∴ ∆x = 1

limn

i ini

v t→∞ =

∆ ∑ = ( )

2

1

t

t

v t dt∫

∴ Displacement in time interval [∆t = t2 – t

1] = ∆x = ( )

2

1

t

t

v t dt∫

Similarly change in velocity ∆v = vf – v

i = ( )

f

i

t

t

a t dt∫



Technique for problems involving non-uniform accele ration

Case 1: Acceleration depends on time.

Steps to find v(t) from a(t)

We have a = dv

dt.

If a is a function of time, then dv = a(t)dt

Integrating both sides with appropriate limits we have ( )

( )

0

v t

v

dv∫ = ( )0

t

a t dt∫ [v(0) = initial velocity at t = 0]

On integrating we get an expression for v(t).

Steps to find x(t) from v(t)

To get x(t), we put v(t) = dx

dt⇒ dx = v(t)dt

Integrating on both sides with appropriate limits we have ( )

( )

0

x t

x

dx∫ = ( )0

t

v t dt∫ (x(0) = position at t = 0)

On integrating we get an expression for x(t).

Case 2: Acceleration depends on position x.

Steps to find v(x) from a(x).

We have a = dv

dt =

dv

dt

dx

dx =

dx

dt

dv

dx = v

dv

dx ⇒ a dx = v dv

On integrating both sides with appropriate units we have 0

x

x

a dx∫ = 0

v

v

v dv∫ [x0 = position when velocity is v

0]

On integrating we get an expression for v(x).

We have v(x) = dx

dt ⇒ ( )

dx

v x = dt

Integrating on both sides with appropriate units we have ( )( )

( )

0

x t

x

dx

v x∫ = 0

t

dt∫


Case 3: Acceleration depends on velocity.

Steps to find v(t) from a(v)

We have a(v) = dv

dt ( )

dv

a v = dt

Integrating on both sides with appropriate limits we have ( )( )

( )

0

v t

v

dv

a v∫ = 0

t

dt∫ .

On integrating we get an expression for v(t).

Steps to find v(x) from a(v)

a(v) = dv

dt =

dv

dt

dx

dx =

dx

dt

dv

dx = v

dv

dx

or a(v) = vdv

dx ( )

vdv

a v = dx

Integrating both sides with appropriate limits we have ( )0

v

v

vdv

a v∫ = 0

x

x

dx∫ [v0 = velocity when x = x

0]


THEORY



Illustration 21:

Acceleration of a particle moving on x-axis depends on time as a = 2t m/sec2. Find velocity and position of particle as afunction of time if the particle is at x = 5 m at t = 0 moving with v = 2 m/sec.

Solution :

Acceleration depends on time [case (a)]

a = 2t = dv

dt∴ dv = 2t dt

Integrating both sides with appropriate units we have 2

v

dv∫ = 0

2t

tdt∫ ⇒ [ ]2

vv = 2

0

tt

⇒ v – 2 = t2

⇒ v = (t2 + 2) m/sec

Now, v = t2 + 2 ⇒ v = dx

dt = t2 + 2

or dx = (t2 + 2)dt On integrating we have

5

x

dx∫ = ( )2

0

2t

t dt+∫ ⇒ [ ]5

xx =

3

0

23

tt

t

+

⇒ x – 5 = 3

3

t + 2t

⇒ x = 3

3

t + 2t + 5

Illustration 22:

Acceleration of particle depends on its positive as a(x) = 2

1

4x + m/s2. Find its velocity as a function of position. If the

particle starts from rest at x = 2m.

Solution :

Acceleration depends on position [case (b)]

a(x) = dv

dt = 2

1

4x +

⇒dx

dt

dv

dx = 2

1

4x + ⇒ v dv = 2 4

dx

x +

On integrating with appropriate units we have

0

v

v dv∫ = 22 4

x dx

x +∫ ⇒2

02

vv

= 1

2

tan2

xx−

⇒2 20

2 2

v − = 1 1 2tan tan

2 2

x− − −

⇒2

2

v = 1tan

2 4

x− π −

⇒ v = 12tan

2 2

x− π −

Illustration 23:

A particle is given an initial speed v0. There is a resisting acceleration k v , where v is instantaneous velocity and k is

some positive constant. Find the time taken to stop and the maximum distance covered by the particle.

Solution :

(a) Stopping time



We have a = dv

dt. As there is resisting acceleration,

dv

dt is negative.

i.e.,dv

dt = –k v ⇒

dv

v = – k dt

On integrating with appropriate limits we have

0

0

v

dv

v∫ = 0

t

k dt− ∫ ⇒0

0

2v

v = [ ]0

tk t−

⇒ 0 – 02 v = – kt

or t = 02 v

k

(b) Stopping distance

a = dv

dt

dx

dx =

dvv

dx

i.e.,dv

vdx

= k v− ⇒v dv

v = – k dx

⇒ v dv = – k dx


0

0

v

v dv∫ = 0

x

k dx− ∫ ⇒3

20

2

3v = kx

⇒ x = 3

202

3

v

k

Illustration 24:

For a particle moving in straight line with constant acceleration if t = x + 3, find the position of the particle when itsvelocity is zero.

Solution :

t = x + 3 ⇒ x = (t – 3)or x = t2 – 6t + 9

Now, velocity of the particle is v = dx

dt = (2t – 6) m/sec. Velocity is zero when 2t – 6 = 0 or t = 3 sec.

∴ Position of the particle when its velocity is zero is x(3) = 32 – 6 × 3 + 9 = 0

i.e., Particle is at origin.

Illustration 25:

Instantaneous velocity of a particle moving in +ve x direction is given as v = 2

3

2x + . At t = 0, particle starts from origin.

Find the average velocity of the particle between the two points P(x = 2) and Q(x = 4) of its path.

Solution :

v< > = average velocity = Displacement

Time Interval = 4 2

Time Interval

−

So, basically we have to find the time interval between the instants when particle is at position Q(x = 4) and position P(x= 2).

Now, v = dx

dt = 2

3

2x + (x2 + 2)dx = 3dt


THEORY



( )4

2

2

2x dx+∫ = 2

1

3t

t

dt∫ or43

2

23

xx

+

= [ ] 2

13

t

tt

342 4

3

+ ×

–

322 2

3

+ ×

= 3(t

2 – t

1)

t2 – t

1 = ∆t =

68

9sec.

∴ v< > = average velocity = 2

t∆ = 2689

= 9

34 m/sec.


1. For a particle moving along x-axis, acceleration is given as a = 2x3. If the speed of the particle is 4 m/sec at x = 0, find speedas a function of x.

2. For a particle moving along x-axis acceleration is given as a = 2v. Find its position as a function of time if at t = 0, it is at x= 0 moving with velocity 5 m/sec.

3. An object moves along the x-axis such that is acceleration is given as a = 3 – 2t. Find the initial speed of the object suchthat the particle will have the same x-coordinate at t = 5.0 s as it had at t = 0. Also find the object’s velocity at t = 5.0 s.

4. Velocity of a particle moving along x-axis is given as v = x2 – 5x + 4 (in m/s) where x denotes the position of the particle inmeters. The magnitude of acceleration of the particle when the velocity of the particle is zero is :

(a) 0 m/s2 (b) 2 m/s2 (c) 3 m/s2 (d) None of these.

5. A particle moves according to the equation t = ax2 + bx, then the acceleration of particle when x = b

a is

(a) 3

a

b− (b) 3

2

8

a

b− (c) 3

2

27

a

b− (d) None of these.

Section - 3.2 Motion in Two and Three Dimension

Consider a particle which move along the curve in the plane of the paper. To locate its position we must specify the origin (thereference point). Here, we have choosen O as the origin.From point O, we draw two perpendicular lines. One of the lines is calledthe x-axis and the other is called the y-axis. If particle is at point P at time t, we have two ways of specifying its position.

1. Rectangular coordinate

( , )x yP

O M

N

x

y

r

θ

From point P we draw two perpendicular segment, PM and PN which areperpendicular to x and y axis respectively. OM and ON are called the x and ycoordinates respectively.

2. Polar coordinates ( r, θ )

Another way to locate the position P uniquely is to specify, the distance ofP from O and angleθ which OP makes with the x-axis.

It is clear from the figure that polar and rectangular coordinates are relatedby the equations

x = r cosθ

y = r sinθ

or, r = 2 2x y+ , θ = tan–1(y/x)



Position vector rr

( ) OPr t =uuurr

is called the position vector of the particle. From the construction in figure, it is clear that OP OM MP= +uuur uuuur uuur

.

If i and j are unit vector (of magnitude unity) along x & y direction then ˆOM xi=uuuur

and ˆMP y j=uuur

.

⇒ ˆ ˆ( ) OP ( ) ( )r t x t i y t j= = +uuurr

The distance of the particle from the origin is

2 2( ) ( )x t y t r+ = r

Displacement

As the particle moves , its position vector changes in such a way that it always extends from the origin to the particle. If particle

is at position 1rr

at time 1t and at 2rr

at time 2t , then its displacement during the time interval is

2 1r r r∆ = −r r r———— (1)

For example, if the particle is at 1

ˆ ˆ2 3r i j= +r at time t = 0 at 2 4 5r i j= +

r rr at t = 1 sec, then its displacement is

2 1ˆ ˆ2 3r r r i j∆ = − = +r r r

.

The magnitude of displacement is 2 22 3 13+ = units.

Velocity

If a particle undergoes a displacement r∆r in time t∆ , then average velocity is

ˆ ˆr x yv i j

t t t

∆ ∆ ∆< >= = +∆ ∆ ∆

rr

———— (2)

The instantaneous velocity vr

is the value <vr

> approaches in the limit as we approach 0t∆ → .

0 0lim limt t

r drv v

t dt∆ → ∆ →

∆= < > = =∆

r rr r

It can be also written as the derivative

( )ˆ ˆ ˆ ˆd dx dyv xi y j i j

dt dt dt= + = +r

———— (3)

which can be rewritten as ˆ ˆx yv v i v j= +r

in which x

dxv

dt= and y

dyv

dt= are called the x-components and y-components of the velocity..

The magnitude of vr

is called the speed of the particle.

Graphical Approach

P

Q

Tangent at P

1r

2r

r∆

Figure shows the path of a particle moving in the plane of the paper. Let the particle

be at P at time t and, at Q at time t t+ ∆ . The position vector at P and Q is 1rr

and

2rr

respectively. The displacement of the particle PQuuur

is 2 1r r r∆ = −r r r. The average

velocity <vr

> is given by eqn (2). It is in the same direction as r∆r . As we approach

t∆ towards zero, the position vector 2rr

moves towards 1rr

so that r∆r approach

towards zero. The direction of r∆r approaches the direction of tangent line. The

average velocity r

vt

∆< > =∆

rr

approaches the instantaneous velocity vr

. In the limit

0t∆ → , we have v v< > =r r and most importantly avgv

rtakes the direction of tangent

line.

Acceleration

If the velocity vr

of a particle changes from 1vr

to 2vr

in a time interval t∆ , its average acceleration a< >r during t∆ is

2 1v v va

t t

− ∆< > = =∆ ∆

r r rr

THEORY



The instantaneous acceleration ar

, is the value a< >r approaches in the limit as we shrink t∆ .It can be written as the derivative.

( )ˆ ˆ ˆ ˆ ˆ ˆyxx y x y

dvdvda v i v j i j a i a j

dt dt dt= + = + = +r

in which xx

dva

dt= and

yy

dva

dt= are called the x-component and y-component of acceleration respectively..

Illustration 26:

A particle moves in the x – y plane in such a way that its x(t) and y(t) coordinates are given by x(t) = – 5t2 + 20 meters andy(t) = 2t + 3 meters. Find

(a) Average velocity in the time interval t = 0 to t = 2 sec.

(b) Average velocity in the time interval t = 0 sec to t = 3 sec.

(c) Instantaneous velocity at t = 2 sec and t = 3 sec.

(d) Average acceletation in the time interval t = 0 to t = 3 sec.

(e) Instantaneous acceleration at time t = 2 sec and t = 3 sec.

Solution :

The position vector of the particle at time t is ˆ ˆ( ) ( ) ( )r t x t i y t j= +r = 2 ˆ ˆ( 5 20) (2 3)t i t j− + + +

(a) The average velocity is given by 2 1

2 1

( ) ( )r t r tv

t t

−< > =

−

r rr

( ) ˆ ˆ0 20 3r i j= +r and ( ) ( )2 ˆ ˆ2 5 2 20 2 2 3r i j= − × + + × +r ˆ ˆ ˆ0 7 7i j j= + =

( )ˆ ˆ ˆ7 20 3ˆ ˆ10 2

2 0

j i jv i j

− +< > = = − +

−r

m/s and 104v< > =rm/s

The angle θ with the x-axis is 1 1 1θ tan tan

5y

x

v

v− −< > = = − < >

(b) Here, we have 1 0t = s and 2 3t = s.

Hence,(3) (0)

3 0

r rv

−< > =−

r rr ( ) ( )ˆ ˆ25 20 9 3

3 0

i j− − + −=

−ˆ ˆ15 2i j= − +

(c) To find the instantaneous velocity, we differentiate ( )r tr

with respect to time. On differentiating, we have

ˆ ˆ10 2dr

v t i jdt

= = − +r

r

Hence, ˆ ˆ ˆ ˆ( 2) 10 2 2 ( 20 2 )v t i j i j= = − × + = − +r

ˆ ˆ ˆ ˆ( 3) 10 3 2 ( 30 2 )v t i j i j= = − × + = − +r

(d) 2 1

2 1

v va

t t

−< > =−

r rr

( ) ( ) ˆ ˆ ˆ3 0 30 2 2 ˆ103 0 3

v t v t i j ja i

= − = − + −< >= = = −−

rr

m/s2

The average acceleration in time interval t = 0 to t = 3 is 10 m/s2 and its direction is in the negative x-direction.

(e) To find the instantaneous acceleration, we differentiate vr

with respect to time.

ˆ ˆ ˆ( 10 2 ) 10dv d

a ti j idt dt

= = − − = −r

rm/s2 (constant)



Illustration 27:

Figure below shows a particle, moving with constant speed v0 m/s in a circle of

radius R meters.

(a) Find the instantaneous velocity at point A, B, and C. x

y

B

R

AC

D

O(b) Find the displacement , average velocity , average acceleration in the timeinterval when particle moves from A to B.

(c) Find the same quantities when particle moves from point A to point C.

Solution :

(a) As we know the speed (magnitude of velocity) of the particle we only need to the direction of velocity, which istangent to the curve along which particle moves.

Thus, A 0

ˆv v j=rm/s,

B 0ˆv v j= −rm/s and

C 0ˆv v j= −rm/s

(b) The distance travelled by the particle from A to B is 2πR πR

4 2= meters

Thus, the time it will take from A to B = 0

πR

2v sec

displacement = A B B Aˆ ˆR Rr r r j i→∆ = − = −

2 2A B R R 2r →∆ = + = R meters

0 0B A

0

ˆ ˆ 2 2R R ˆ ˆtime taken πR / 2 π π

v vr r j iv j i

v

− −< > = = = −r r

rm/s

0 0B A

0

ˆ ˆ

time taken πR / 2

v i v jv va

v

− −−< >= =

r rr

= 202 ˆ ˆ( )

πR

vi j− + m/s

(c) We leave it as an exercise for you. Verify that

A Cˆ2Rr i→∆ = − , 02 ˆ

π

vv i

−< > =r

m/s and 202 ˆ

πR

va j

−< > =r

m/s

Illustration 28:

Position vector of a point relative to origin varies with time t as 2ˆ ˆ,r ati bt j= −rwhere a and b are positive constants. Find

the equation of the point’s trajectory.

Solution :

It is given that 2ˆ ˆ,r at i bt j= −r therefore x = at and y = – bt2

Eliminating t, we get

y = – b2

x

a

y = 2

2

bx

a−

3.2.1 Motion in Two Dimension with Constant Accele ration

We consider the special case of motion in a plane (two dimension) with constant acceleration. As the particle moves, theacceleration a

rdoes not vary either in magnitude or direction. Consequently, the components of acceleration a

x and a

y remain

constant. We then have a situation which can be described as the sum of two component motion occurring simultaneously withconstant acceleration along each of two perpendicular directions independently.

THEORY



The equations

v = u + at

s = ut + 1

2at2

v2 – u2 = 2a s

can be applied separately for x-axis and y-axis

The two sets of equations are

For motion along x-axis For motion along y-axis

vx = u

x + a

xt v

y = u

y + a

yt

x – x0 = u

xt +

1

2a

xt2 y – y

0 = u

yt +

1

2a

yt2

vx2 – u

x2 = 2a

x(x – x

0) v

y2 – u

y2 = 2a

y(y – y

0)

Illustration 29:

For a particle moving in the x - y plane, if ˆ ˆ(3 2 )a i j= −rm/s2 and ˆ ˆ(2 4 )mv i j= +r

m/s, then find position of particle when its

y-coordinate is maximum.

Solution :

We have, ax = 3 m/s2, a

y = – 2 m/s2 and u

x = 2m/s, u

y = 4 m/s.

therefore, vy = u

y + a

yt = (4 – 2t) and y = u

yt +

1

2a

yt2 = 4t – t2

Similarly vx = u

x + a

xt = 2 + 3t and x = u

xt +

1

2a

xt2 = 2t +

3

2t2

When y is maximum, dy

dt = 0

⇒ 4 – 2t = 0 ⇒ t = 2s.

At t = 2s, x = (2 × 2) + 23

22 ×

= 10 m and y = (4 × 2) – (2)2 = 4 m.

Hence, position vector of the particle is ˆ ˆ ˆ ˆ(10 4 )r xi yj i j= + = +rm


1. A particle travels with speed 50 m/s from the point (3, – 7) in direction ˆ ˆ7 24i j− . The position vector of particle after 3

seconds will be ............ .

2. The position vector rr

of a moving particle at time t after the start of the motion is given by rr

= (5 + 20t) i + (95 + 10t –

5t2) j . At time t = T, the particle is moving at right angles to tis initial direction of motion. The value of T is ............. andthe distance of the particle from its initial position at this time is ............ .

3. Read the passage carefully and answer the following questions.“In a three dimensional co-ordinate system a particle moves such that during motion at time t the position vector of aparticle of mass m = 3 kg is given by r

r = 6t i – t3 j + cos t k ”



(a) The net force acting on the particle is :

(i)– 18t j – 3 sint k (ii) – 12t2 j – 3 cost k (iii) – 18t2 j – 3 cost k (iv) None of these

(b) At t = 2

π the acceleration of the particle is :

(i)π (ii) 2π (iii) 3 π (iv) None of these.

(c) At t = π the speed of particle is :

(i) 23 4 π+ (ii) 23 9 π+ (iii) 25 4 π+ (iv) None of these.

Section - 3.3 Projectile Motion

The most common example of a particle moving with (nearly) uniform acceleration is the motion of particle near the surface ofearth . Such a motion is called projectile motion .

We know that when a particle is given an initial velocity in the verticaldirection, it moves along a vertical line. Suppose the particle is projected

with initial velocity ur

which is directed in direction which makes a certainangle with the horizontal. Such a particle has two dimensional motionand will move in a curve as we know from experience v0

Figure below illustrates the motion of a particle (say stone) with initialvelocity v

0 at angle θ above the horizontal.

Note :

• The acceleration of the particle is constant. Its magnitude is g = 9.8 m/s2, and direction of acceleration is directed alongthe vertical downward direction.

• Note that velocity vector is changing with time both in magnitude and direction. The direction of velocity is always alongthe tangent to the curve. There is no fixed relation between the direction of velocity and acceleration.

• The acceleration being in the vertical direction, the horizontal component of velocity remains constant.

• When the particle is at the highest point, velocity is directed towards the horizontal. This means that at the highest point,vertical component of velocity = 0.

• When particle again hits the ground its y-coordinate is zero.

Analysis of a Projectile

The motion of a particle as projectile can be imagined as being made up of two parts : Horizontal and Vertical which areindependent of each other. We can directly apply all 3 equations of motion which we used in the study of uniformly acceleratedstraight line motion to horizontal and vertical components separately.

Consider a particle thrown from origin with speed u making anangle θ with horizontal direction. Here we choose x-axisalong horizontal and y-axis along vertical.

(vertical direction)

x0

ug

gg g g

g

g

y

(Horizontal direction)

During whole motion of the particle its acceleration in ‘g’

downwards. i.e., a = – g j

Vertical component of initial velocity = (u sinθ) j

x

u

y

u sinθ u

u cos

vy

u cos

u cos

u cos u cos

u cos

vy

vy

vy

As the particle moves up, its vertical component ofvelocity decreases and finally becomes zero at the topdue to acceleration ‘g’ downwards.

THEORY



The horizontal component of velocity remains constanti.e., u

x = u cosθ as there is no component of acceleration

along the horizontal.

As the particle comes downwards from topmost point, the verticalcomponent becomes –ve and increases in magnitude.

x0

u

y

( , )x y

Originx

y

Position vector at time ( ) = t

For Horizontal Component of Motion For Vertical Comp onent of Motion

ax = 0 a

y = – g

vx = u cosθ (constant) v

y = u sinθ – gt

x-coordinate at time t is y-coordinate at time t is

x = vx0t +

1

2a

xt2 y = v

y0t +

1

2a

yt2

= (u cosθ)t = (u sinθ)t – 1

2gt2

Equation of trajectory

Assuming that particle is at origin at t = 0.

We have x = (u cosθ)t or t = cos

x

u θ

y = (u sinθ)t – 1

2gt2 = (u sinθ)

cos

x

u θ – 2

1

2 cos

xg

u θ

y = x tanθ – 2

2 2

1

2 cos

xg

u θ

(a) Projectile Fired from Ground

Time of flight (T)

Time interval for which particle remains in air.

u

R

Hmax

When the particle returns back to ground its verticaldisplacement is zero.

Lets assume that the particle returns to ground at t = Twhen it was projected at t = 0.

Then putting y = 0 in y = (u sinθ)t – 1

2g t2 we get

0 = (u sinθ)T – 1

2gT2 T =

2 sinu

g

θ =

2u

a

⊥⊥

Range (R)

Horizontal distance covered by the projectile.



Range (R) = (Horizontal Speed) × (Time of flight) = (u cosθ) × 2 sinu

g

θ

∴ Range (R) = 22 sin cosu

g

θ θ =

2 sin 2u

g

θ

1. R = 22 sin cosu

g

θ θ =

22 cos sin2 2

u

g

π π − θ − θ

=

22 sin cos2 2

u

g

π π − θ − θ

∴ For same speed of projection.

Range for angle of projection (θ) = Range for angle of projection 2

π − θ

2. Range is maximum for angle of projection = 45°

3. Equation of trajectory y = x tanθ – 2

2 2

1

2 cos

gx

u θ = x tanθ

2 2

1sin

2 coscos

gx

u

−

θ θ θ

= x tanθ 1R

x −

Maximum Height (H max)

When particle is at maximum height its vy = 0

∴ Applying v2 = u2 + 2as for vertical direction. we get

02 = (u sinq)2 + 2 × (– g) × Hmax

or Hmax

=

2 2sin

2

u

g

θ

At Hmax

particle reaches after time = sinu

g

θ from the time of projection and has only horizontal speed equal to u cosθ.

Illustration 30:

A particle is projected from ground with speed 502 m/sec at an angle of 45° with horizontal. Find its speed at a height70 m above ground. (Take g = 10 m/s2)

Solution :

Let v be the speed of the particle at height 70 m above ground.

If vx and v

y are horizontal and vertical components of velocity at height 70 m above

ground then v = 2 2x yv v+ vy

70 m

v = ??

250Now, vx = 50 2 cos 45° = 50 m/sec (constant)

For vy we can apply v2 = u2 + 2as along vertical.

i.e., vy2 = (u sinθ)2 – 2 × 10 × 70 = ( )50 2 sin 45° – 1400 = 1100 (m/sec)2

∴ Speed v = 2 2x yv v+ = ( )2

50 1100+ = 3600 = 60 m/sec

THEORY



Illustration 31:

A particle is projected with speed 60 m/sec at an angle of 60° with horizontal. Find position vector and velocity vector ofthe particle 8 sec after the time of projection. Assume x-axis along horizontal, y-axis along vertical and origin as point ofprojection. (g = 10 m/s2)

Solution :

Let ˆ ˆr xi yj= + be the position vector and ˆ ˆx yv v i v j= +

be the velocity vector of particle 8 sec after time ofprojection.

For position vector :

x

y ( , )x yv

r60 m

/sec

60°x = (u cosθ) × t = (60 cos 60°) × 8 = 240 m.

y = (u sinθ)t – 1

2gt2 = (60 sin 60°) × 8 –

1

2× 10 × 82 = 240 3 – 320 = 80(3 3 – 4) m

∴ Position vector ˆ ˆr xi yj= + = ( )ˆ ˆ240 80 3 3 4i j+ − meter

For velocity vector

vx= u cosθ = 60 cos 60° = 30 m/sec

vy= (u sinθ) – gt = (60 sin 60°) – 10 × 8

= (30 3 – 80) m/sec

∴ ( )ˆ ˆ30 30 3 80v i j= + − m/sec.

Illustration 32:

In the figure shown two projective are fired simultaneously. Whatshould be the initial speed u of the right side projectile for the twoprojectiles to collide in mid-air? Also find time when they collide.

(g = 10 m/s2)

u

20 m/se

c

60°45°

10 m

Solution :

Consider origin at point of projection of left side projectile, x-axis along horizontal and y-axis along vertical.

Let us assume that particles collide at point (x, y).

∴ Equating y-coordinate for both the particle we have

y = (20 sin 45°)t – 1

2gt2 = (u sin 60°)t –

1

2gt2

⇒ 20 × 1

2t = u ×

3

2t ⇒ u = 20

2

3 m/sec

For time : Let it be ‘T’. Equating x-coordinate for both the particles we have

(u cos 60°)T + 10 = (20 cos 45°)T

On solving T = 3

6 2−sec

3.3.1 Projectile Fired from a Height



Case 1.

Again we can break the motion its two components:

Range (R)

H

u

Along Horizontal Along Vertical

acceleration = 0 acceleration = g (downwards)

velocity = vx = u (constant) Initial velocity = 0

vy(t) = 0 + gt = gt

Time of flight (T)

Time for which particle remains in air.

Particle will remain in motion until its vertical displacement becomes ‘H’.

∴ Along vertical – H = 0 × T – 1

2gT2 or T =

2H

g

Range (R)

Displacement along horizontal direction till the time particle is flying.

Range (R) = (Horizontal velocity) × (Time of flight)

= u2H

g

Case 2.

Range (R)

H

u

Again break the motion along horizontal and vertical components.

Time of flight ( T)

Vertical displacement = – H

∴ – H = (u sinθ) T – 1

2gT2 (solving it we can get T)

For Range (R) : Range = (u cosθ) (Time of flight)

Hmax

above ground =

2 2sin

2

u

g

θ + H.

Case 3.

Time of flight H

u

Range (R)

– H = (– u sinθ)T – 1

2gT2

For Range (R) : Range = (u cosθ) (Time of flight)

Illustration 33:

From the top of a 80 m high tower a stone is projected with speed 50 m/sec,at an angle of 37° as shown. Find

(a) Speed after 2 sec

Range (R)

80 m

37°50

m/se

c

(b) Time of flight

(c) Maximum height above ground level attained by the particle

(d) Horizontal range

(e) Speed just before striking the ground. (Take g = 10 m/s2)

THEORY



Solution :

Initial velocity in horizontal direction = 50 cos 37° = 40 m/sec

Initial velocity in vertical direction = 50 sin 37° = 30 m/sec

(a) Horizontal component of velocity after 2 sec = 40 m/sec

Vertical component of velocity after 2 sec = 30 – 10 × 2 = 10 m/sec

∴ Speed = 2 2x yv v+ = ( ) ( )2 2

40 10+ = 10 17 m/sec

(b) Vertical displacement = – 80

∴ – 80 = 30 T – 1

2 × 10 T2. On solving T = 8 sec

(c) Max. height above level of projection = 2 2sin

2

u

g

θ =

( )230

20 = 45 m

∴ Max height above ground level = 80 + 45 = 125 m.

(d) Horizontal Range = (Horizontal speed) × Time of flight

= 40 × 8 = 320 m.

(e) Vertical velocity just before hitting vy

= uy + at

= 30 – 10 × 8

= – 50 m/sec.

∴ speed just before hitting = 2 2x yv v+ = ( ) ( )2 2

40 50+ − = 10 41 m/sec

Illustration 34:

H

30°

60°

vB

vT

100 m

200 3

Shots fired simultaneously from the top and floor of a vertical cliffat elevations of 30° and 60° respectively, strike an objectsimultaneously which is at a height of 100 meters from the ground

and at a horizontal distance of 2003 meters from the cliff. Find theheight of the cliff, the velocities of projection of the shot and thetime taken by the shots to hit the object.

(g = 10 m/sec2)

Solution :

Let T = Time taken by shots to hit the object

vT = Velocity of projectile fired from top

vB = Velocity of projectile fired from bottom

H = Height of cliff.

For shot fired from the bottom :

(vB cos 60°) T = 200 3

or vB T = 400 3 –––––– (1)

100 = (vB sin 60°)T –

1

2 × 10T2 ⇒ 100 =

3

2(v

B T) – 5T2 –––––– (2)

Substitutions (1) in (2) we get

100 = 3

2(v

B T) – 5T2 ⇒ T = 10 sec



∴ vB =

400 3

10 = 40 3 m/sec

For shot fired from the top :

(vT cos 30°)T = 200 3 ⇒ v

T ×

3

2 × 10 = 200 3 . ∴ v

T = 40 m/sec

Vertical displacement = 100 – H = (vT sin 30°) × 10 –

1

2 × 10 × (10)2

On solving H = 400 m.

Illustration 35: v1v2

g

Two particles move in a uniform gravitational field with anacceleration g. At the initial moment of time, the particles werelocated at one point and moved with velocities 3 m/sec and 4 m/sechorizontally in opposite directions. Find the distance between themat the moment when their velocity vector become mutuallyperpendicular.(g = 10 m/s2).

Solution :

Let us assume origin at the point of projection and x – y axes as shown in the figure.

Initial velocity of ball (1) is 1

ˆ3v i= ,

Initial velocity of ball (2) is 2

ˆ4v i= −Velocity vector of ball (1) and (2) at some time (t) are

v1(t) = ˆ ˆ3i gt j− = ˆ ˆ3 10i t j− and v

2(t) = ˆ ˆ4i gt j− − = ˆ ˆ4 10i t j− −

When these vectors become perpendicular 1 2. 0v v =

( ) ( )ˆ ˆ ˆ ˆ3 10 . 4 10i t j i t j− − − = 0 t = 12

10 sec

Now we have find distance between them at t = 12

10 sec

x1(t) = 3t x

2(t) = – 4t

y1(t) = 0 × t –

1

2 × 10 × t2, y

2(t) = 0 × t –

1

2 × 10 × t2

∴ Distance between them is ( ) ( )( ) ( ) ( )( )2 2

1 2 1 2x t x t y t y t− + −

= ( )2 27 0t + = 7t = 7 12

10 m

Illustration 36:

From a point on the ground at a distance 12 m from the foot of a vertical wall, a ball is thrown at an angle of 45° which justclears the top of the wall and afterwards strikes the ground at a distance 6 m on the other side. Find the height of the wall.

Solution :

Range = 12 m + 6 m = 18 m.

45°

h

12 m 6 m

Consider origin at the point of projection. The ball passthrough the coordinate (12, h).

So, this point must satisfy e.g. of trajectory.

y = x tanθ 1R

x −

h = 12 tan 45°12

118

−

= 4 m

THEORY



Illustration 37:

Two seconds after projection, a projectile is moving at an angle 30° above horizontal. After one more second it is movinghorizontally. Find its initial velocity of projection. (g = 10 m/s2)

Solution :

Let initial velocity of projection = u at an angle ‘θ’ with horizontal.

Initially horizontal component = u cosθInitially vertical component = u sinθ.

After 2 sec horizontal component = u cosθAfter 2 sec vertical component = u sinθ – gt = u sinθ – 10 × 2 = u sinθ – 20

After 2 sec velocity vector make an angle of 30° with horizontal.

∴ tan 30° = 1

3 =

y

x

v

v =

sin 20

cos

u

u

θ −θ

or u cosθ = (u sinθ – 20) 3 ––––– – (1)

After 3 sec. from the time of projection, the particle is moving horizontally, so total time of flight = 6 sec

i.e.2 sinu

g

θ =

2 sin

10

u θ = 6 sec ⇒ u sinθ = 30 m/sec

Putting u sinθ = 30 m/sec in eq. (1) we get u cosθ = 10 3 m/sec

∴ Initial velocity of projection = 2 2x yv v+ = ( ) ( )2 2

cos sinu uθ + θ

= ( ) ( )2 2

10 3 30+

= 20 3 m/sec

3.3.2 Projectile on an Inclined Plane

Case 1.

Particle is projected up the incline

u

Hmax

y

x

Let a particle be thrown with speed u at an angle a with inclined plane ofinclination b. Assuming x-axis parallel to the plane and the y-axisperpendicular to the plane, we will find range, time of flight and maximumheight.

Component of g are g cosβ and g sinβ, perpendicular and along the inclined plane.

x

y

g sin

g

g cos

i.e., ax = – g sinβ u

x = u cosα

ay = – g sinβ u

y = u sinα

Time of flight (T) = when the particle strikes the inclined plane its y-coordinate becomes zero

y = uy t +

1

2a

yt2

⇒ 0 = (u sinα)T – 1

2(g cosβ)T2

T = 2 sin

cosu

g

αβ

= 2u

a

⊥⊥



Maximum Height (H max)At y = H

max, v

y = 0

vy2 = u

y2 + 2a

ys

y ⇒ 02 = (u sinα)2 – 2 × g cosβ × H

max ⇒ H

max =

2 2sin

2 cos

u

g

αβ

Range (R)

When the particle strikes the inclined plane x-coordinate is equal to range of the particle.

x = uxt +

1

2a

xt2

R = (u cosα)2 sin

cos

u

g

α β

– 1

2(g sinβ)

22 sin

cos

u

g

α β

R = ( )2

2

2 sin cos

cos

u

g

α α + ββ Maximum Range at

4 2

π β α = −

Case 2.

Particle is projected down the incline

u

Hmax

y

x

ax = g sinβ u

x = u cosα

ay = – g cosβ u

y = u sinα

Time of flight (T)

y = uyt +

1

2a

yt2

0 = (u sinα)T – 1

2(g cosβ)T2

⇒ T = 2 sin

cos

u

g

αβ

Maximum Height (H max)

vy = 0 at H

max.

vy2 = u

y2 + 2a

ys

y⇒ 02 = u2 sin2α – 2g cosβ H

max⇒ H

max =

2 2sin

2 cos

u

g

αβ

Range (R)

Range = x-coordinate at the time of strike.

x = uxt +

1

2a

xt2 ⇒ R = (u cosα)

2 sin

cos

u

g

α β

+ 1

2g sinβ

22 sin

cos

u

g

α β

⇒ R = ( )2

2

2 sin cos

cos

u

g

α α −ββ

Maximum Range at α = 4 2

π β+

Illustration 38:

On an inclined plane of inclination 30°, a ball is thrown at an angle of 60° with the horizontal from the foot of the incline witha velocity of 10 3 m/sec. Find

(a) Time of flight (b) Maximum height above inclined plane.

(c) Range (d) Time after which velocity vector is horizontal. (Take g = 10 m/s2)

THEORY



Solution :

(a) Time of flight = 2 sin30

cos30

u

g

°° = 2 sec

x

Hmax

30°30°

y

10 cos 30°3

10sin 30°

3 10 3

(b) Hmax

= 2 2sin 30

2 cos30

u

g

°°

= 5 3

2 m

g cos 30°

g sin 30°

g

(c) Range = (10 3 cos 30°) × 2 – 1

2 × (g sin 30°) × 22 = 20 m.

(d) It can also be seen as a projectile fired from ground with speed 103 m/sec and an angle 60° with horizontal.

So, velocity vector will become horizontal at half the time of flight = 1 2 sin 60

2

u

g

°

= 1 2 10 3 3

2 10 2

× ×

= 1.5 sec

Illustration 39:

A particle is projected from the foot of the inclined plane at an angle a with inclined plane of inclination β. If the particlestrikes the plane perpendicularly then prove that cotβ = 2 tanα.

Solution :

Let O be the point of projection initial velocity u and P be the point of strike.

If the particle strikes the inclined plane perpendicularly it means x-component of velocity at the time of strike is zero.

Particle will strike after t = 2 sin

cos

u

g

αβ

u

yx

O

PApplying vx = u

x + a

xt we get

0 = u cosα – g sinb × 2 sin

cos

u

g

αβ

or cosα = 2sin sin

cos

β αβ cotβ = 2 tanα

Illustration 40:

Two inclined planes OA and OB having inclinations 30° and 60° withhorizontal respectively intersect each other at O. A particle projectedperpendicularly from point P with speed u = 10 3 m/sec strikes the planeOB perpendicularly at Q. Calculate

(a) Time of flight.

(b) Speed with which particle strikes the plane OB.

(c) Distance PQ. Take (g = 10 m/s2)

A

P

u

60°30°

O

90°

B90°

Q

Solution :

Let u be the initial speed of projection, v be the final speed and t be the time of flight.

y

xAcceleration along horizontal is zero. Therefore, equating thehorizontal component of u and v we get

u cos 60° = v cos 30°



∴ v = cos60

cos30

u °° =

10 3cos60

cos30

°°

= 10 m/sec.

60°30° u cos 60°

u sin 60°

P

u

B

60°30°

vv sin 30°

v cos 30°

60°30°O

A

Q

Initial velocity along y-direction = u sin 60° = 10 3 × 3

2 = 15 m/sec

Final velocity along y-direction = – v sin 30° = –10 × 1

2 = – 5 m/sec.

Acceleration along y-direction = – g = – 10 m/s2.

Applying vy = u

y + a

yt for time of flight we have – 5 = 15 – 10 × T ⇒ T = 2 sec

Now PQ = 2 2OP OQ+For the projectile, Horizontal displacement = OP cos 30° + OQ cos 60°

= u cos 60° × Time of flight

∴ OP3

2 +

OQ

2 = 10 3 ×

1

2 × 2 ⇒ OP 3 + OQ = 20 3 ––––––– (1)

Vertical displacement = OQ sin 60° – OP sin 30° = vyT +

1

2a

yT2

⇒ OQ3

2 –

OP

2 = (u sin 60°) × 2 –

1

2 × 10 × 22

⇒ OQ 3 – OP = 20 ––––––– (2)

From (1) and (2) we have OP = 10 m and OQ = 103 m

∴ PQ = 2 2OP OQ+ = ( ) ( )2210 10 3+ = 20 m


1. Fill in the blanks

(a) A particle is projected with an initial speed u and at an angle θ with the horizontal. The average velocity of a projectilebetween the instants it crosses half the maximum height is ............. .

(b) The average speed of a projectile is ............. (more/less) than its average velocity.

(c) At the highest point of projectile, the angle between its acceleration and velocity is ............ .

2. A boat is moving directly away from a gun on the shore with speed v1. The gun fires a shell with speed v

2 at an angle of

elevation α and hits the boat. The distance of the boat from the gun at the moment it is fired is :

(a)22 sinv

g

α(v

2 cosa – v

1) (b)

2 sinv

g

α(v

2 cosa – v

1) (c)

22 sinv

g

α(v

2 cosa + v

1) (d) None of these

3. A staircase contains three steps each 10 cm high and 20 cm wide(figure). What should be the minimum horizontal velocity of a ballrolling off the uppermost plane so as to hitdirectly the lowest plane :

(a) 1 m/s (b) 2 m/s (c) 3 m/s (d) None of these

4. A ball is thrown from ground level so as to just clear a wall 4 m high at a distance of 4 m and falls at a distance of 14 m fromthe wall. Find the speed of projection.

5. Read the passage carefully and answer the following questions.

“A rocket is initially at rest on the ground. When its engines fire, the rocket flies off in a straight line at an angle of 53° abovethe horizontal with a constant acceleration of magnitude g. The engines stop at a time T after launch, after which the rocketis in projectile motion. You can ignore air resistance and assume g is independent of altitude.”

THEORY



(a) The maximum altitude reached by the rocket is :

(i) (7/8)gT2 (ii) (18/25)gT2 (iii) (7/25)gT2 (iv) None of these

(b) The horizontal distance from the launch point to where the rocket hits the ground is :

(i) (3/2)gT2 (ii) (4/5)gT2 (iii) (5/6)gT2 (iv) None of these

6. Two towers AB and CD are situated at a distance d apart as shown infigure. AB is 20 m high and CD is 30 m high from the ground. An object isthrown from the top of AB horizontally with a velocity of 10 ms– 1 towardsCD. Simultaneously, another object is thrown from the top of CD at anangle of 60° to the horizontal towards AB with the same magnitude of theinitial velocity as that of the first object. The two objects move in thesame vertical plane and collide in mid-air.

The distance d between the towers is ......... .

60°

2 m

mA

B D

C

d

7. A building 4.8 m high 2b meters wide has a flat roof. A ball is projected from a point on the horizontal ground 14.4 m awayfrom the building along its width. If projected with velocity 16 m/s at an angle of 45° with the ground, the ball hits the roofin the middle, find the width 2b. Also find the angle of projection so that the ball just crosses the roof if projected with

velocity 10 3 m/s. (g = 10 m/s)

8. A particle is projected with speed u and an angle a with horizontal. Find the time after which its velocity vector will becomeperpendicular to initial velocity vector.

Section - 3.4 Relative Motion in One and Two Dimensi on

Suppose, Rahul standing on the highway (frame A) sees car P, moving towards north at a speed of 60 km/hr. There is another carQ following this at a speed of 50 km/hr.

1. According to the observation of A (frame A). Two cars Q and P are moving due north at a speed of 50 km/hr and 60km/hr.

2. According to the observation of a passenger in car Q (frame Q), car P moves due north at 10 km/hr and Rahul moves duesouth at a speed of 50 km/hr.

3. According to the observation of passenger in car P, Rahul moves due south with a speed of 60 km/hr while car Q moves duesouth with speed of 10 km/hr.

We notice that velocity of an object depends on the reference frame of the person who is doing the measurement

Let x-axis point along the north. Then, in symbolic notation, we can write.

1. vQ/A

= 50 km/hr; vP/A

= 60 km/hr

2. vP/Q

= 10 km/hr; vA/Q

= – 50 km/hr

3. vA/P

= 60 km/hr; vQ/P

= – 10 km/hr

in which vp/q

= velocity of p with respect q.

Following two identities can easily be established

vP/A

= vP/Q

+ vQ/A

vQ/A

= – vA/Q

In general when P/Qvr

and Q/Avr

are not in the same direction, we have

P/A P/Q Q/Av v v= +r r r––––––– (1)

Q/A A/Qv v= −r r––––––– (2)

Differentiating the eqn (1) and eqn (2) with respect to time yields

P/A P/Q Q/Aa a a= +r r r––––––– (3)



Q/A A/Qa a= −r r––––––– (4)

Note that if Q/Avr

is constant then , eqn (3) reduces to

P/A P/Qa a=r r

Illustration 41:

Rain drop are falling vertically downward with a speed of 3 m/s. A car moves due east at a speed of 4 m/s. Find the velocityof rain drops with respect to car.

Solution :

Let us choose x-axis along the direction of velocity of car and let y-axis point vertically upwards.

We have

4 m/s

3 m/s

x

y

C/Gˆ4v i=r

R/Gˆ3v j= −r

in which G→ ground frame R → Rain drops

C→ frame fixed to the car

We know (see eqn (1))

R/G R/C C/Gv v v= +r r r

⇒ R/C R/G C/Gˆ ˆ3 4v v v j i= − = − −r r r

2 2R/C 3 4 5v = + =r m/s

and angle θ with the vertical is θ = tan–1(4/3) ≅ 53°

θ

y

x

Illustration 42:

A motor boat going downstream in a river overcome a raft at a point A. After one hour it turned back and after some timepassed the raft at point B at a distance equal 8 km from point A . Find the flow velocity of the river assuming the duty ofthe engine to be constant.

Solution :

The velocity of the raft is equal to the flow velocity as it moves with the water. Let us take X-axis along the direction of theflow velocity.

Let the velocity of flow, vR/G

= v and let the velocity of boat with respect to river,

|vB/R

| = u

|vB/R

| = u when it goes downstream

= – u when it moves upstream

Since the boat moves with the same speed with respect to raft (river) whether it goes downstream or upstream it will passthe raft 60 minutes after turning. In the meantime, displacement of the ground with respect to raft is equal to – 8 km. Thus;

we have

– 8 = 2 × – v

⇒ v = 4 km/hr.

Try yourself. Solve the same problem in a frame fixed to the ground.

THEORY



Illustration 43:

An elevator whose floor to ceiling distance is 5.0 m, starts ascending with acceleration of 2.5 m/s2. After one second, a boltstarts falling freely from the ceiling of the elevator. Calculate.

(a) Free fall time of the bolt.

(b) The distance and displacement cover by the bolt during the free fall in the reference frame fixed to the ground.

(Take g = 10 m/s2)

Solution :

(a) Let us choose the x-axis along vertical upward direction with the origin at the position of the floor when bolt startfalling.

Let xF and x

B refer to the coordinates of the floor and bolt in the ground frame.

After 1 sec , speed of floor and bolt

= 0 + 2.5 × 1 = 2.5 m/s

The coordinates of the bullet and the floor at time t (after bolt starts falling)

xF = u

Ft +

1

2a

Ft2 = 2.5 × t +

1

2 × 2.5t2

xB = x

0 + u

Bt +

1

2a

Bt2 = 5 + 2.5t –

1

2 × 10t2

When bullet strikes the floor,

xF = x

B

⇒ 2.5 × t + 1

2 × 2.5 × t2 = 5 + 2.5t – 5t2

⇒ 6.25t2 = 5 ⇒ t = 0.89 sec

(b) Displacement of the bullet with respect to ground.

⇒2

B

1

2 Bs ut a t= +

= 2.5 × 0.89 – 5(0.89)2 = – 1.74 m A

C

H

As the velocity of bullet changes direction during motion, displacement will not be equal to distance.The situation is shown below.

Let A and C be initial and final position and let H be the point where velocity changes direction. Then

distance = 2(AH) AC+

= ( )2

0 2.52 1.74

2 10

−+ −

− × = 2.37 m

It is instructive to solve part (a) in the frame fixed to the elevator.

Let t = 0 , when bullet starts falling.

B/E B/G G/E B/G E/Ga a a a a= + = −r r r r r

= ˆ ˆ ˆ2.5 12.5gi i i− − = −So, a = – 12.5 m/s2

The initial velocity of bolt with respect to elevator is zero. So, u = 0.

Applying,

s = ut + 1

2at2

⇒ – 5 = 0 – 1

2(12.5)t2

⇒ t = 5

6.25 = 0.89 sec



Illustration 44:

A man going north with a velocity of 4 km/hr finds that wind flows from east. He double his speed and finds that windcomes from North-east. In what direction and with what velocity is wind flowing with respect to ground.

Solution :

Let us choose +ve x-axis along East and +ve y-axis along North as shown.

Let the velocity of wind with respect to ground be

E

N

y-axis

Wx-axis

S

W/Gˆ ˆ

x yv v i v j= +r

Substituting this in W/M W/G M/Gv v v= −r r r––––––– (2)

we have W/M M/Gˆ ˆ

x yv v i v j v= + −r r

Case 1. M/G

ˆ4v j=r

W/Mˆ ˆ( 4)y xv v j v i= − +r

As wind appears to flow from east, the y-component of M/Gvr

is zero . Hence

⇒ vy – 4 = 0 ⇒ v

y = 4 km/hr

Case 2. M/G

ˆ8v j=r

In this case eqn (2) reduces to

M/Gˆ ˆ( 8)y xv v j v i= − +r

In this case wind flows from North-East toward South-West. 45°

45°W E

N

S

W/Gv

W/Mv

i.e. x-component and y-component of M/Gvr

are equal.

⇒ vy – 8 = v

x

⇒ vx = – 4 km/hr

Hence

M/Gˆ ˆ4 4v i j= − +r

From this we conclude, W/G 4 2v =rkm/hr in North-West direction.

Illustration 45:

αv

CBy

Ax

A man in a boat crosses a river from point A (shown in the figure). If he rowsperpendicular to the banks then, 10 minutes after he starts, he will reach point C lyingat a distance s = 120 m downstream from point B. If the man heads at a certain angle αto the straight line AB(AB is perpendicular to the banks) against the current he willreach point B after 12.5 minutes. Find (a) the width of the river ‘d’ (b) the velocity ofboat u relative to the water, (c) the speed of the current v and angle α . Assume thevelocity of the boat relative to the water to be constant and of same magnitude in boththe cases.

Solution :

Let x-axis point along the direction of current and y-axis along AB.

B/R R/G,v u v v= =r r

Case 1. (When he rows perpendicular to the banks)

THEORY



B/Rˆv uj=r

R/Gˆv vi=r

B/G B/R R/Gˆ ˆv v v vi uj= + = +r r r

BC = vx × 10 = v × 10 ––––––– (1)

⇒ v = BC

10 = 12 m/min AB = v

y × 10 = u × 10

⇒AB

10 10

du = = ––––––– (2)

Case 2. (When he rows at an angle α )

B/Rˆ ˆsinα cosαv u i u j= − +r

R/Gˆv vi=r

⇒ B/Gˆ ˆ( sinα) cosαv v u i u j= − +r

Since B/Gvr

points along y-axis, x-component of the velocity is zero in this case.

⇒ v = u sinα ––––––– (3) and

d = uy × t = u cosα × 12.5 ––––––– (4)

Combining eqns (2) and (4) , we get

10u = u cosα × 12.5

⇒ cosα = 4/5 ⇒ α ≅ 37°

and from (3) , we 12 5

20sin 3

vu

α×= = = m/min d = u × 10 = 200 m


1. An object A is moving with 10 m/s and B is moving with 5 m/s in the same direction. A is 100 m behind B. Find the taken byA to meet B.

2. Car A is moving towards north and car B towards East as shown in the figure. A

B

54 kmph

72 kmph

Calculate (a) Velocity of ground w.r.t car B

(b) Velocity of car A w.r.t car B.

(c) Rate of separation of two cars.

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

4. A man moving with 5 m/sec towards East observes rain falling vertically at the rate of 10 m/sec. Find the speed of rain w.r.tground.

5. A man can swim at the rate of 5 km/hr in still water. A 1 km wide river flows at the rate of 3 km/hr. The man wishes to swimacross 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 will he take to cross the river



Section - 3.5 Circular Motion

When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant, then its motion is knowas circular motion with respect to that fixed (or moving) point. The fixed point is called centre, and the distance of particle from itis called radius.

1. Angular Position : Suppose a particle P is moving in a circle of radius rand centre O. The angular position of the particle P at a given instant may bedescribed by the angle θ between OP and the reference line OX. Thisangle θ is called the angular position of the particle. P

XO

Y

θr

P'

∆θ2. Angular Displacement : Angle through which the position vector of themoving particle rotates in a given time interval is called its angulardisplacement. As the particle moves on above circle its angular positionθ changes. Suppose the point rotates through an angle ∆ θ in time ∆t, thenis ∆ θ angular displacement.

Distance covered = arc PP' = r ∆ θAngular displacement is a dimensionless quantity. Its SI unit is radian.

Infinitesimally small angular displacement is a vector quantity, but finite angular displacement is a scalar, because while theaddition of the Infinitesimally small angular displacements is commutative, addition of finite angular displacement is not.

1 2 2 1θ θ θ θd d d d+ = +r r r r

but 1 2 2 1θ θ θ θ+ ≠ +

Direction of small angular displacement is decided by right hand thumb rule. When the fingers are curled along the motionof the point then thumb will represent the direction of angular displacement.

3. Angular Velocity ( ωωωω )

(a)Angular displacement

Total time takenavgω =

2 1

2 1

θ θ θavg t t t

ω − ∆= =− ∆

Average angular velocity is a scalar.

(b) Instantaneous Angular Velocity

It is the limit of average angular velocity as ∆t approaches zero. i.e.

0

θ θlimt

d

t dtω

∆ →

∆= =∆

r rr

Instantaneous angular velocity ωr is a vector, whose direction is given by right hand thumb rule.

Angular velocity has dimension of [T–1] and SI unit rad/s.

If T is the period and ‘f’ the frequency of uniform circular motion2π

2πTavg fω = =

4. Angular Acceleration α :

(a) Average Angular Acceleration :

Let 1ω and 2ω be the instantaneous angular speeds at times t1 and t

2 respectively, then the average angular

acceleration αavg is defined as

2 1

2 1

αavg t t t

ω ω ω− ∆= =− ∆

r r r

(b) Instantaneous Angular Acceleration :

It is the limit of average angular acceleration as ∆t approaches zero, i.e.,

THEORY



0α lim

t

d

t dt

ω ω∆ →

∆= =∆

r rr

Since θ

,d

dtω =

rr

∴ 2

2

θα ,

d d

dt dt

ω= =rr

rAlso α

d

d

ωωθ

=r

r

If α = 0, circular motion is said to be uniform.

Motion with constant angular velocity

θ , α 0tω= =

Motion with constant angular acceleration

0ω ⇒ Inital angular velocity ω ⇒ Final angular velocity

α ⇒ Constant angular acceleration θ ⇒ Angular displacement

Circular motion with constant angular acceleration is analogous to one dimensional translational motion with constant acceleration.Hence even here equation of motion have same form.

0 αtω ω= +2

0

1θ α

2t tω= +

2 20 2 θω ω α= + O

v rω= ×rr r

Here, vr

is velocity of the particle, ωr is angular velocity about centre of circular motion and ‘ rr

’ is position of particle w.r.t. centerof circular motion.

Since rω ⊥r r

v = rω for circular motion.

Velocity and acceleration in uniform circular motio n

The magnitude of velocity does not change with time but the direction of the velocity (along tangent) keeps on changing frommoment to moment. As the velocity vector changes direction with time, the acceleration is non-zero in uniform circular motion.

Let v = speed of moving particle (same at all points) and r = radius of the circle.

Let the particle start from A at t = 0. After time t, it reaches point P where the position vector is :

ˆ ˆ( ) OP cos sinr t r t i r t jω ω= = +

⇒ ˆ ˆ( ) cos cosd r

v t r t i r t jdt

ω ω ω ω= = +

tωr

P(time )t

A = 0tO

Acceleration = 2 2ˆ ˆ( ) cos sin

d va t r t i r t j

dtω ω ω ω= = − −

= ( )2 ˆ ˆcos sinr t i r t jω ω ω− +

⇒ 2a rω= −

a

aa

a

a

aa

a

v

v

vv

v

v

vv

The magnitude of the acceleration is 2rω and it is directed opposite to r andhence towards the centre O.

The figure shows the direction of velocity and acceleration for different positionsof moving particle on the circle. As the acceleration is directed towards the centre,it is known as centripetal acceleration or radian acceleration (along the radius).

⇒ Centripetal acceleration = 2rω = 2v

r



Note : So far we have observed that in uniform circular motion, the magnitude of velocity (v) and magnitude of acceleration 2

v

r

are constant, while the direction of the velocity (along the tangent) and the direction of acceleration (along the radius) keepon changing with time.

Non-Uniform Circular Motion

If the speed of the particle rotating in the circle changes with time, it is said to be in non-uniform circular motion. The accelerationof the particle in that case has two components :

1. A radial (centripetal) component which causes the changes in the direction of velocity. It is directed towards the centre and

has a magnitude ar, given as : a

r =

2v

r. As the radial component is perpendicular to velocity, it is also called normal

component of acceleration denoted as an =

2v

r.

Note that this component is also present in uniform circular motion.

2. A tangential component which causes the change in magnitude of velocity. It is directed along the tangent and itsmagnitude is decided by the net tangential force acting on the particle. Its magnitude is given by a

t as :

at =

dv

dtwhere v is the speed of the particle.

The tangential acceleration is in the direction of motion if the particle speeds up and opposite to the direction of motion ifthe particle slows down.

Note :1. In uniform circular motion, tangential component a

t = 0 m/s2 because speed does not change.

2. The concept of radial (normal) and tangential acceleration can be applied to motion in curves other than circles.

3. The net acceleration is :

a = 2 2r ta a+

and it makes an angle θ with tangent :

at

ar

θa

θ = tan–1 r

t

a

a

Illustration 46:y

xA particle is moving with constant speed on a circular path shown in figure. The

instantaneous velocity of the particle is ˆ ˆ( )v i j= − +rm/s. Through which quadrant is the

particle moving when it is travelling

(a) Clockwise

(b) Anticlockwise.

Solution :

(a) Clockwise : Fourth quadrant45°

(b) Anticlockwise : Second quadrant

45°

THEORY



Illustration 47:

A particle travels in a circle of radius 20 cm with uniform angular acceleration. If the speed changes from 5 m/sec to6 m/sec in 2 sec, find the magnitude of angular acceleration.

Solution :

We know that v = ω r.

When v = 5 m/sec, ω = v

r =

5

0.2 = 25 rad/sec

When v = 6 m/sec, ω = v

r =

6

0.2 = 30 rad/sec.

As the angular accn is uniform, so by applying f i tω ω= + α we have 30 = 25 + α × 2 ⇒ α = 2.5 rad/s2.

Illustration 48:

A particle moving on a circular path of radius 2 m covers 15 revolutions in 5 sec with uniform speed. Find magnitude of itsacceleration.

Solution :

Particle is moving with uniform speed. Therefore it has only centripetal acceleration.

15 revolution in 5 sec ⇒ 15 × 2π radians in 5 sec.

So, ω = 15 2π

5

×rad/sec = 6π rad/sec.

∴ Centripetal accn = 2rω = (6π)2 × 2 = 72π2 m/sec2.

Illustration 49:A particle moves in a circle of radius 2 m with time varying speed given by v = 2t2 m/sec. Find

(a) Magnitude of tangential accn of particle at t = 2 sec

(b) Magnitude of total accn of particle at t = 1 sec

(c) Angle between velocity vector of the particle and its total accn at t = 2 sec

(d) Angle through which it turned from t = 1 sec to t = 3 sec.

Solution :

(a) Tangential accn = dv

dt = ( )22

dt

dt = 4t = 8 m/s2

(b) Total accn = 2 2t ca a+

Now at = 4t = 4 m/sec2. a

c =

( )222

2

t = 2t4 = 2 m/sec2

∴ Total accn = 2 24 2+ = 2 5 m/sec2. θ ac

at

atotal

(c) Angle both total accn and velocity vector is 4

1 1 12θ tan tan tan (4)

4c

t

a t

a t− − −

= = =

.

(d) To find angle swept between t = 1 sec to t = 3 sec, first we will find distance moved on circular path during this interval.

ds

dt = 2t2 ⇒ ds = 2t2dt ⇒


s = Distance travelled on circular path during t : [1, 3] = 33

32 3 3

11

2 2 2 522 3 1

3 3 3 3

tt dt

= = × − × =

∫ m



Angle swept =

52Arc 263

Radius 2 3

= = radians.


1. A particle is moving with constant speed V m/s along the circumference of a circle of radius R meter as shown. A, B and Care three points on periphery of the circle and ∆ABC is equilateral. The magnitude of average velocity of particle, as itmoves from A to C in clockwise sense, will be :

(a)3V

3π(b)

3V

4πR

A

C B

V

(c)3 3V

4π(d)

3 3V

2π

2. A car is moving with speed 30 m/sec on a circular path of radius 500 m. Its speed is increasing at the rate of 2 m/sec2. Whatis the acceleration of the car at that moment?

3. In a uniform circular motion, assuming vr

= velocity, rr

= radius vector, , ω = angular velocity relative to the centre of thecircle, a

r = acceleration, which of the following is/are correct?

(a) | | 0v∆ ≠r but | | 0v∆ =r

(b) | | 0ω∆ = but | | 0ω∆ =

(c) | | 0r∆ ≠r but | | 0r∆ =r

(d) | | 0a∆ ≠r but | | 0a∆ =r

4. A particle revolves in circle of radius R = 2 m, in x-y plane, in clockwise direction asshown (the centre C lies on x-axis. The y-axis is tangent to the circle). Its centripetalacceleration is of magnitude 18 m/s2. At time t = 0, particle is at (R, R). Then at later time‘ t’ the x and y coordinates of the particle are :

(a) x = 2 cos 3t (b) y = 2 cos 3t

x

y

C

(c) x = 2(1 + sin 3t) (d) y = 2(1 + sin 3t)

5. A fan is rotating with ω = 100 rad/sec. It is switched off and its angular velocity becomes 50 rad/sec after sometime.Find out the ω of the fan after it rotated half the revolutions as it rotated when its speed changes from 100 rad/sec to50 rad sec.

6. A particle starts moving with constant angular acceleration of magnitude 2 rad/s2. Find out the time after which particle’stotal accn vector makes angle of 37° with velocity vector.

7. A particle moves clockwise in a circle of radius 1 m with centre at (x, y) = (1 m, 0). It starts at rest at the origin at time t = 0.

Its speed increases at the constant rate of π

2

m/s2.

(a) How long does it take to travel halfway around the circle?

(b) What is the speed at that time?

(c) What is the net acceleration at that time?

8. The velocity and acceleration vectors of a particle undergoing circular motion are ˆ2v i=rm/s and ˆ ˆ2 4a i j= +r

m/s2

respectively at an instant of time. The radius of the circle is

(a) 1 m (b) 2 m (c) 3 m (d) 4 m

9. The tape in a standard cassette has length l = 120 m; the tape plays at a constant linear speed for t = 2 hrs. As the tape starts,

the full reel has an outer radius of R1 = 12 2cm, and an inner radius of R

2 = 5 2 cm. At some point during the play, both

reel have the same angular speed. Calculate this angular speed in rad/sec.

(a)1

500(b)

5

39(c)

1

510 2(d) None

THEORY



Solved Examples

Example 1:

Water drops are falling at regular intervals from a roof. At an instant when a drop is about to leave the roof find the ratioof separations between 3 successive drops below the roof.

Solution :

Let the drops fall at a regular interval of ‘t0’ from the roof.

When a drop is about to leave the roof the 1st, 2nd and 3rd drop below it have travelled for time interval t0, 2t

0 and 3t

0

respectively.

Distance below the roof

1st drop →1

2g t

02

g t(2 )02

gt02

g t(3 )02

2nd drop → 2g t02

3rd drop →2

09

2

gt

∴ Separation of 1st drop from roof = 2

0

2

g t

Separation of 2nd drop and 1st drop = 2g t02 –

1

2g t

02

= 2

03

2

g t

Separation of 3rd drop and 2nd drop = 2

09

2

gt – 2g t

02

= 2

05

2

g t

∴ Ratio = 1 : 3 : 5

Example 2:

A body is in straight line motion with an acceleration given by a = 32 – 4v. Find the velocity of the body at t = ln 2 sec ifinitially its velocity was 4 m/sec.

Solution :

At t = 0, v = 4 m/sec.

We have a = dv

dt = 32 – 4v ⇒

32 4

dv

v− = dt


4 32 4

v dv

v−∫ = ln 2

0

dt∫

⇒( )

4

ln 32 4

4

vv− −

= [ ]ln 2

0t ⇒

( )ln 32 4

4

v− − +

( )ln 32 4 4

4

− × = ln 2 – 0

⇒1 16

ln4 32 4v

−

= ln 2 ⇒16

32 4v− = 16 ⇒ v = 31

4 m/sec



SOLVED

EXAMPLES

Example 3:

a (m/s )2

t (sec)

– 5.0

O

5.0The acceleration of a point started at t = 0, varies with time as shown ingiven graph. Find the distance travelled in 30 seconds and draw thevelocity-time and position-time graphs if initial velocity and position arezero.

Solution : a (m/s )2

t (sec)

– 5.0

0

O

5.0

As given in the acceleration-time graph,

for time interval (0, 10) : a is constant and +ve

⇒ velocity increases linearly.

⇒ slope of x – t graph must be increasing

for time interval (10, 20) : a is zero

⇒ velocity is constant

⇒ slope of x – t graph is constant

for time interval (20, 30) : a is constant and –ve

⇒ velocity decreases linearly.

⇒ slope of x – t graph is decreasing

At t = 10 s, v = u + at = 0 + 5 × 10 = 50 m/s.

As we see from v – t graph, velocity is always nonnegative for timeinterval (0, 30), therefore distance travelled is equal to displacement.

s= displacement = area under v – t graph

= 1

10 502 × ×

+ (50 × 10) + 1

10 502 × ×

= 250 + 500 + 250 = 1000 m.

Note : • At t = 30 sec, v = 0 ∴ at t = 30 sec, slope of x – t graph also becomes zero.

• For time interval (0, 30) area under a – t graph is zero ⇒ change in velocity = 0 ⇒ vfin

= vin = 0 m/s.

Example 4:

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coinat the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The personobserves that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.(g = 32 ft/s2)

Solution :

Ground frame

6 fta

gt2

2at2

2

g

Initial velocities of both elevator and coin are zero. Suppose that the coinfalls on floor in t seconds, then the distance travelled by coin = distancetravelled by floor + initial distance between coin and floor

1

2gt2 = 6 +

1

2at2

Substituting t = 1

we get g = 12 + a

⇒ a = 32 – 12 = 20 ft/s2.



Example 5:

4 m/s

3 m/s

200 mB O

120 m

1

A2

Two ships, 1 and 2, move with constant velocities 3 m/s and 4 m/s along twomutually perpendicular straight tracks towards the intersection point O. Atthe moment t = 0 the ships were located at the distances 120 m and 200 fromthe point). How soon will the distance between the ships become the shortestand what is it equal to?

Solution :

Suppose that ship 1 starts moving from point B (at t = 0) andship 2 from point A (at t = 0 only), then after time ‘t’, 1 must havetravelled 4t units along –ve x–axis. Therefore, at some ‘t’ ship 1is at point (– (200 – 4t), 0) and ship 2 is at point (0, 120 – 3t).Therefore, distance between 1 and 2 is

d = ( ) ( )2 2

1 2 1 2x x y y− + −4t

A

(– (200 – 4 ), 0)t2 O

2 (0, 120 – 3 )t

3t

B 1

= ( )( ){ } ( ){ }2 20 200 4 120 3t t− − − + − = ( ) ( )2 2

200 4 120 3t t− + − ———— (1)

When d is minimum, d2 is also minimum.Therefore, derivative of d2 w.r.t. time must be zero.∴ 2(200 – 4t)(– 4) + 2(120 – 3t)(– 3) = 0

⇒ – 1600 + 32t – 720 + 18t = 0

⇒ 50t = 2320 ⇒ t = 2320/50 = 46.4 sec.

Substituting t = 46.4 sec in equation (i), we will get the minimum distance, which comes out to be 24 m.

Example 6:

A rubber ball is released from a height of 490 m above the floor. It bounces repeatedly, always rising to 81/100 of the heightthrough which it falls.

(a) Ignoring the practical fact that the ball has a finite size (in other words, treating the ball as point mass that bounces aninfinite number of times), show that its total distance of travel is 46.7 m.

(b) Determine the time required for the infinite number of bounces.

(c) Determine the average speed. (Take g = 9.8 m/s2)

Solution :

Let h = 4.9 m

(a) Distance travelled = h + 22

81 81..........

100 100h h

+ +

= h + 20.81

1 0.81

h −

= 4.9 + 20.81 4.9

0.19

×

= 46.7 m

(b) Time required to fall through height h = 2h

g

Total time = 2h

g + 2

22 2...........

nh n h

g g

+ +

where n = 81

100

Total time = 2 2

21

h h n

g g n

+ −

= 2h

g(1 + 18) = 19 sec



(c) Average speed = distance travelled

time taken =

46.7

19 = 2.46 m/s

Example 7:

α

βv

u

uA particle is projected with speed u at an angle ‘α’ with horizontal.Find the speed of the particle at the moment when the velocitymakes an angle ‘β’ with horizontal.

Solution :

Let the speed of the particle at the instant when the velocity vector makes an angle ‘β’ be v.

Horizontal component of velocity at that instant = v cosβHorizontal component of intial velocity = u cosα.

Equating the horizontal components we get

u cosα = v cosβ [Q ax = 0]

or v = cosα

cosβ

u

Example 8:

A ship is approaching a cliff of height 105 m above sea level. A gun fitted on the ship can fire shots with a speed of110m/sec. Find the maximam distance from the foot of the cliff from where the gun can hit an object on the top of thecliff. [g = 10m/s2]

Solution :

Let that distance be ‘D’, θ be the angle of projection and ‘T’ be the time taken to hit the cliff.

For Horizontal Motion :

(110 cosθ)T = D ———— (1)

For Vertical Motion :

105 m

D

110 m

/sec

Q

105 = (110 sinθ)T – 1

2 × 10 × T2 ———— (2)

Substituting ‘T’ From eqution (1) we have

105 = 2

D D(110sinθ) 5

110cosθ 110cosθ −

⇒ 105 = D tanθ – 2

5

(110) D2 sec2θ ———— (3)

For ‘D’ to be maximum D

θ

d

d = 0.

Differentiating the above equation w.r.t ‘θ’ we get,

0 = 2 2 2 2

2

D 5 Dtanθ Dsec θ 2D sec θ D (2sec θ tanθ)

θ θ(110)

d d

d d + − × +

Putting D

θ

d

d = 0 in above equation we get

D tanθ = 1210.

Putting D tanθ = 1210 in equation (3) we have

105 = 1210 – ( )2 22

5D (1210 )

(110)× + ⇒ D = 1100 m

SOLVED

EXAMPLES



Example 9:

A vertical pole has a red mark at some height. A stone is projected from a fixed point on the ground. When projected at anangle of 45° it hits the pole perpendicularly 1 m above the mark. When projected with a different velocity at an angle of 37°,it hits the pole perpendicularly 1.5 m below the mark. Find the velocity and angle of projection so that it hits the markperpendicularly. [g = 10 m/s2]

Solution :

The stone hits the pole perpendicularly means that the stone was at highest point of trajectory.

When angle of projection = 45°

2D = Range and H + 1 = Max. Height

i.e., 2D = 2

1 sin90u

g

° and H + 1 =

21 sin 45

2

u

g

°

⇒ 2(H + 1) = D ———— (1)

When angle of projection = 37°.37° 45°

1 m1.5 m

H

D2D = Range and H – 1.5 = Max. Height

i.e., 2D = 2

22 sin37 cos37u

g

° ° and H – 1.5 =

2 22 sin 37

2

u

g

°⇒ 9D = 24H – 36 ———— (2)

From (1) and (2) we have H = 9m and D = 20 m.

So the stone that will hit the red mark has range = 40 m and Hmax

= 9 m.

∴ 40 = 2 sin 2θu

g and 9 =

2 2sin θ

2

u

g

Solving these equation we get u = 3620

3m/sec and θ = tan–1

9

10

Example 10:

θ

90°

R

Find the range ‘R’ of a particle projected with speed u perpendicularto the plane of inclination θ with horizontal.

Solution :

Here ux = 0, u

y = u

ax = g sinθ, a

y = –g cosθ

Let Range be ‘R’ and time of flight be ‘T’.

y-coordinate will become zero when projectile will land on the plane.

applying sy = u

yt +

1

2a

yt2 we get

0 = uT – 1

2g cosθ T2 ⇒ T =

2

cosθ

u

g

For Range, applying sx = u

xT +

1

2a

xT2 we get

R = ( )2

2 1 20 sinθ

cosθ 2 cosθ

u ug

g g

× + ×

Range = 2 22 sinθsec θu

g



Example 11:

A ferrari, mercedes and a toy rocket are moving as shown in thefigure. Assuming x-y plane to be the plane of motion of objects.Find

(a) Velocity of Ferrari w.r.t Rocket.

FERRARI MERCEDES

15 m/sec 20 m/sec

ROCKET

37°

30 m

/sec y

x

(b) Velocity of Mercedes w.r.t Ferrari.

(c) Velocity of tree w.r.t rocket.

(d) Velocity of mercedes w.r.t rocket.

Solution :

Vur

Ferrari, ground = ˆ15i−

Vur

Mercedes, ground = ˆ20i

Vur

Rocket, ground = ˆ ˆ30cos37 30sin37i j° + ° = ˆ ˆ24 18i j+

(a) Vur

Ferrari, Rocket= V

ur

Ferrari, ground – V

ur

Rocket, ground ⇒ ˆ ˆ ˆ15 (24 18 )i i j− − + = ˆ ˆ39 18i j− −

(b) Vur

Mercedes, Ferrari = V

ur

Mercedes, ground – V

ur

Ferrari, ground ⇒ ˆ ˆ20 (15 )i j− = ˆ35i

(c) Vur

Tree, Rocket = V

ur

Tree, ground – V

ur

Rocket, ground ⇒ ˆ ˆ0 (24 18 )i j− + = – ˆ ˆ24 18i j−

(d) Vur

Mercedes, Rocket = V

ur

Mercedes, ground – V

ur

Rocket, ground ⇒ ˆ ˆ ˆ20 (24 18 )i i j− + = ˆ ˆ4 18i j− −

Example 12:

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

(a) Find the velocity with which the man should move so that rain appears to fall vertically to him.

(b) Now if he further increases his speed, rain again appears to fall at 30° with the vertical. Find his new velocity.

Solution :

R,GVur

= ˆ ˆ20sin30 20cos30i j− ° − °

= ˆ ˆ10 10 3i j− − .

(a) Let the man is moving with speed V0 towards negative x-axis.

(In this case only he will feel the rain coming down vertically)

∴ M,GVur

= 0ˆV i−ur

∴ R, MVur

= Velocity of rain w.r.t man

= R,GVur

– M,GVur

= ( ) ( )0ˆ ˆ ˆ10 10 3 Vi j i− − − −

= ( )0ˆ ˆ10 V 10 3i j− + −

V0 V =

20

m/s

ec

R

30°

y

x

Rain feels to come down vertically to man.

∴ – 10 + V0 = 0

⇒ V0 = 10 m/sec

∴ M,GVur

= ˆ10i− m/sec

(b) Let the new increased speed of man is V'.

SOLVED

EXAMPLES



∴ M,GVur

= ˆV i′

R, MVur

= R,GVur

– M,GVur

= ( ) ( )ˆ ˆ ˆ10 10 3 Vi j i′− − − −30° 30°

R, GV 20=R,MV

M,GV V′=

= ˆ ˆ(V –10) 10 3i j′ −

Now we know that R, MVur

makes an angle = 30° with vertical.

⇒ tan 30° = V 10

10 3

′ −⇒ V' = 20 m/sec

Example 13:

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 theparticles moves with constant speed V. A always has its velocity along AB, B along BC and C along CA. At what time willthe particle meet each other?

Solution :

The motion of the particles is roughly sketched in figure. By symmetrythey will meet at the centroid O of the triangle. At any instant the particleswill from an equilateral triangle ABC with the same Centroid O. All theparticles will meet at the centre. Concentrate on the motion of any oneparticle, say B. At any instant its velocity makes angle 30° with BO. Thecomponent of this velocity along BO is V cos 30°. Thiscomponent is the rate of decrease of the distance BO.

OC3 C2

C1

CB

A

B1

B2

B3

A3

A2

A1

BO = / 2

cos30

a

° = 3

a = displacement of each particle. Therefore, the time taken for BO to become zero

CB

A

Q

30° a/2

/ 3a=

/ 3

V cos30

d

° =

2

3V 3

d

× =

2

3V

d.

Example 14:

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with anacceleration of 1 m/s2 and the projection speed in the vertical direction is 9.8 m/sec. How far behind the boy will the ballfall on the car?

Solution :

Let the initial velocity of car be u. u

9.8 m/secu m/sec

xc

a = 1m/s2Xball

V =9.8 M/sec

ball, boy

V = boy, ground u

Vball, ground

Time of flight of the ball = 2u

a

⊥⊥ =

2 9.8

9.8

× = 2 sec

Distance travelled by car in Horizontal direction = xc

= u × 2 + 1

2 × 1 × 22 = 2u + 2

Distance travelled by ball in Horizontal direction = xBall

= 2 × u = 2u

∴xc – x

B = 2m



Example 15:

An elevator is going up with an upward acceleration of 1m/s2. At the instant when its velocity is 2m/sec, a stone isprojected upward from its floor with a speed of 2m/sec relative to the elevator, at an elevation of 30°.

(a) Calculate the time taken by the stone to return to floor.

(b) If the elevator was moving with a downward acceleration equal to g, how would the motion be.

Solution :

(a) Initial velocity of Ball w.r.t lift = ˆ ˆ2cos30 2sin30i j° + °

= ˆ ˆ3i j+

Initial velocity of Lift = ˆ2 j

Initial velocity of ball w.r.t ground = , , ,V V VB G B L L G= +ur ur ur

= ˆ ˆ3 3i j+

The ball will return to floor when y displacement of floor of lift becomes equal to y displacement of ball.

yBall

= uyT + a

yT2 = 3T –

1

2 × 10 × T2

ylift

= 2T

Equating both we get 3T – 5T2 = 2T ⇒ T = 0.2 sec

If lift is moving with downward acceleration ‘g’ then relative acceleration of stone w.r.t lift = S,G L,Ga a−

= ˆ ˆ( )gj gj− − − = 0

Also B,LV = ˆ ˆ3 3i j+

∴ Ball will keep on separating from floor of lift with velocity B,LV = ˆ ˆ3 3i j+

Motion of ball w.r.t someone sitting in lift will be straight line.

SOLVED

EXAMPLES



Exercises

Level - 1

Single Choice Questions

1. In the following velocity-time graph of a body, the distance anddisplacement travelled by the body in 5 seconds in meters will be :

(a) 70, 110

(b) 110, 70

40

30

20

10

0

–10

–20

–30

–40

1 2 3

4 5

v (m

/s)

t (sec)(c) 40, 70

(d) 90, 50

2. A body thrown vertically up from the ground passes the height 10.2 m twice at an interval of 10s. What was its initialvelocity (g = 10 m/s2)?

(a) 52 m/s (b) 26 m/s (c) 35 m/s (d) 60 m/s

3. If the angle of projection θ corresponds to horizontal range being equal to the maximum height then tanθ equals :

(a) 1 (b) 1/ 3 (c) 3 (d) 4

4. If y = x – x2 is the equation of the path of a projectile, then which of the following is incorrect :

(a) Range = 1 m (b) Maximum height = 0.25 m

(c) Time of flight = 0.5 sec. (b) Angle of projection = 45°

5. Two particles, one with constant velocity 50 m/s and the other with uniform acceleration 10 m/s2, start moving simultaneouslyfrom the same place in the same direction. They will be at a distance of 125 m from each other for the first time after

(a) 5 sec. (b) ( )5 1 2+ sec. (c) 10 sec. (d) ( )10 2 1+ sec.

6. A river is flowing from west to east at a speed of 5 meter per minute. A man on the south bank of the river, capable ofswimming at 10 metre per minute in still water, want to swim across the river in the shortest time. He should swim in thedirection :

(a) Due north (b) 30° east of north (c) 30° north of west (d) 60° east of north

7. A projectile is fired with a velocity u making an angle θ with the horizontal. What is the change in velocity when it is at thehighest point?

(a) u cosθ (b) u (c) u sinθ (d) u cosθ – u

8. A man moves in x-y plane along the path shown. At what point is the average velocityvector in the same direction as his instantaneous velocity vector. The man starts frompoint P :

(a) A (b) B

y

A

BP

CD

x(c) C (d) D

9. When a particle moves in a circle with a uniform speed

(a) Its velocity and acceleration are both constant (b) Its velocity is constant but the acceleration changes

(c) Its acceleration is constant but the velocity changes (d) Its velocity and acceleration both change



10. A particle moves along a circle of radius 20

π

m with constant tangential acceleration. If the speed of the particle is 80 m/

s at the end of the second revolution after motion has begun, the tangential acceleration is :

(a) 160πm/s2 (b) 40πm/s2 (c) 40 m/s2 (d) 640πm/s2

Multiple Choice Questions

11. An object follows a curved path. The following quantities may remain constant during the motion

(a) speed (b) velocity (c) acceleration (d) magnitude of acceleration

12. Two particles are thrown from the same point in the same vertical plane, as shown in figure simultaneously. Then indicatethe correct statements :

(a) Time of flight for B is less than that of A

θ1θ2 AB

x

y

(b) Projection speed of B is greater than that of A

(c) Horizontal component of velocity for B is greater than that of A

(d) The vertical component of velocities of both A and B are always equalthroughout the duration for which both the particles are in air.

13. An aeroplane at a constant speed releases a bomb. As the bomb drops away from the aeroplane,

(a) It will always be vertically below the aeroplane

(b) It will always be vertically below the aeroplane only if the aeroplane was flying horizontally

(c) It will always be vertically below the aeroplane only if the aeroplane was flying at an angle of 45° to the horizontal

(d) It will gradually fall behind the aeroplane if the aeroplane was flying horizontally.

14. A particle is moving with uniform acceleration along a straight line AB. Its speed at A and B are 2 m/s and 14 m/srespectively. Then :

(a) Its speed at the mid-point of AB is 10 m/s

(b) Its speed at a point P such that AP : PB = 1 : 5 is 6 m/s

(c) The time to go from A to the mid-point of AB is double of that to go from mid-point to B

(d) None of these

15. The figure shows the velocity (v) of a particle plotted against time (t).

(a) The particle changes its direction of motion at some point. a

2TO

T

t

v

(b) The acceleration of the particle remains constant.

(c) The displacement of the particle is zero.

(d) The initial and final speeds of the particle are the same.

Comprehension Type

Comprehension 1

A particle is moving along x-axis at t = 0. Its initial velocity is 40 m/s along positive x-axis and an acceleration of 10 m/s2 alongnegative x-axis. Particle starts from x = 10 m.

16. Velocity of particle is zero at t equal to

(a) 6 sec (b) 4 sec (c) 8 sec (d) 2 sec

17. Maximum x co-ordinate of particle (in positive direction) is

(a) 90 m (b) 60 m (c) 120 m (d) 30 m

18. Velocity of particle at origin is

(a) 30 2m/sec (b) 20 2 m/sec (c) 20 2− m/sec (d) 30 2− m/sec

EXERCISES



Comprehension 2

01 2 3 4 5

v(m/s)

t

2

4

6 7

– 28

The velocity time graph of a particle moving along a straight line isshown in the figure. In the time interval from t = 0 to t = 8 second,answer the following three questions.

19. The distance traveled by the particle is :

(a) 18 m (b) 6 m

(c) 8 m (d) None of these

20. The value of maximum speed

average speed for the given time interval is :

(a) 16/9 (b) 8/9 (c) 8/3 (d) 16/3

21. The value of magnitudeof maximum velocity

magnitudeof average velocity for the given time interval is :

(a) 8/9 (b) 16/3 (c) 16/5 (d) 8/3

Matrix Match

22. A ball is projected from the ground with velocity v such that its range is maximum.

Column - I Column - II

(a) Velocity at half of the maximum height (p) 3 / 2v

(b) Velocity at the maximum height (q)2

v

(c) Change in its velocity when it returns to the ground (r) 2v

(d) Average velocity when it reaches the maximum height (s)5

2 2

v

23. A balloon rises up with constant net acceleration of 10 m/s2. After 2s a particle drops from the balloon, After further 2s matchthe following : (Take g = 10 m/s2)


(a) Height of particle from ground (p) Zero

(b) Speed of particle (q) 10 SI units

(c) Displacement of particle (r) 40 SI units

(d) Acceleration of particle (s) 20 SI units

Level - 2

Single Choice Questions

1. A balloon starts rising from the ground with an acceleration of 1.25 ms–2. After 8 seconds, a stone is released from theballoon. After releasing, the stone will :

(a) Cover a distance of 40 m till it strikes ground

(b) Have a displacement of 50 m till it reaches ground

(c) Reach the ground in 4 second

(d) Begin to move down instantaneously.



2. Three particles start from origin at the same time with a velocity 2 ms–1 along positive x-axis, the second with a velocity 6ms–1 along negative y-axis. Find the speed of the third particle along x = y line so that the three particles may always lie ina straight line :

(a) 3 3− (b) 4 2 (c) 3 2 (d) 2 2

3. Rain is falling with a speed of 4 m/s in a direction making an angle of 30° with vertical towards south. What should be themagnitude & direction of velocity of cyclist to hold his umbrella exactly vertical, so that rain does not wet him:

(a) 2 m/s towards north (b) 4 m/s towards south (c) 2 m/s towards south (d) 4 m/s towards north

4. A stone falls from rest. The total distance covered by it in the last second of its motion is equal to the distance covered inthe first three seconds of its motion. How long does the stone remains in the air?

(a) 4 s (b) 5 s (c) 6 s (d) 7 s

5. When a ceiling fan is switched off, its angular velocity falls to half while it makes 36 rotations. How many more rotationswill it make before coming to rest? (Assume uniform angular retardation.)

(a) 36 (b) 24 (c) 18 (d) 12

6. Take the z-axis as vertical and the xy plane as horizontal. A particle A is projected at 4 2 m/s at an angle of 45° to thehorizontal, in the xz plane. Particle B is projected at 5 m/s at an angle θ = tan–1(4/3) to the y-axis, in the yz plane. Which ofthe following is not correct for the velocity of B with respect to A?

(a) Its inital magnitude is 5 m/s.

(b) Its magnitude will change with time.

(c) It lies in the xy plane.

(d) It will initially make an angle (θ + π /2) with the positive x-axis.

7. A particle starts from the origin at t = 0 and moves in the xy plane with constant acceleration a in the y-direction. Itsequation of motion is y = bx2. The x-component of its velocity is

(a) variable (b)2a

b

(c)2

a

b(d) 2

a

b

8. Two particles start moving on the same circle of radius 2m, from the same point P at t = 0, with constant tangentialaccelerations 2 m/s2 and 6 m/s2, clockwise, respectively. The point where they meet for the first time is Q. The smaller anglesubtended by PQ at center of circle is :

(a) 120° (b) 60° (c) 135° (d) 90°

9. A point moves along an are of a circle of radius R. Its velocity depends upon the distance covered s as v = a x , where a

is constant. The angle θ between the vector of total acceleration and tangential acceleration is :

(a) tanθ = R

s(b) tanθ =

2R

s(c) tanθ =

2R

s(d) tanθ =

2

R

s

10. The velocity of a graph varies as v = 1t

b −

. If the particle starts from origin, it comes back to origin ofter time

(a) 2b (b) b (c) a (d) a2/b

Multiple Choice Questions

11. A body is projected vertically upwards from the ground. Taking air resistance into account :

(a) The time of rise to maximum height is less than that in vacuum

(b) The time of rise is greater than the time of fall

(c) The time of rise is less than the time of fall

(d) The speed of the particle when it returns to the ground is less than the speed of projection.

EXERCISES



12. A man throws a stone vertically up with a speed of 20 ms–1 from top of a high rise building. Two seconds later, an identicalstone is thrown vertically downward with the same speed 20 ms–1. Then :

(a) The relative velocity between the two stones remains constant till one hits the ground

(b) Both will have the same kinetic energy, when they hit the ground

(c) The relative acceleration between the two is equal to zero, till one hits the ground

(d) The time interval between their hitting the ground is 2 seconds.

13. The upper end of the string of a simple pendulum is fixed to a vertical z-axis, and set in motion such that the bob moves along a horizontal circularpath of radius 2 m, parallel to the xy plane, 5 m above the origin. The bobhas a speed of 3 m/s. The string breaks when the bob is vertically above thex-axis, and it lands on the xy plane at a point (x, y).

5m

2m

z

O

x

y

(a) x = 2 m (b) x > 2 m

(c) y = 3 m (d) y = 5 m

14. A particle is projected from origin such that its range is 4a. It just clears a wall at x = 3a of length 2a.

(a) The time of flight is 8

3

a

g(b) The time taken to reach x =

2

3

a

g

(c) The maximum height is 8

3a (d) The time taken to travel from x = 2a to 3a is 13

ag .

15. Two people A and B start from origin with their positions varying as x1 = t2 – 2t and x

2 = 4t – t2. When they meet

(a) A is moving in positive x direction. (b) A is changing his direction.

(c) B is moving in negative x direction. (d) B is changing his direction.

Comprehension Type

Comprehension 1

A particle is thrown from origin with a velocity ˆ5 5 3 .i j+ At the same instant, an inclined plane of inclination 60° whose left most

end is at x = a starts moving towards origin with speed v. The particle hits the incline plane.

60° 60°

P

u

( , 0)ax

y

16. The particle does so in time

(a)2

3s (b)

1

3s (c) 3s (d) 2 3s

17. The coordinates of the particle at the time it hits the incline

(a)10 20

,3 3

(b)10 10

,33

(c)10

, 10 33

(d)10 15

,3 3

18. If a = 10 3m/s, what is the value of v?

(a)35

3m/s (b) 5 m/s (c) 10 3m/s (d) 20 m/s



Comprehension 2

A particle undergoes uniform circular motion. The velocity and angular velocity of the particle at an instant of time is ˆ ˆ3 4v i j= +rm/

sec and ˆ ˆ6xi jω = +r rad/sec

19. The value of x is

(a) 8 (b) – 8 (c) 6 (d) None

20. The radius of circle in meters is

(a) 0.5 m (b) 1 m (c) 2 m (d) None

21. The acceleration of particle at that instant is

(a) – 50k (b) – 42k (c) ˆ ˆ2 3i j+ (d) None

Comprehension 3

An observer having a gun observes a remotely controlled balloon. When he first noticed the balloon, it was at an altitude of 800m and moving vertically upward at a constant velocity of 5 m/sec. The horizontal displacement of balloon from the observer is

1600 m. Shells fired from on gun have an initial velocity of 400 m/sec at a fixed angle 3 4

θ sinθ and cosθ5 5

= =

. The observer

having gun waits (for some time after observing balloon) and fired so as to destroy the balloon. (Assume g = 10 m/s2)

22. The flight time of the shell before it strikes the balloon is :


23. The altitude of the collision above ground level is :


24. After noticing the balloon, the time for which observer having gun waits before firing the shell is :


Matrix Match

25. Two balls A and B are projected from the top of towers of height 30 m and 20 m respectively at the same instant. If theycollide mid-way between the towers after 1 sec from the time of projection, then match the entries in column-I to appropriateentries in column-II. (Take g = 10 m/s2)

30m

30°

v1A

60°

v2

B

20m

D


(a) Height above ground at which they collide (in meter) (p) 10 2

(b) Initial speed of particle A in (m/sec) (q) 10

(c) Initial speed of particle B in m/sec (r) 20

(d) Distance between the towers in meter (s) 30

EXERCISES



26. The velocity time graph for a particle moving along a straight line is given in each situation of column-I. In the time interval

0 ,t< < ∞ match the graph in column-I with corresponding statements in column-II.


(a)

v

t

(p) Speed of particle is continously decreasing.

(b)

v

t(q) Magnitude of acceleration of particle is decreasing with time.

(c)

v

t

(r) Direction of acceleration of particle does not change.

(d)

v

t

(s) Magnitude of acceleration of particle does not change.



Previous Years’ IITJEE Questions

1. A particle of mass 10–2 kg is moving along the positive x-axis under the influence of a force F(x) = – 2

K

2x where K = 10–

2 N m2. At time t = 0 it is at x = 1.0 m and its velocity is v = 0. Find [JEE 98, 8]

(a) Its velocity when it reaches x = 0.50 m

(b) The time at which it reaches x = 0.25 m.

2. In 1.0 sec. a particle goes from point A to point B moving in a semicircle of radius 1.0 m. The magnitude of average velocityis : [JEE 99, 2]

(a) 3.14 m/sec

(b) 2.0 m/sec1 m

A

B(c) 1.0 m/sec

(d) zero

3. The coordinates of a particle moving in a plane are given by x(t) = a cos (pt) and y(t) = b sin (pt), where a, b (<a) and p arepositive constants of appropriate dimensions then - [JEE 1999, 3/200]

(a) The path of the particle is an ellipse

(b) The velocity and acceleration of the particle are normal to each other at t = π/2p

(c) The acceleration of the particle is always directed towards a focus

(d) The distance travelled by the particle in time interval t = 0 to t = π/2p is a.

4. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h above the ground as

[JEE 2000, 3]

(a)

v

hd

(b)

v

hd (c)

v

hd

(d)

v

hd

5. An object A is kept fixed at the point x = 3 m and y = 1.25 m on a plank P raised abovethe ground. At time t = 0 the plank starts moving along the + x direction with anacceleration 1.5 m/s2. At the same instant a stone is projected from the origin with avelocity u as shown. A stationary person on the ground observes the stone hittingthe object during its downward motion at an angle of 45° to the horizontal. All themotions are in x-y plane. Find ‘u’ and the time after which the stone hits the object.

[Take g = 10 m/s2] [JEE 2000, 10/100]

y

x

1.25 m

3.0 m

A

u

6. On a frictionless horizontal surface, assumed to be the x-y plane a small trolley A ismoving along a straight line parallel to the y-axis as shown in the figure with a constant

velocity of ( )3 1− m/s. At a particular instant, when the line OA makes an angle of

45° with the x-axis, a ball is thrown along the surface from the origin O. Its velocity

makes an angle φ with the x-axis when it hits the trolley..

y

x

A

45°O

(a) The motion of the ball is observed from the frame of the trolley. Calculate the angle θ made by the velocity of the ballwith x-axis in this frame.

(b) Find the speed of the ball with respect to the surface, if φ = 4θ/3. [JEE 2002, 2 + 3/60]



7. A particle is initially at rest, it is subjected to a linear acceleration a, as shown in the figure. The maximum speed attained bythe particle is [JEE Scr. 2004; 3]

(a) 605 m/s

(b) 110 m/s

(c) 55 m/s

11time ( )s

a (m/s )2

10

(d) 550 m/s

8. 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 1

S

Sn

n+ is [JEE Scr. 2004 ,3]

(a)2 1

2

n

n

−(b)

2 1

2 1

n

n

−+ (c)

2 1

2 1

n

n

+− (d)

2

2 1

n

n−

9. The velocity displacement graph of a particle moving along a straight line is shown.

v

x

v0

x0

The most suitable acceleration-displacement graph will be [JEE Scr. 2005; 3]

(a)

a

x

(b)

a

x

(c)

a

x

(d)

a

x

10. Statement-1 : For an observer looking out through the window of a fast moving train, the nearby objects appear to movein the opposite direction to the train, while the distant objects appear to be stationary.

Statement-2 : If the observer and the object are moving at velocities 1Vr

and 2Vr

respectively with reference to a laboratory

frame, the velocity of the object with respect to the observer is 1 2V V−r r

. [JEE 2008, 3/163]

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(b) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(c) Statement-1 is True, Statement-2 is False

(d) Statement-1 is False, Statement-2 is True.



Answer Key


1. (c) 2. 12 sec 3. 2.5 m/s2 4. (a) 60 miles/hr, (b) zero

5. (a) → (ii), (b) → (i), (c) → (iii)


1. 48 m 2. 5 m/sec 3. 0.5 m 4. (a) 5. (b) 6. (a)7. 50 m, 2 m/s2, 20 m

8. (a) 10 m, (b) 10( )2 1− m/sec upwards and 10 m/s2 downwards, (c) ( )54 2 1

2− m/sec upwards.

9. 4 + 2 6 sec after the stone was thrown, 129 m below the top of the shaft.10. (a) yes, (b) (i) yes, (ii) no, (iii) yes


1. t = 15 sec 2. (a) 3. a — [t2, t

3], b — [t

3, t

4], c — [t

1, t

2], d — [t

4, t

5] 4. (d)

5. (b) 6. (c)


1. v = 4 16x + ) 2. x = 5

2(e2t – 1) 3. v

0 =

5

6m/sec, v

5 =

55

6− m/sec 4. (a)

5. (c)


1. ˆ ˆ45 151i j− 2. 5, 125 3. (a) → (iii), (b) → (iii), (c) → (iv)


1. (a) u cosθ , (b) more, (c) 2

π2. (a) 3. (b) 4. 182 m/sec

5. (a) → (ii), (b) → (i) 6. 10 3 7. Width = 9.6 m, Angle = 45° or tan– t3

2

8. t = cosecu

g

α


1. 20 sec 2. (a) 20 m/sec toward west, (b) 25 m/sec at 37° North of west, (c) 25 m/sec.

3. ˆˆ ˆ2i j k+ + where i → East, j → North, k → Upwards 4. (5 5 m/sec)

5. (a) 127° w.r.t river flow, (b) 4 kmph perpendicular to river flow, (c) 15 min


1. (c) 2. (a), (b), (c) & (d) 3. (a), (b), (c) & (d)

4. (b), (c) 5. ω= 505

2rad/sec 6. t =

1 3

2 2sec

7. (a) t = 2 sec, (b) πm/sec, (c) 2π1 4π

2+ m/sec2 9. (b)



Level - 1

1. (b) 2. (a) 3. (d) 4. (c) 5. (a) 6. (a)

7. (c) 8. (c) 9. (d) 10. (c) 11. (a), (b), (c) 12.(b), (c), (d)

13. (a) 14. (a), (b), (c) 15. (b) 16. (a) 17. (b) 18. (a)

19. (a) 20. (b) 21. (c)

22. (a) → (p), (b) → (q), (c) → (r), (d) → (s) 23. (a) → (r), (b) → (p), (c) → (s), (d) → (q)

Level - 2

1. (c) 2. (c) 3. (c) 4. (d) 5. (d) 6. (b)

7. (d) 8. (d) 9. (d) 10. (a) 11. (a), (c), (d) 12. (a), (b), (c), (d)13. (a), (c) 14. (a), (c) 15. (a), (d) 16. (a) 17. (a) 18. (d)

19. (b) 20. (a) 21. (b) 22. (b) 23. (a) 24. (b)25. (a) → (s), (b) → (q), (c) → (p), (d) → (p)26. (a) → (r, s), (b) → (r, s), (c) → (p, q, r), (d) → (p, q, r)

Answer Key ; Previous Years’ IITJEE Questions

1. (a) ˆ1v i= −r m/sec, (b) π 3

3 4t = + 2. (b) 3. (a), (b) 4. (a)

5. u = 425

8m/sec, t = 1 sec 6. (a) 45°, (b) 2 m/sec 7. (c) 8. (b)

9. (b) 10. (b)

kinematics

Documents

hand thumb

v2 cosa v1

mouth opening

comprehensiontypecomprehension

equilateral

finite angular

uniform circular

particle turns