CONTENTS

2.1

Position

2.2

Rest and motion

2.3

Types of motion

2.4

Point mass

2.5

Distance and displacement

2.6

Speed and velocity

2.7

Acceleration

2.8

Position-time graph

2.9

Velocity-time graph

2.10

Equations of kinematics

2.11

Motion under gravity

2.12

Motion with variable acceleration

Sample Problems

Practice Problems (Basic and Advance Level)

Answer Sheet of Practice Problems

Free fall of Feather and Apple (in vacuum)

from the Pisa tower are the examples of one

dimensional motion under gravity.

But interesting thing is that they both will

reach the ground simultaneously. Because

time of fall (in vacuum) does not depend upon

the mass of the falling body.

This was the result of famous Galileo’s Pisa

tower experiment.

Chapter

2

86 Motion In One Dimension

2.1 Position.

Any object is situated at point

O

and three observers from three different places are looking for same

object, then all three observers will have different observations about the position of point

O

and no one will be

wrong. Because they are observing the object from their different positions.

Observer ‘

A

’ says : Point

O

is 3

m

away in west direction.

Observer ‘

B

’ says : Point

O

is 4

m

away in south direction.

Observer ‘

C

’ says : Point

O

is 5

m

away in east direction.

Therefore position of any point is completely expressed

by two factors: Its distance from the observer and its direction

with respect to observer.

That is why position is characterised by a vector known as position vector.

Let point

P

is in a

xy

plane and its coordinates are (

x

,

y

). Then position vector

)(r



of point will be

jyix ˆˆ 

and if the point

P

is in a space and its coordinates are (

x

,

y

,

z

) then position vector can be expressed as

.

ˆ

ˆˆ kzjyixr 



2.2 Rest and Motion.

If a body does not change its position as time passes with respect to frame of reference, it is said to be at

rest.

And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion.

Frame of Reference : It is a system to which a set of coordinates are attached and with reference to which observer

describes any event.

C

4

m

3

m

E

W

N

5

m

O

A

B

S

Motion In One Dimension 87

A passenger standing on platform observes that tree on a platform is at rest. But when the same passenger

is passing away in a train through station, observes that tree is in motion. In both conditions observer is right.

But observations are different because in first situation observer stands on a platform, which is reference frame

at rest and in second situation observer moving in train, which is reference frame in motion.

So rest and motion are relative terms. It depends upon the frame of references.

2.3 Types of Motion.

One dimensional

Two dimensional

Three dimensional

Motion of a body in a straight line is

called one dimensional motion.

Motion of body in a plane is called

two dimensional motion.

Motion of body in a space is called

three dimensional motion.

When only one coordinate of the

position of a body changes with time

then it is said to be moving one

dimensionally.

When two coordinates of the position

of a body changes with time then it is

said to be moving two dimensionally.

When all three coordinates of the

position of a body changes with time

then it is said to be moving three

dimensionally.

e.g.

. Motion of car on a straight road.

Motion of freely falling body.

e.g

. Motion of car on a circular turn.

Motion of billiards ball.

e.g.

. Motion of flying kite.

Motion of flying insect.

2.4 Particle or Point Mass.

The smallest part of matter with zero dimension which can be described by its mass and position is defined

as a particle.

Tree is at

rest

Platform (Frame of reference)

Moving train (Frame of reference)

Tree is in

motion

88 Motion In One Dimension

If the size of a body is negligible in comparison to its range of motion then that body is known as a

particle.

A body (Group of particles) to be known as a particle depends upon types of motion. For example in a

planetary motion around the sun the different planets can be presumed to be the particles.

In above consideration when we treat body as particle, all parts of the body undergo same displacement

and have same velocity and acceleration.

2.5 Distance and Displacement.

(1) Distance : It is the actual path length covered by a moving particle in a given interval of time.

(i) If a particle starts from

A

and reach to

C

through point

B

as shown in the figure.

Then distance travelled by particle

7 BCAB

m

(ii) Distance is a scalar quantity.

(iii) Dimension : [

M

0

L

1

T

0]

(iv) Unit :

metre

(S.I.)

(2) Displacement : Displacement is the change in position vector

i

.

e

., A vector joining initial to final position.

(i) Displacement is a vector quantity

(ii) Dimension : [

M

0

L

1

T

0]

(iii) Unit :

metre

(S.I.)

(iv) In the above figure the displacement of the particle

BCABAC 



|| AC

o

BCABBCAB 90cos)()(2)()( 22 

= 5

m

(v) If

n

SSSS  ........,, 321

are the displacements of a body then the total (net) displacement is the vector sum

of the individuals.

n

SSSSS   ........

321

(3) Comparison between distance and displacement :

(i) The magnitude of displacement is equal to minimum possible distance between two positions.

So distance



|Displacement|.

4

m

3

m

B

C

A

Motion In One Dimension 89

(ii) For a moving particle distance can never be negative or zero while displacement can be.

(zero displacement means that body after motion has came back to initial position)

i.e.

, Distance > 0 but Displacement > = or < 0

(iii) For motion between two points displacement is single valued while distance depends on actual path

and so can have many values.

(iv) For a moving particle distance can never decrease with time while displacement can. Decrease in

displacement with time means body is moving towards the initial position.

(v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is

along a straight line without change in direction.

(vi) If

A

r



and

B

r



are the position vectors of particle initially and finally.

Then displacement of the particle

ABAB rrr  

and s is the distance travelled if the particle has gone through the path

APB

.

Sample problems based on distance and displacement

Problem

1. A man goes 10

m

towards North, then 20

m

towards east then displacement is

[KCET (Med.) 1999; JIPMER 1999; AFMC 2003]

(a) 22.5

m

(b) 25

m

(c) 25.5

m

(d) 30

m

Solution

: (a) If we take east as

x

axis and north as

y

axis, then displacement

ji ˆ

10

ˆ

20 

So, magnitude of displacement

5101020 22 

= 22.5

m

.

Problem

2. A body moves over one fourth of a circular arc in a circle of radius

r.

The

magnitude of distance travelled

and displacement will be respectively

(a)

2,

2r

r



(b)

r

r,

4



(c)

2

,r

r



(d)

rr,



Solution

: (a) Let particle start from

A

, its position vector

irr A

Oˆ





Y

X

A

B

AB

r



B

r



A

r



s

Y

X

O

A

B

90 Motion In One Dimension

After one quarter position vector

.

ˆ

jrrOB 



So displacement

irjr ˆˆ 

Magnitude of displacement

.2r

and distance = one fourth of circumference

24

2rr





Problem

3. The displacement of the point of the wheel initially in contact with the ground, when the wheel roles

forward half a revolution will be (radius of the wheel is

R

)

(a)

4

2



R

(b)

4

2



R

(c)

R



2

(d)

R



Solution

: (b) Horizontal distance covered by the wheel in half revolution =

R

So the displacement of the point which was initially in contact with a

ground =

22 )2()( RR 



.4

2



R

2.6 Speed and Velocity.

(1) Speed : Rate of distance covered with time is called speed.

(i) It is a scalar quantity having symbol



.

(ii) Dimension : [

M

0

L

1

T

–1]

(iii) Unit :

metre/second

(S.I.),

cm

/

second

(C.G.S.)

(iv) Types of speed :

(a) Uniform speed : When a particle covers equal distances in equal intervals of time, (no matter how small

the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal

distance (= 5

m

) in each second. So we can say that particle is moving with uniform speed of 5

m

/

s

.

Distance

Time

Uniform Speed

5

m

5

m

1

sec

1

sec

1

sec

1

sec

1

sec

5

m

5

m

5

m

5m/

s

5m/

s

5m/s

5m/

s

5m/

s

5

m

5

m/s

1

m/s

2

R

P

new

R

P

initial

Motion In One Dimension 91

(b) Non-uniform (variable) speed : In non-uniform speed particle covers unequal distances in equal

intervals of time. In the given illustration motorcyclist travels 5

m

in 1st second, 8

m

in 2nd second, 10

m

in 3rd

second, 4

m

in 4th second

etc

.

Therefore its speed is different for every time interval of one second. This means particle is moving with

variable speed.

(c) Average speed : The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of

distance travelled to the time taken.

Average speed

takenTime

travelledDistance



;

t

s

vav 







Time average speed

: When particle moves with different uniform speed

1



,

2



,

3



...

etc

in different

time intervals

1

t

,

2

t

,

3

t

, ...

etc

respectively, its average speed over the total time of journey is given as

elapsed Total time

covered distanceTotal



av

v

......

321



ttt

ddd

=

......

321

332211



ttt



Special case : When particle moves with speed

v

1 upto half time of its total motion and in rest time it is

moving with speed

v

2 then

221 vv

vav





Distance

Time

Variable Speed

5

m

8

m

1

sec

1

sec

1

sec

1

sec

1

sec

10

m

4

m

6

m

5

m

/

s

8

m

/

s

10

m

/

s

4

m

/

s

6

m

/

s

7

m

1

sec

7

m

/

s

92 Motion In One Dimension



Distance averaged speed

: When a particle describes different distances

1

d

,

2

d

,

3

d

, ...... with different

time intervals

1

t

,

2

t

,

3

t

, ...... with speeds

......,, 321 vvv

respectively then the speed of particle averaged over the

total distance can be given as

elapsed Total time

covered distanceTotal



av



......

321



ttt

ddd

......

3

2

1

321







ddd

 When particle moves the first half of a distance at a speed of

v

1 and second half of the distance at speed

v

2

then

21

2

vv

vav 



 When particle covers one-third distance at speed

v

1, next one third at speed

v

2 and last one third at speed

v

3,

then

133221

321

3

vvvvvv

vvv

vav 



(d) Instantaneous speed : It is the speed of a particle at particular instant. When we say “speed”, it usually

means instantaneous speed.

The instantaneous speed is average speed for infinitesimally small time interval (

i

.

e

.,

0t

). Thus

Instantaneous speed

t

s

vt



 0

lim

dt

ds



(2) Velocity : Rate of change of position

i.e.

rate of displacement with time is called velocity.

(i) It is a scalar quantity having symbol

v

.

(ii) Dimension : [

M

0

L

1

T

–1]

(iii) Unit :

metre/second

(S.I.),

cm

/

second

(C.G.S.)

(iv) Types

(a) Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its

velocity remains same and this is possible only when the particles moves in same straight line without reversing

its direction.

Motion In One Dimension 93

(b) Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or

direction of velocity changes (or both changes).

(c) Average velocity : It is defined as the ratio of displacement to time taken by the body

takenTime

ntDisplaceme

velocityAverage 

;

t

r

vav 







(d) Instantaneous velocity : Instantaneous velocity is defined as rate of change of position vector of

particles with time at a certain instant of time.

Instantaneous velocity

t

r

vt







0

lim

dt

rd



(v) Comparison between instantaneous speed and instantaneous velocity

(a) instantaneous velocity is always tangential to the path followed by

the particle.

When a stone is thrown from point

O

then at point of projection the

instantaneous velocity of stone is

1

v

, at point

A

the instantaneous velocity

of stone is

2

v

, similarly at point

B

and

C

are

3

v

and

4

v

respectively.

Direction of these velocities can be found out by drawing a tangent on the trajectory at a given point.

(b) A particle may have constant instantaneous speed but variable instantaneous velocity.

Example

: When a particle is performing uniform circular motion then for every instant of its circular motion

its speed remains constant but velocity changes at every instant.

(c) The magnitude of instantaneous velocity is equal to the instantaneous speed.

(d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are

always equal.

(e) If displacement is given as a function of time, then time derivative of displacement will give velocity.

Let displacement

2

210 tAtAAx 



Instantaneous velocity

)( 2

210 tAtAA

dt

d

dt

xd

v



tAAv 21 2





2

X

O

v

1

Y



3



4

C

B

A

94 Motion In One Dimension

For the given value of

t

, we can find out the instantaneous velocity.

e.g.

for

0t

,Instantaneous velocity

1

Av 



and Instantaneous speed

1

|| Av 



(vi) Comparison between average speed and average velocity

(a) Average speed is scalar while average velocity is a vector both having same units (

m

/

s

) and

dimensions

][ 1

LT

.

(b) Average speed or velocity depends on time interval over which it is defined.

(c) For a given time interval average velocity is single valued while average speed can have many values

depending on path followed.

(d) If after motion body comes back to its initial position then

0





av

v

(as

0r



) but

0





av

v

and finite

as

)0( s

.

(e) For a moving body average speed can never be negative or zero (unless

)t

while average velocity

can be

i

.

e

.

0

av

v

while



a

v



= or < 0.

Sample problems based on speed and velocity

Problem

4. If a car covers 2/5th of the total distance with

v

1 speed and 3/5th distance with

v

2 then average speed is

[MP PMT 2003]

(a)

21

2

1vv

(b)

2

21 vv 

(c)

21

2

vv



(d)

21

23

5

vv



Solution

: (d) Average speed =

takenTotal time

travelleddistanceTotal

21 tt

x





=

21

)5/3(

)5/2(

v

x

v

x



12

21 32

5

vv





Problem

5. A car accelerated from initial position and then returned at initial point, then [AIEEE 2002]

(a) Velocity is zero but speed increases (b) Speed is zero but velocity increases

(c) Both speed and velocity increase (d) Both speed and velocity decrease

Solution

: (a) As the net displacement = 0

Hence velocity = 0 ; but speed increases.

t

1

t

2

(2/5)

x

(3/5)

x

Motion In One Dimension 95

Note

: 

1

|speed Average|

| velocityAverage| 

|velocityAv.||speedAv.| 

Problem

6. A man walks on a straight road from his home to a market 2.5

km

away with a speed of 5

km

/

h

. Finding

the market closed, he instantly turns and walks back home with a speed of 7.5

km

/

h

. The average speed of

the man over the interval of time 0 to 40

min

. is equal to [AMU (Med.) 2002]

(a) 5

km

/

h

(b)

4

25

km

/

h

(c)

4

30

km

/

h

(d)

8

45

km

/

h

Solution

: (d) Time taken in going to market

.min30

2

1

5

5.2  hr

As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10

min.

So distance travelled in remaining 10 min =

.25.1

60

10

5.7 km

Hence, average speed =

Total time

distanceTotal

=

hrkm

hr

km /

8

45

.)60/40(

)25.15.2( 



.

Problem

7. The relation

633  xt

describes the displacement of a particle in one direction where

x

is in

metres

and

t

in sec. The displacement, when velocity is zero, is [CPMT 2000]

(a) 24

metres

(b) 12

metres

(c) 5

metres

(d) Zero

Solution

: (d)

633  xt



)63(3  tx



2

)63(3  tx



121232 ttx



v

=

126)12123( 2 ttt

dt

d

dt

dx

If velocity = 0 then,

0126t

sect 2

Hence at

t

= 2,

x

= 3(2)2 – 12 (2) + 12 = 0

metres

.

Problem

8. The motion of a particle is described by the equation

2

btax 

where

15a

cm

and

3b

cm

. Its

instantaneous velocity at time 3

sec

will be [AMU (Med.) 2000]

(a) 36

cm/sec

(b) 18

cm/sec

(c) 16

cm/sec

(d) 32

cm/sec

Solution

: (b)

2

btax 



v

=

bt

dt

dx 20 

At

t

= 3

sec

,

v

=

332 

=

seccm /18

(As

cmb3

)

96 Motion In One Dimension

Problem

9. A train has a speed of 60

km/h

for the first one hour and 40

km/h

for the next half hour. Its average speed

in

km/h

is [JIPMER 1999]

(a) 50 (b) 53.33 (c) 48 (d) 70

Solution

: (b) Total distance travelled =

km80

2

1

40160 

and Total time taken =

hrhrhr 2

3

2

1

1

 Average speed

33.53

23

80 

km/h

Problem

10. A person completes half of its his journey with speed

1



and rest half with speed

2



. The average speed

of the person is [RPET 1993; MP PMT 2001]

(a)

221









(b)

21

2









(c)

21









(d)

21





Solution

: (b) In this problem total distance is divided into two equal parts. So

2

1

21





dd dd

av 





=

21

2/2/ 22



dd





21

11 2









av

=

21

2





Problem

11. A car moving on a straight road covers one third of the distance with 20

km/hr

and the rest with 60

km/hr

.

The average speed is [MP PMT 1999; CPMT 2002]

(a) 40

km/hr

(b) 80

km/hr

(c)

hrkm /

3

2

46

(d) 36

km/hr

Solution

: (d) Let total distance travelled =

x

and total time taken

t

1 +

t

2

60

3/2

20

3/ xx 

 Average speed

180

2

60

11

60

)3/2(

20

)3/1( 





xx

x

=

hrkm /36

2.7 Acceleration.

The time rate of change of velocity of an object is called acceleration of the object.

(1) It is a vector quantity. It’s direction is same as that of change in velocity (Not of the velocity)

(2) There are three possible ways by which change in velocity may occur

When only direction of velocity

changes

When only magnitude of velocity

changes

When both magnitude and direction

of velocity changes

Acceleration perpendicular to

Acceleration parallel or anti-parallel

Acceleration has two components one

Motion In One Dimension 97

velocity

to velocity

is perpendicular to velocity and

another parallel or anti-parallel to

velocity

e.g.

Uniform circular motion

e.g.

Motion under gravity

e.g.

Projectile motion

(3) Dimension : [

M

0

L

1

T

–2]

(4) Unit :

metre/second

2 (S.I.);

cm

/

second

2 (C.G.S.)

(5) Types of acceleration :

(i) Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the

acceleration remains constant during particle motion.

Note :  If a particle is moving with uniform acceleration, this does not necessarily imply that particle is

moving in straight line.

e.g.

Projectile motion.

(ii) Non-uniform acceleration : A body is said to have non-uniform acceleration, if magnitude or direction

or both, change during motion.

(iii) Average acceleration :

t

vv

t

v

aa







12







The direction of average acceleration vector is the direction of the change in velocity vector as

t

v

a







(iv) Instantaneous acceleration =

dt

vd

t

v

at







 0

lim

(v) For a moving body there is no relation between the direction of instantaneous velocity and direction of

acceleration.

X

O

Y



2



3



g



1



a





a

98 Motion In One Dimension

e.g.

(a) In uniform circular motion



= 90º always

(b) In a projectile motion



is variable for every point of trajectory.

(vi) If a force

F

acts on a particle of mass

m

, by Newton’s 2nd law, acceleration

m

F

a





(vii) By definition

2

dt

xd

dt

vd

a















dt

xd

v



As

i.e.,

if

x

is given as a function of time, second time derivative of displacement gives acceleration

(viii) If velocity is given as a function of position, then by chain rule











 dt

dx

v

dx

d

v

dt

dx

dv

dt

dv

aas.



(ix) If a particle is accelerated for a time

t

1 by acceleration

a

1 and for time

t

2 by acceleration

a

2 then average

acceleration is

21

2211 tt

tata

aa







(x) If same force is applied on two bodies of different masses

1

m

and

2

m

separately then it produces accelerations

1

a

and

2

a

respectively. Now these bodies are attached together and form a

combined system and same force is applied on that system so that

a be the acceleration of the combined system, then

 

ammF 21 



21 a

F

a

F

a

F

So,

21

111

aaa 



21

21 aa

aa

a



(xi) Acceleration can be positive, zero or negative. Positive acceleration means velocity increasing with time,

zero acceleration means velocity is uniform constant while negative acceleration (retardation) means velocity is

decreasing with time.

(xii) For motion of a body under gravity, acceleration will be equal to “

g

”, where g is the acceleration due to

gravity. Its normal value is

2

m/s8.9

or

2

cm/s980

or

2

feet/s32

.

Sample problems based on acceleration

a

1

F

m

1

a

2

F

m

2

m

2

m

1

a

F

Motion In One Dimension 99

Problem

12. The displacement of a particle, moving in a straight line, is given by

422 2 tts

where

s

is in

metres

and

t

in seconds. The acceleration of the particle is [CPMT 2001]

(a) 2

m

/

s

2 (b) 4

m

/

s

2 (c) 6

m

/

s

2 (d) 8

m

/

s

2

Solution

: (b) Given

422 2 tts

 velocity (

v

)

24  t

dt

ds

and acceleration (

a

)

2

/40)1(4 sm

dt

dv 

Problem

13. The position

x

of a particle varies with time

t

as

.

32 btatx

The acceleration of the particle will be zero

at time

t

equal to [CBSE PMT 1997; BHU 1999; DPMT 2000; KCET (Med.) 2000]

(a)

b

a

(b)

b

a

3

2

(c)

b

a

3

(d) Zero

Solution

: (c) Given

32 btatx

 velocity

2

32)( btat

dt

dx

v

and acceleration (

a

) =

.62 bta

dt

dv 

When acceleration = 0 

062  bta



b

a

b

a

t36

2

.

Problem

14. The displacement of the particle is given by

.

42 dtctbtay 

The initial velocity and acceleration are

respectively [CPMT 1999, 2003]

(a)

db 4,

(b)

cb 2,

(c)

cb 2,

(d)

dc 4,2 

Solution

: (c) Given

42 dtctbtay 



v

=

3

420 dtctb

dt

dy 

Putting

,0t

v

initial =

b

So initial velocity =

b

Now, acceleration (

a

)

2

1220 tdc

dt

dv 

Putting

t

= 0,

a

initial

= 2

c

Problem

15. The relation between time

t

and distance

x

is

,

2xxt





where



and



are constants. The retardation

is (

v

is the velocity) [NCERT 1982]

(a)

3

2v



(b)

3

2v



(c)

3

2v



(d)

32

2v



Solution

: (a) differentiating time with respect to distance



 x

dx

dt 2







 xdt

dx

v2

1

So, acceleration (

a

) =

dt

dx

dv

dt

dv .

=

32

22..2

)2(

2. vvv

x

v

dx

dv

v















Problem

16. If displacement of a particle is directly proportional to the square of time. Then particle is moving with

[RPET 1999]

(a) Uniform acceleration (b) Variable acceleration

(c) Uniform velocity (d) Variable acceleration but uniform velocity

Solution

: (a) Given that

2

tx 

or

2

Ktx

(where

K

= constant)

100 Motion In One Dimension

Velocity (

v

)

Kt

dt

dx 2

and Acceleration (

a

)

K

dt

d2



It is clear that velocity is time dependent and acceleration does not depend on time.

So we can say that particle is moving with uniform acceleration but variable velocity.

Problem

17. A particle is moving eastwards with velocity of 5

m

/

s

. In 10

sec

the velocity changes to 5

m

/

s

northwards.

The average acceleration in this time is [IIT-JEE 1982]

(a) Zero (b)

2

m/s

2

1

toward north-west

(c)

2

m/s

2

1

toward north-east (d)

2

m/s

2

1

toward north-west

Solution

: (b)

12



 

255590cos2 22

21

2

1 o



25



Average acceleration

2

m/s

2

1

10

25 



t



toward north-

west (As clear from the figure).

Problem

18. A body starts from the origin and moves along the

x

-axis such that velocity at any instant is given by

)24( 3tt 

, where

t

is in second and velocity is in

m

/

s

. What is the acceleration of the particle, when it is 2

m

from the origin? [IIT-JEE 1982]

(a)

2

28 m/s

(b)

2

22 m/s

(c)

2

/12 sm

(d)

2

/10 sm

Solution

: (b) Given that

tt 24 3





dtx



,

Cttx  24

, at

00,0  Cxt

When particle is 2

m

away from the origin

24

2tt 



02

24  tt



0)1()2( 22  tt



2t

sec

 

21224 23  ttt

td

d

td

d

a





212 2 ta

for

2t

sec



 

2212 2a



2

/22 sma 

1

v





sm /5

2











90o

sm /5

1





Motion In One Dimension 101

Problem

19. A body of mass 10

kg

is moving with a constant velocity of 10

m

/

s

. When a constant force acts for 4

sec

on

it, it moves with a velocity 2

m

/

sec

in the opposite direction. The acceleration produced in it is [MP PET 1997]

(a)

2

3m/s

(b)

2

3m/s

(c)

2

3.0 m/s

(d)

2

3.0 m/s

Solution

: (b) Let particle moves towards east and by the application of constant force it moves towards west

sm/10

1





and

sm/2

2





. Acceleration

Time

velocityin Change



t12









4

)10()2( 

a

4

12



2

/3 sm

2.8 Position Time Graph.

During motion of the particle its parameters of kinematical analysis (

u

,

v

,

a

,

r

) changes with time. This can

be represented on the graph.

Position time graph is plotted by taking time

t

along

x

-axis and position of the particle on

y

-axis.

Let

AB

is a position-time graph for any moving particle

As Velocity =

12

takenTime

positionin Change

tt

yy





…(i)

From triangle

ABC

12

tan tt

yy

AC

AD

AC

BC







….(ii)

By comparing (i) and (ii) Velocity = tan



v

= tan



It is clear that slope of position-time graph represents the velocity of the particle.

Various position – time graphs and their interpretation

Time

A

B

C

D

x

y

2

y

1

O

t

1

t

2

Position



102 Motion In One Dimension



= 0o so

v

= 0

i

.

e

., line parallel to time axis represents that the particle is at rest.



= 90o so

v

= 

i

.

e

., line perpendicular to time axis represents that particle is changing its position

but time does not changes it means the particle possesses infinite velocity.

Practically this is not possible.



= constant so

v

= constant,

a

= 0

i

.

e

., line with constant slope represents uniform velocity of the particle.



is increasing so

v

is increasing,

a

is positive.

i

.

e

., line bending towards position axis represents increasing velocity of particle.

It means the particle possesses acceleration.



is decreasing so

v

is decreasing,

a

is negative

i

.

e

., line bending towards time axis represents decreasing velocity of the particle.

It means the particle possesses retardation.



constant but > 90o so

v

will be constant but negative

i

.

e

., line with negative slope represent that particle returns towards the point of

reference. (negative displacement).

T

P

O

T

P

O

T

P

O

T

P

O

T

P

O

T

P



Motion In One Dimension 103

Straight line segments of different slopes represent that velocity of the body

changes after certain interval of time.

This graph shows that at one instant the particle has two positions. Which is not

possible.

The graph shows that particle coming towards origin initially and after that it is

moving away from origin.

Note :  If the graph is plotted between distance and time then it is always an increasing curve and it

never comes back towards origin because distance never decrease with

time. Hence such type of distance time graph is valid up to point A

only, after point A it is not valid as shown in the figure.

 For two particles having displacement time graph with slopes



1 and



2

possesses velocities

v

1 and

v

2 respectively then

2

1

2

1tan

tan







Sample problems based on position-time graph

Problem

20. The position of a particle moving along the

x

-axis at certain times is given below :

t

(

s

)

0

1

2

3

x

(

m

)

– 2

0

6

16

Which of the following describes the motion correctly [AMU (Engg.) 2001]

(a) Uniform, accelerated (b) Uniform, decelerated

(c) Non-uniform, accelerated (d) There is not enough data for generalisation

Solution

: (a) Instantaneous velocity

t

x

v





, By using the data from the table

O

T

P

S

C

B

A

O

T

P

O

T

P

A

Distance

O

Time

104 Motion In One Dimension

smvsmv /6

1

06

,/2

1

)2(0 21 









and

smv /10

1

616

3





i.e.

the speed is increasing at a constant

rate so motion is uniformly accelerated.

Problem

21. Which of the following graph represents uniform motion [DCE 1999]

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

Solution

: (a) When distance time graph is a straight line with constant slope than motion is uniform.

Problem

22. The displacement-time graph for two particles

A

and

B

are straight lines inclined at angles of 30o and 60o

with the time axis. The ratio of velocities of

vA

:

vB

is [CPMT 1990; MP PET 1999; MP PET 2001]

(a) 1 : 2 (b)

3:1

(c)

1:3

(d) 1 : 3

Solution

: (d)



tanv

from displacement graph. So

3

1

33

1

3

3/1

60tan

30tan 



 o

o

B

A

v

Problem

23. From the following displacement time graph find out the velocity of a moving body

(a)

3

1

m

/

s

(b) 3

m

/

s

(c)

3

m

/

s

(d)

3

1

Solution

: (c) In first instant you will apply



tan

and say,

3

1

30tan  o



m

/

s

.

But it is wrong because formula



tan

is valid when angle is measured with time axis.

Here angle is taken from displacement axis. So angle from time axis

ooo 603090 

.

Now

360tan  o



Problem

24. The diagram shows the displacement-time graph for a particle moving in a straight line. The average

velocity for the interval

t

= 0,

t

= 5 is

(a) 0

(b) 6

ms

–1

O

30o

Time (

sec

)

Displacement (meter)

2

4

5

t

O

– 10

10

20

x

t

s

t

s

t

s

t

s

Motion In One Dimension 105

(c) – 2

ms

–1

(d) 2

ms

–1

Solution

: (c) Average velocity =

timeTotal

ntdisplacemeTotal

=

     

5

102020 

= –2

m/s

Problem

25. Figure shows the displacement time graph of a body. What is the ratio of the speed in the first second and

that in the next two seconds

(a) 1 : 2

(b) 1 : 3

(c) 3 : 1

(d) 2 : 1

Solution

: (d) Speed in first second = 30 and Speed in next two seconds = 15. So that ratio 2 : 1

2.9 Velocity Time Graph.

The graph is plotted by taking time

t

along

x

-axis and velocity of the particle on

y

-axis.

Distance and displacement : The area covered between the velocity time graph and time axis gives the

displacement and distance travelled by the body for a given time interval.

Then Total distance

|||||| 321 AAA 

= Addition of modulus of different area.

i.e.



dts||



Total displacement

321 AAA 

= Addition of different area considering their sign.

i.e.



dtr



here

A

1 and

A

2 are area of triangle 1 and 2 respectively and

A

3 is the area of trapezium .

Acceleration : Let

AB

is a velocity-time graph for any moving particle

As Acceleration =

12

takenTime

velocityin Change

tt

vv





…(i)

From triangle

ABC

,

12

tan tt

vv

AC

AD

AC

BC







….(ii)

By comparing (i) and (ii)

Acceleration (

a

) =



tan

Time

A

B

C

D

x

y

v

2

v

1

O

t

1

t

2

Velocity



+



–



t

1

2

3

1

2

3

0

10

20

30

Y

X

Displacement

Time

106 Motion In One Dimension

It is clear that slope of velocity-time graph represents the acceleration of the particle.

Motion In One Dimension 107

Various velocity – time graphs and their interpretation



= 0,

a

= 0,

v

= constant

i

.

e

., line parallel to time axis represents that the particle is moving with constant

velocity.



= 90o,

a

= ,

v

= increasing

i

.

e

., line perpendicular to time axis represents that the particle is increasing its velocity,

but time does not change. It means the particle possesses infinite acceleration.

Practically it is not possible.



=constant, so

a

= constant and

v

is increasing uniformly with time

i

.

e

., line with constant slope represents uniform acceleration of the particle.



increasing so acceleration increasing

i

.

e

., line bending towards velocity axis represent the increasing acceleration in the

body.



decreasing so acceleration decreasing

i

.

e

. line bending towards time axis represents the decreasing acceleration in the body

Positive constant acceleration because



is constant and < 90o but initial velocity of the

particle is negative.

Time

O

Velocity

Time

O

Velocity

Time

O

Velocity

Time

O

Velocity

O

Time

Velocity

O

Time

108 Motion In One Dimension

Positive constant acceleration because



is constant and < 90o but initial velocity of

particle is positive.

Negative constant acceleration because



is constant and > 90o but initial velocity of

the particle is positive.

Negative constant acceleration because



is constant and > 90o but initial velocity of

the particle is zero.

Negative constant acceleration because



is constant and > 90o but initial velocity of

the particle is negative.

Sample problems based on velocity-time graph

Problem

26. A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the

ball during its flight if the air resistance is not ignored [AIIMS 2003]

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

Solution

:

(c)

In first half of motion the acceleration is uniform & velocity gradually decreases, so slope will be negative

but for next half acceleration is positive. So slope will be positive. Thus graph '

C

' is correct.

Not ignoring air resistance means upward motion will have acceleration (

a

+

g

) and the downward motion

will have

).(ag 

O

Time

Velocity

O

Time

Velocity

O

Time

Velocity

O

Time

Velocity

Speed

Time

Speed

Time

Speed

Time

Speed

Time

Motion In One Dimension 109

Problem

27. A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is

shown in the figure. The maximum acceleration during the journey is [Kerala (Engg.) 2002]

(a) 140

km h

–2 (b) 160

km h

–2 (c) 100

km h

–2 (d) 120

km h

–2

Solution

:

(b) Maximum acceleration means maximum slope in speed – time graph.

that slope is for line

CD

. So,



max

a

slope of

CD

=

2

160

25.0

40

00.125.1

2060 





hkm

.

Problem

28. The graph of displacement

v

/

s

time is

Its corresponding velocity-time graph will be [DCE 2001]

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

S

t

A

0.25

0.75

1.00

1.5

2.00

E

M

N

B

C

L

D

100

80

60

40

20

Time in hours

Speed in

km

/

hours

V

t

V

t

V

t

110 Motion In One Dimension

Solution

:

(a) We know that the velocity of body is given by the slope of displacement – time graph. So it is clear that

initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of

the graph) and then it will be negative.

Problem

29. In the following graph, distance travelled by the body in

metres

is [EAMCET 1994]

(a) 200

(b) 250

(c) 300

(d) 400

Solution

:

(a) Distance = The area under

v

–

t

graph

2

1

S

10)1030(

= 200 metre

Problem

30. For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its

motion is what fraction of the total distance covered by it in all the seven seconds [MP PMT/PET 1998; RPET 2001]

(a)

2

1

(b)

4

1

(c)

3

1

(d)

3

2

Solution

:

(b) Distance covered in total 7 seconds = Area of trapezium

ABCD

10)62(

2

1

= 40

m

Distance covered in last 2 second = area of triangle

CDQ=

10102

2

1

m

So required fraction

4

1

40

10 

Problem

31. The velocity time graph of a body moving in a straight line is shown in the figure. The displacement and

distance travelled by the body in 6

sec

are respectively [MP PET 1994]

(a)

mm 16,8

(b)

mm 8,16

10

20

30

0

5

10

15

Y

X

Time (

s

)

v

(

m/s

)

3

5

7

0

5

10

Time (

s

)

v

(

m/s

)

1

A

B

C

D

P

Q

4

2

O

– 4

– 2

2

4

6

t

(

sec

)

v (m/s

)

A

B

C

D

E

F

G

H

I

Motion In One Dimension 111

(c)

mm 16,16

(d)

mm 8,8

Solution

:

(a) Area of rectangle

ABCO

= 4  2 = 8

m

Area of rectangle

CDEF

= 2 × (– 2) = – 4

m

Area of rectangle

FGHI

= 2×2 = 4

m

Displacement = sum of area with their sign = 8 + (– 4) + 4 = 8

m

Distance = sum of area with out sign = 8 + 4 + 4 = 16

m

Problem

32. A ball is thrown vertically upward which of the following graph represents velocity time graph of the ball

during its flight (air resistance is neglected) [CPMT 1993; AMU (Engg.) 2000]

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

Solution

: (d) In the positive region the velocity decreases linearly (during rise) and in negative region velocity increase

linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in

the negative region.

Problem

33. A ball is dropped vertically from a height

d

above the ground. It hits the ground and bounces up vertically

to a height

2

d

. Neglecting subsequent motion and air resistance, its velocity



varies with the height

h

above the ground as. [IIT-JEE (Screening) 2000]

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

Solution

: (a) When ball is dropped from height

d

its velocity will be zero.

d

h



d

/2

d

h



d

/2

d

h



d

/2

d

h



d

/2

Velocity

Time

Velocity

Time

Velocity

Time

Velocity

Time

112 Motion In One Dimension

As ball comes downward

h

decreases and



increases just before the rebound from the earth

h

= 0 and

v

= maximum and just after rebound velocity reduces to half and direction becomes opposite.

As soon as the height increases its velocity decreases and becomes zero at

2

d

h

.

This interpretation is clearly shown by graph (a).

Problem

34. The acceleration-time graph of a body is shown below –

The most probable velocity-time graph of the body is

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

Solution

: (c) From given

ta 

graph acceleration is increasing at constant rate



k

dt

da 

(constant) 

kta

(by integration)



kt

dt

dv 



ktdtdv 



 tdtkdv



2

kt

v

i.e.

,

v

is dependent on time parabolically and parabola is symmetric about

v

-axis.

and suddenly acceleration becomes zero. i.e. velocity becomes constant.

Hence (c) is most probable graph.

Problem

35. Which of the following velocity time graphs is not possible

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

t

a

t



t



t



t



t

O

v

t

O

v

t

O

v

t

O

v

Motion In One Dimension 113

Solution

: (d) Particle can not possess two velocities at a single instant so graph (d) is not possible.

Problem

36. For a certain body, the velocity-time graph is shown in the figure. The ratio of applied forces for intervals

AB

and

BC

is

(a)

2

1



(b)

2

1



(c)

3

1



(d)

3

1



Solution

: (d) Ratio of applied force = Ratio of acceleration

=

 

3

31

120tan

30tan





BC

AB

a

=

31

Problem

37. Velocity-time graphs of two cars which start from rest at the same time, are shown in the figure. Graph

shows, that

(a) Initial velocity of

A

is greater than the initial velocity of

B

(b) Acceleration in

A

is increasing at lesser rate than in

B

(c) Acceleration in

A

is greater than in

B

(d) Acceleration in

B

is greater than in

A

Solution

: (c) At a certain instant

t

slope of

A

is greater than

B

(

BA





), so acceleration in

A

is greater than

B

Problem

38. Which one of the following graphs represent the velocity of a steel ball which fall from a height on to a

marble floor? (Here



represents the velocity of the particle and

t

the time)

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

Velocity

Time

D

B

C

t

A



30o

60o

O

Time

Velocity

A

B

B

A

t

Velocity

Time

Velocity

Time

Velocity

Time

114 Motion In One Dimension

Solution

: (a) Initially when ball falls from a height its velocity is zero and goes on increasing when it comes down. Just

after rebound from the earth its velocity decreases in magnitude and its direction gets reversed. This

process is repeated untill ball comes to at rest. This interpretation is well explained in graph (a).

Problem

39. The adjoining curve represents the velocity-time graph of a particle, its acceleration values along

OA

,

AB

and

BC

in

metre/sec

2 are respectively

(a) 1, 0, – 0.5

(b) 1, 0, 0.5

(c) 1, 1, 0.5

(d) 1, 0.5, 0

Solution

: (a) Acceleration along

OA

2

12 /1

10

010 sm

t

vv 









Acceleration along

OB

0

10

0

Acceleration along

BC





20

100

– 0.5

m

/

s

2

2.10 Equations of Kinematics.

These are the various relations between

u

,

v

,

a

,

t

and

s

for the moving particle where the notations are used

as :

u

= Initial velocity of the particle at time

t

= 0

sec

v

= Final velocity at time

t

sec

a

= Acceleration of the particle

s

= Distance travelled in time

t

sec

sn

= Distance travelled by the body in

n

th

sec

(1) When particle moves with zero acceleration

(i) It is a unidirectional motion with constant speed.

5

10

20

30

40

A

B

O

Velocity (

m/sec

)

Time (

sec

)

Motion In One Dimension 115

(ii) Magnitude of displacement is always equal to the distance travelled.

(iii)

v

=

u

,

s

=

u t

[As

a

= 0]

(2) When particle moves with constant acceleration

(i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain

constant.

(ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti-parallel to

each other.

(iii) Equations of motion in scalar from Equation of motion in vector from

atu



tauv  

2

1atuts

2

1tatus  

asu2

22 



sauuvv  .2.. 

t

vu

s











2

tvus )(

2

1 

)12(

2 n

a

usn

)12(

2 n

a

usn





(3) Important points for uniformly accelerated motion

(i) If a body starts from rest and moves with uniform acceleration then distance covered by the body in

t

sec

is proportional to

t

2 (

i.e.

2

ts 

).

So we can say that the ratio of distance covered in 1

sec

, 2

sec

and 3

sec

is

222 3:2:1

or 1 : 4 : 9.

(ii) If a body starts from rest and moves with uniform acceleration then distance covered by the body in

n

th

sec is proportional to

)12( n

(

i.e.

)12(  nsn

So we can say that the ratio of distance covered in I

sec

, II

sec

and III

sec

is 1 : 3 : 5.

(iii) A body moving with a velocity

u

is stopped by application of brakes after covering a distance

s

. If the

same body moves with velocity

nu

and same braking force is applied on it then it will come to rest after

covering a distance of

n

2

s

.

As

asu2

22 





asu20 2



a

u

s2

2



,

2

us 

[since

a

is constant]

116 Motion In One Dimension

So we can say that if

u

becomes

n

times then

s

becomes

n

2 times that of previous value.

(iv) A particle moving with uniform acceleration from

A

to

B

along a straight line has velocities

1



and

2



at

A

and

B

respectively. If

C

is the mid-point between

A

and

B

then velocity of the particle at

C

is equal to

2

1









Sample problems based on uniform acceleration

Problem

40. A body

A

moves with a uniform acceleration

a

and zero initial velocity. Another body

B

, starts from the

same point moves in the same direction with a constant velocity

v

. The two bodies meet after a time

t

.

The value of

t

is [MP PET 2003]

(a)

a

v2

(b)

a

v

(c)

a

v

2

(d)

a

v

2

Solution

: (a) Let they meet after time '

t

' . Distance covered by body A =

2

1at

; Distance covered by body

B

=

vt

and

vtat 

2

1



a

v

t2



.

Problem

41. A student is standing at a distance of 50metres from the bus. As soon as the bus starts its motion with an

acceleration of 1

ms

–2, the student starts running towards the bus with a uniform velocity

u

. Assuming the

motion to be along a straight road, the minimum value of

u

, so that the students is able to catch the bus

is

[KCET 2003]

(a) 5

ms

–1 (b) 8

ms

–1 (c) 10

ms

–1 (d) 12

ms

–1

Solution

: (c) Let student will catch the bus after

t

sec

. So it will cover distance

ut

.

Similarly distance travelled by the bus will be

2

1at

for the given condition

2

50

2

1

50 2

2t

atut 



u

2

50 t

t

(As

2

/1 sma 

)

To find the minimum value of

u

,

0

dt

du

, so we get

t

= 10

sec

then

u

= 10

m/s

.

Problem

42. A car, moving with a speed of 50 km/

hr

, can be stopped by brakes after at least 6

m

. If the same car is

moving at a speed of 100

km

/

hr

, the minimum stopping distance is [AIEEE 2003]

(a) 6

m

(b) 12

m

(c) 18

m

(d) 24

m

Motion In One Dimension 117

Solution

: (d)

asuv 2

22 



asu20 2



2

2us

a

u

s

(As

a

= constant)

.241244

50

100 12

2

1

2

1

2mss

u

s

s





























Problem

43. The velocity of a bullet is reduced from 200

m

/

s

to 100

m

/

s

while travelling through a wooden block of

thickness 10

cm

. The retardation, assuming it to be uniform, will be [AIIMS 2001]

(a)

4

1010 

m

/

s

2 (b)

4

1012 

m

/

s

2 (c)

4

105.13 

m

/

s

2 (d)

4

1015 

m

/

s

2

Solution

: (d)

smu /200

,

smv /100

,

ms 1.0

24

2222 /1015

1.02

)100()200(

2sm

s

vu

a











Problem

44. A body A starts from rest with an acceleration

1

a

. After 2 seconds, another body

B

starts from rest with an

acceleration

2

a

. If they travel equal distances in the 5th second, after the start of

A

, then the ratio

21 :aa

is equal to [AIIMS 2001]

(a) 5 : 9 (b) 5 : 7 (c) 9 : 5 (d) 9 : 7

Solution

: (a) By using

 

12

2 n

a

uSn

, Distance travelled by body

A

in 5th second =

)152(

2

01 a

Distance travelled by body B in 3rd second is =

)132(

2

02 a

According to problem :

)152(

2

01 a

=

)132(

2

02 a



9

5

59

2

1

21  a

a

aa

Problem

45. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06

m

is 0.34

ms

–1.

If the change in velocity of the body is 0.18

ms

–1 during this time, its uniform acceleration is [EAMCET (Med.) 2000]

(a) 0.01

ms

–2 (b) 0.02

ms

–2 (c) 0.03

ms

–2 (d) 0.04

ms

–2

Solution

: (b)

sec9

34.0

06.3

velocityAverage

Distance

Time 

and

Time

velocityin Change

onAccelerati 

2

/02.0

9

18.0 sm

.

Problem

46. A particle travels 10

m

in first 5

sec

and 10

m

in next 3

sec

. Assuming constant acceleration what is the

distance travelled in next 2

sec

[RPET 2000]

(a) 8.3

m

(b) 9.3

m

(c) 10.3

m

(d) None of above

Solution

: (a) Let initial

)0( t

velocity of particle =

u

for first 5 sec of motion

metres 10

5

, so by using

2

1tauts

2

)5(

2

1

510 au 



452  au

…. (i)

for first 8 sec of motion

metres 20

8

118 Motion In One Dimension

2

)8(

2

1

820 au 



582  au

.... (ii)

By solving (i) and (ii)

2

/

3

1

/

6

7smasmu 

Now distance travelled by particle in total 10

sec

.

 

2

10 10

2

1

10 aus 

by substituting the value of

u

and

a

we will get

3.28

10 s

m

So the distance in last 2

sec

=

s

10 –

s

8

3.8203.28 

m

Problem

47. A body travels for 15

sec

starting from rest with constant acceleration. If it travels distances

21,SS

and

3

S

in the first five seconds, second five seconds and next five seconds respectively the relation between

21,SS

and

3

S

is [AMU (Engg.) 2000]

(a)

321 SSS 

(b)

321 35 SSS 

(c)

321 5

1

3

1SSS 

(d)

321 3

1

5

1SSS 

Solution

: (c) Since the body starts from rest. Therefore

0u

.

2

25

)5(

2

12

1a

aS 

2

100

)10(

2

12

21 a

aSS 



12 2

100 S

a

S

=

2

75 a

2

225

)15(

2

12

321 a

aSSS 



123 2

225 SS

a

S

=

2

125 a

Thus Clearly

321 5

1

3

1SSS 

Problem

48. If a body having initial velocity zero is moving with uniform acceleration

,sec/8 2

m

the distance travelled

by it in fifth second will be [MP PMT 1996; DPMT 2000]

(a) 36

metres

(b) 40

metres

(c) 100

metres

(d) Zero

Solution

: (a)

)12(

2

1 nauSn

metres36]152[)8(

2

1

0

Problem

49. The engine of a car produces acceleration 4m/sec2 in the car, if this car pulls another car of same mass,

what will be the acceleration produced [RPET 1996]

(a) 8

m/s

2 (b) 2

m/s

2 (c) 4

m/s

2 (d)

2

/

2

1sm

Solution

: (b)

F

=

ma

m

a1



if

F

= constant. Since the force is same and the effective mass of system becomes double

m

a

2

1

2

,

2

1

2m/s2

2 a

a

Problem

50. A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd

second.

[CBSE PMT 1993]

(a) 7/5 (b) 5/7 (c) 7/3 (d) 3/7

Motion In One Dimension 119

Solution

: (a) As

)12(  nSn

,

5

7

3

4

S

2.11 Motion of Body Under Gravity (Free Fall).

The force of attraction of earth on bodies, is called force of gravity. Acceleration produced in the body by

the force of gravity, is called acceleration due to gravity. It is represented by the symbol

g

.

In the absence of air resistance, it is found that all bodies (irrespective of the size, weight or composition)

fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth

from a small altitude (

h

<<

R

) is called free fall.

An ideal one-dimensional motion under gravity in which air resistance and the small changes in

acceleration with height are neglected.

(1) If a body dropped from some height (initial velocity zero)

(i) Equation of motion : Taking initial position as origin and direction of motion (

i.e.

, downward direction) as

a positive, here we have

u

= 0 [As body starts from rest]

a

= +

g

[As acceleration is in the direction of motion]

v

=

g t

…(i)

2

1gth

…(ii)

gh2

2



…(iii)

)12(

2 n

g

hn

...(iv)

(ii) Graph of distance velocity and acceleration with respect to time :

u

= 0

v

h

g

v

g

h

t 2

ghv2

g

v

h2

2



s

t

a

g

t

v



t

tan



=

g

120 Motion In One Dimension

(iii) As

h

= (1/2)

gt

2,

i.e.

,

h



t

2, distance covered in time

t

, 2

t

, 3

t

,

etc

., will be in the ratio of 12 : 22 : 32,

i.e.

,

square of integers.

(iv) The distance covered in the

nth sec

,

)12(

2

1 nghn

So distance covered in I, II, III

sec

,

etc

., will be in the ratio of 1 : 3 : 5,

i.e.

, odd integers only.

(2) If a body is projected vertically downward with some initial velocity

Equation of motion :

tgu 



2

1tguth

ghu2

22 



)12(

2 n

g

uhn

(3) If a body is projected vertically upward

(i) Equation of motion : Taking initial position as origin and direction of motion (

i.e.

, vertically up) as

positive

a

= –

g

[As acceleration is downwards while motion upwards]

So, if the body is projected with velocity

u

and after time

t

it reaches up to height

h

then

tgu 



;

2

1tguth

;

ghu2

22 



;

)12(

2 n

g

uhn

(ii) For maximum height

v

= 0

So from above equation

u

=

gt

,

2

1gth

and

ghu2

2

g

u

g

h

t 2

ghu2

g

u

h2

2



u

v

= 0

h

Motion In One Dimension 121

(iii) Graph of distance, velocity and acceleration with respect to time (for maximum height) :

It is clear that both quantities do not depend upon the mass of the body or we can say that in absence of

air resistance, all bodies fall on the surface of the earth with the same rate.

(4) In case of motion under gravity for a given body, mass, acceleration, and mechanical energy remain

constant while speed, velocity, momentum, kinetic energy and potential energy change.

(5) The motion is independent of the mass of the body, as in any equation of motion, mass is not involved.

That is why a heavy and light body when released from the same height, reach the ground simultaneously and

with same velocity

i.e.

,

)/2( ght 

and

ghv2

.

(6) In case of motion under gravity time taken to go up is equal to the time taken to fall down through the

same distance. Time of descent (

t

1) = time of ascent (

t

2) =

u/g

 Total time of flight

T

=

t

1 +

t

2

g

u2



(7) In case of motion under gravity, the speed with which a body is projected up is equal to the speed with

which it comes back to the point of projection.

As well as the magnitude of velocity at any point on the path is same

whether the body is moving in upwards or downward direction.

(8) A ball is dropped from a building of height

h

and it reaches after

t

seconds on earth. From the same building if two ball are thrown (one upwards

s

t

(

u/g

)

(

u

2/2

g

)

(

u/g

)

(2

u/g

)

t

v

O

–

v

+

–

t

a

O

g

+

–

a

u

=0

t

1

t

2

t

122 Motion In One Dimension

and other downwards) with the same velocity

u

and they reach the earth surface after

t

1 and

t

2 seconds

respectively then

21 ttt 

(9) A body is thrown vertically upwards. If air resistance is to be taken into account, then the time of ascent

is less than the time of descent.

t

2 >

t

1

Let

u

is the initial velocity of body then time of ascent

ag

u

t



1

and

)(2

2

ag

u

h



where

g

is acceleration due to gravity and

a

is retardation by air resistance and for upward motion both

will work vertically downward.

For downward motion

a

and

g

will work in opposite direction because

a

always work in direction opposite

to motion and

g

always work vertically downward.

So

2

)(

2

1tagh 



2

2)(

2

1

)(2 tag

ag

u





))((

2agag

u

t



Comparing

t

1 and

t

2 we can say that

t

2 >

t

1 since (

g

+

a

) > (

g

–

a

)

(10) A particle is dropped vertically from rest from a height. The time taken

by it to fall through successive distance of 1

m

each will then be in the ratio of the

difference in the square roots of the integers

i.e.

.),........34).......(23(),12(,1 

Sample problems based on motion under gravity

Problem

51. If a body is thrown up with the velocity of 15

m/s

then maximum height attained by the body is (

g

= 10

m/s

2)

u

= 0

1

1t

12

2t

23

3t

34

4t

1

m

1

m

1

m

1

m

Motion In One Dimension 123

[MP PMT 2003]

(a) 11.25

m

(b) 16.2

m

(c) 24.5

m

(d) 7.62

m

Solution

: (a)

m

g

u

H25.11

102

)15(

2

22

max 





Problem

52. A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its

motion is

)/10(2

smg 

[MP PET 2003]

(a) 25

m

(b) 45

m

(c) 90

m

(d) 125

m

Solution

: (b)

   

152

2

10

12

25 thnhn

g

h

45

m

.

Problem

53. If a ball is thrown vertically upwards with speed

u

, the distance covered during the last

t

seconds of its

ascent is

[CBSE 2003]

(a)

2

1gt

(b)

2

1gtut 

(c)

tgtu)( 

(d)

ut

Solution

: (a) If ball is thrown with velocity

u

, then time of flight

g

u



velocity after

:sec











t

g

u











 t

g

u

guv

=

gt

.

So, distance in last '

t

'

sec

:



2

0

.)(2)( 2hggt 



.

2

12

gth

Problem

54. A man throws balls with the same speed vertically upwards one after the other at an interval of 2

seconds. What should be the speed of the throw so that more than two balls are in the sky at any

time (Given

)/8.9 2

smg 

[CBSE PMT 2003]

(a) At least 0.8

m

/

s

(b) Any speed less than 19.6

m

/

s

(c) Only with speed 19.6

m

/

s

(d) More than 19.6

m

/

s

Solution

: (d) Interval of ball throw = 2

sec

.

If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be

greater than 4

sec

.

i.e.

secT 4

or

smusec

g

U/6.194

2

It is clear that for

6.19u

First ball will just strike the ground (in sky), second ball will be at highest point (in

sky), and third ball will be at point of projection or on ground (not in sky).

Problem

55. A man drops a ball downside from the roof of a tower of height 400

meters

. At the same time another ball

is thrown upside with a velocity 50

meter/sec

. from the surface of the tower, then they will meet at which

height from the surface of the tower [CPMT 2003]

(a) 100

meters

(b) 320

meters

(c) 80

meters

(d) 240

meters

sect

g

u











t sec

h

124 Motion In One Dimension

Solution

: (c) Let both balls meet at point

P

after time

t

.

The distance travelled by ball

A

(

2

12

1

)gth

.....(i)

The distance travelled by ball

B

2

22

1

)( gtuth

.....(ii)

By adding (i) and (ii)

uthh  21

= 400 (Given

.400

21  hhh

)

sect 850/400 

and

mhmh 80 ,320 21 

Problem

56. A very large number of balls are thrown vertically upwards in quick succession in such a way that the next

ball is thrown when the previous one is at the maximum height. If the maximum height is 5

m

, the number

of ball thrown per minute is (take

2

10 

msg

) [KCET (Med.) 2002]

(a) 120 (b) 80 (c) 60 (d) 40

Solution

: (c) Maximum height of ball = 5

m,

So velocity of projection

ghu2

sm /105102

time interval between two balls (time of ascent) =

.

60

1

1minsec

g

u

So no. of ball thrown per min = 60

Problem

57. A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10

m/s

, then

maximum height attained by it is (Take

10g

m

/

s

2) [CBSE PMT 2001]

(a) 8

m

(b) 10

m

(c) 12

m

(d) 16

m

Solution

: (b) Let particle thrown with velocity

u

and its maximum height is

H

then

g

u

H2

2



When particle is at a height

2/H

, then its speed is

sm /10

From equation

ghuv 2

22 

,

 

g

u

gu

H

gu 4

2

210 2

22

2

















200

2u



Maximum height

102

200

2



 g

u

H

= 10

m

Problem

58. A stone is shot straight upward with a speed of 20

m/sec

from a tower 200

m

high. The speed with which it

strikes the ground is approximately [AMU (Engg.) 1999]

(a) 60

m/sec

(b) 65

m/sec

(c) 70

m/sec

(d) 75

m/sec

Solution

: (b) Speed of stone in a vertically upward direction is 20

m/s

. So for vertical downward motion we will consider

smu /20

smvghuv /65200102)20(2 222 

Problem

59. A body freely falling from the rest has a velocity ‘

v

’ after it falls through a height ‘

h

’. The distance it has to

fall down for its velocity to become double, is [BHU 1999]

A

h

1

h

2

B

P

400

m

Motion In One Dimension 125

(a)

h2

(b)

h4

(c)

h6

(d)

h8

Solution

: (b) Let at point A initial velocity of body is equal to zero

For path

AB

:

ghv20

2

… (i)

For path

AC

:

gxv20)2( 2



gxv24 2

… (ii)

Solving (i) and (ii)

hx 4

Problem

60. A body sliding on a smooth inclined plane requires 4

seconds

to reach the bottom starting from rest at the

top. How much time does it take to cover one-fourth distance starting from rest at the top [BHU 1998]

(a) 1

s

(b) 2

s

(c) 4

s

(d) 16

s

Solution

: (b)

2

1atS



st 

 

constant a As

 2

14/

1

2

1

2s

s

t

s

t

t2

2

4

2

1

2

Problem

61. A stone dropped from a building of height

h

and it reaches after

t

seconds on earth. From the same

building if two stones are thrown (one upwards and other downwards) with the same velocity

u

and they

reach the earth surface after

1

t

and

2

t

seconds respectively, then [CPMT 1997; UPSEAT 2002; KCET (Engg./Med.) 2002]

(a)

21 ttt 

(b)

221 tt

t



(c)

21ttt 

(d)

2

1ttt 

Solution

: (c) For first case of dropping

.

2

12

gth

For second case of downward throwing

2

11 2

1gtuth

2

1gt



)(

2

12

1

2

1ttgut 

......(i)

For third case of upward throwing

22

22 2

1

2

1gtgtuth



)(

2

12

2

2ttgut 

.......(ii)

on solving these two equations :

2

1

2

1

tt

t







.

21ttt 

Problem

62. By which velocity a ball be projected vertically downward so that the distance covered by it in 5th second is

twice the distance it covers in its 6th second (

2

/10 smg 

) [CPMT 1997; MH CET 2000]

(a)

sm /8.58

(b)

sm /49

(c)

sm /65

(d)

sm /6.19

Solution

: (c) By formula

)12(

2

1 nguhn



]}162[

2

10

{2]152[

2

10  uu



)55(245  uu



./65 smu 

x

A

B

C

u =

0

h

v

2

v

126 Motion In One Dimension

Problem

63. Water drops fall at regular intervals from a tap which is 5

m

above the ground. The third drop is leaving

the tap at the instant the first drop touches the ground. How far above the ground is the second drop at

that instant [CBSE PMT 1995]

(a) 2.50

m

(b) 3.75

m

(c) 4.00

m

(d) 1.25

m

Solution

: (b) Let the interval be

t

then from question

For first drop

5)2(

2

12tg

.....(i) For second drop

2

1gtx

.....(ii)

By solving (i) and (ii)

4

5

x

and hence required height

h

.75.3

4

5

5m

Problem

64. A balloon is at a height of 81

m

and is ascending upwards with a velocity of

./12 sm

A body of 2

kg

weight

is dropped from it. If

,/10 2

smg 

the body will reach the surface of the earth in [MP PMT 1994]

(a) 1.5

s

(b) 4.025

s

(c) 5.4

s

(d) 6.75

s

Solution

: (c) As the balloon is going up we will take initial velocity of falling body

,/12 sm

,81mh 

2

/10 smg 

By applying

2

1gtuth

;

2

)10(

2

1

1281 tt 



0811252 tt



10

162014412 

t

10

176412 



.sec4.5

Problem

65. A particle is dropped under gravity from rest from a height

)/8.9( 2

smgh 

and it travels a distance 9

h/

25

in the last second, the height

h

is [MNR 1987]

(a) 100

m

(b) 122.5

m

(c) 145

m

(d) 167.5

m

Solution

: (b) Distance travelled in

n

sec =

2

1gn

=

h

.....(i)

Distance travelled in

sec

th

n

25

9

)12(

2

h

n

g

.....(ii)

Solving (i) and (ii) we get.

mh 5.122

.

Problem

66. A stone thrown upward with a speed

u

from the top of the tower reaches the ground with a velocity 3

u

.

The height of the tower is [EAMCET 1983; RPET 2003]

(a)

gu /3 2

(b)

gu /4 2

(c)

gu /6 2

(d)

gu /9 2

Solution

: (b) For vertical downward motion we will consider initial velocity = –

u.

By applying

ghuv 2

22 

,

ghuu 2)()3( 22 



g

u

h2

4



.

Problem

67. A stone dropped from the top of the tower touches the ground in 4

sec.

The height of the tower is about

[MP PET 1986; AFMC 1994; CPMT 1997; BHU 1998; DPMT 1999; RPET 1999]

Motion In One Dimension 127

(a) 80

m

(b) 40

m

(c) 20

m

(d) 160

m

Solution

: (a)

.80410

2

1

2

122 mgth

Problem

68. A body is released from a great height and falls freely towards the earth. Another body is released from

the same height exactly one second later. The separation between the two bodies, two seconds after the

release of the second body is [CPMT 1983; Kerala PMT 2002]

(a) 4.9

m

(b) 9.8

m

(c) 19.6

m

(d) 24.5

m

Solution

: (d) The separation between two bodies, two second after the release of second body is given by :

s

=

)(

2

12

2

1ttg 

=

5.24)23(8.9

2

122 

m

.

2.12 Motion With Variable Acceleration.

(i) If acceleration is a function of time

)(tfa 

then



 tdttfuv 0)(

and

 

dtdttfuts 

 )(

(ii) If acceleration is a function of distance

)(xfa 

then



 x

xdxxfuv

0

)(2

22

(iii) If acceleration is a function of velocity

a

=

f

(

v

) then



v

uvf

dv

t)(

and



 v

uvf

vdv

xx )(

0

Sample problems based on variable acceleration

Problem

69. An electron starting from rest has a velocity that increases linearly with the time that is

,ktv

where

.sec/2 2

mk 

The distance travelled in the first 3

seconds

will be [NCERT 1982]

(a) 9

m

(b) 16

m

(c) 27

m

(d) 36

m

Solution

: (a)

x

2

1

t

tdtv















 

3

0

2

3

02

22 t

dtt

9

m

.

Problem

70. The acceleration of a particle is increasing linearly with time

t

as

bt.

The particle starts from the origin with

an initial velocity

.

0

v

The distance travelled by the particle in time

t

will be [CBSE PMT 1995]

(a)

2

03

1bttv 

(b)

3

03

1bttv 

(c)

3

06

1bttv 

(d)

2

02

1bttv 

Solution :

(c)

2

1

v

dv

=

2

1

t

tdta

=

2

1

)(

t

tdtbt

128 Motion In One Dimension

2

1

2

12

t

bt

vv 

















22

2

0

2

12 bt

v

bt

vv

t



















3

0

2

06

1

2bttvdt

bt

dtvS  

Problem

71. The motion of a particle is described by the equation

atu

. The distance travelled by the particle in the

first 4 seconds [DCE 2000]

(a)

a4

(b)

a12

(c)

a6

(d)

a8

Solution

: (d)

atu



at

dt

ds 



4

0

2

4

02













 t

adtats

= 8

a.