class 9 science chapter 7 motion pdf download

MOTION
1.1 INTRODUCTION
Motion is change in position of an object with time.
The branch of physics which deals with the study of the motion of
the object is known as mechanics.
Mechanics consists of three branches:
(i) Statics, (ii) Kinematics and (iii) Dynamics
(i) Statics: The branch of mechanics which deals with the study of
the objects at rest.
(ii) Kinematics: The branch of mechanics which describe motion
without going into the cause of motion.
(iii) Dynamics: The branch of mechanics which deals with the
study of the motion of the objects by taking into the account of the
force which cause the motion in the objects.
Question based on basic knowledge required to understand
this chapter
1. If the speed of an object moving alonga straight line keeps changing,
its motion is said to be
(A) uniform (B) non-uniform
(C) circular (D) None of these
2. Which of the following relations is correct?
(A) speed =
time
distance
(B) speed = distance × time
(C) speed =
distance
time
(D) None of these
3. The basic unit of time is
(A) an hour (B) a minute
(C) a second (D) None of these
4. The symbol of kilometre is
(A) kilo (B) km
(C) kilom (D) None of these
1.1 Introduction
*1.2 Point Object
1.3 Motion in one
dimensions
1.4 Uniform and non-
uniform motion
*1.5 Scalar and Vector
quantities
1.6 Distance
1.7 Displacement
1.8 Comparative Study of
Displacement &
Distance
1.9 Speed
1.10 Distance by speed-time
graph
1.11 Velocity
1.12 Comparative Study of
Speed and Velocity
1.13 Acceleration
1.14 Distance Time Graph
1.15 Velocity Time Graph
1.16 Equations of Motion by
Graphical Method
1.17 Circular Motion
VAVA CLASSES/PHY/9TH
All right copy reserved. No part of the material can be produced without prior permission

5. The symbols of all units are written in
(A) plural (B) singular
(C) both singular and plural (D)None of these
6. A speedometre records the speed directly in
(A) m/s (B) km/s (C) km/h (D) m/min
7. A device that measures the distance by the vehicle is known as
(A) a speedometre (B) an odometer (C) a barometer (D) None of these
8. Fig. shows the distance-time graph for the motion of two vehiclesAand B. Which one of them is moving
faster?
(A) A (B) B
(C) Both A and B are moving with same speed (D)None of these
9. Which of the following distance-time graphs shows a truck moving with speed which is not constant?
(A) (B) (C) (D)
10. Motion of a pendulum is
(A) Along a straight line (B) circular (C) periodic (D) None of these
1.2 POINT OBJECT
If the position of an object changes by distance much greater than its own size in a reasonable duration of
time then the object may be regarded as a point object.
Examples: (i) Moving persons on earth as seen from a distant aircraft in space or a satellite are called
point objects.
(ii) Size of earth moving in space is taken as a point object w.r.t. large size of universe.
1.3 MOTION IN ONE DIMENSION
Motion in one dimension: An object moving along a straight line or path is said to have one-dimensional
motion.
Suppose, an object moves in a particular direction then its position at any time can be described by
knowing its distance from the starting point of the motion. It means, only one position coordinate (say, x)
is required to describe the position of the object in one dimensional motion.
Example: Motion of a bus on a straight road and motion of a train on a straight track, an object dropped
from a certain height above the ground etc.

1.4 UNIFORM AND NON-UNIFORM MOTION
Uniform: If a particle covers equal displacement in equal time interval them the motion of the particle is
termed as uniform.
Non-uniform: If a particle does not cover equal displacement in equal time interval then the motion of the
particle is termed as non-uniform.
1.5 SCALAR AND VECTOR QUANTITIES
Physical quantities maybe divided into two groups accordingto their properties required for fullyexpressing
them.
(i) Scalar quantities (ii) Vector quantities
Scalar Quantities: Such physical quantities, as are completely represented only by their magnitude and
unit are known as scalar quantities. Thus mass, time, distance, volume, temperature, etc. are scalar
quantities. Scalar quantities may be positive or negative. These quantities obey the general algebraic rules
of addition, subtraction, multiplication, division etc.
Vector Quantities: The physical quantities which need the knowledge of direction also together with
their magnitude and unit for their full representation, are known as Vector quantities. Thus, displacement,
velocity, area, force, acceleration, momentum, torque, angular momentum, current density, etc. are Vector
quantities.
For example if Ram moves 100 m. eastward then the displacement of Ram is possible only in one direction.
Hence, the displacement is a vector quantity.
*1.5.1 Representation of Vectors
A vector quantity is represented by a segment having an arrow at its end point.
In figure from where the arrow start in known as initial point (tail) and the point P where the arrow exits
is known as the top (head) of the vector. The length of segment is proportional to the magnitude of vector
taken in a proper scale. The direction of arrow represents the direction of vector quantity.
For example if a particle displaced by 15 metre in north direction
then it can be represented proportionately as shown in figure. In
this way the displacement 15 metre in north direction represented
by vector AB and the magnitude of the vector by |AB | and it is
also termed as modulus or mode of the vector. it is mearly a number
so the mode (magnitude) is a scalar quantity. In this way the vector
AB can be represent in magnitude and in direction as following.
AB = |AB |n̂
*1.5.2 Types of Vectors
The main type of vector quantities are as follow–
(1) Equivalent Vector –
If the magnitude, unit and direction of two or more than two vectors
is same then they are termed as equal vectors.
It is not necessary that the position of all equal vectors in space, to be

same. In figure three vector A , B and C are shown. These vectors
are of same length and parallel to each other also. Hence all of them are
equal vectors. If a given vector is displaced parallely to it then it remains
unchanged. For example if Rahim feels that the air is blowing from
north direction then by parallel displacement the direction of air can be
represented towards south as shown in figure.
(2) Unequal Vector:
If the magnitude or the direction or the both, are different for two vectors
then they are termed as unequal vectors. In figure all the three states, both vector A and B are unequal
vectors.
(3) Unit Vector –
Such a vector which has unit magnitude is known as unit vector. In general, the unit vector of vector A
is represented by Â . Hence
Â =
|
A
|
A
.....(i)
(4) Opposite or Negative vector –
If the magnitudes of two vectors are same but their directions are opposite to each other then they are
termed as negative vector. As in figure if two vectors A and B are mutually opposite to each other than
A = – B
(5) Zero Vector –
The vector which has zero magnitude and has no definite direction is known as zero vector and it is
represented by zero (0).
Remember – (i) If we multiply a vector quantity with zero then the resultant which we get is a zero vector.
(ii) If A = B then difference ( A – B ) will be the zero vector.
.
(6) Colinear Vector –
If two or more than two vectors are parallel to a straight line or opposite to each other are known as
colinear vectors. In figure colinear vectors in two different states are shown.
1.6 DISTANCE
The length of the actual path between intial and final position of a particle in a given interval of time is
called distance covered by the particle. Distance is the actual length of the path. Distance between two
points is path dependent.

1.6.1 Characteristics of Distance
(i) It is a scalar quantity
(ii) It depends on the path
(iii) It never reduces with time
(iv) Distance covered by a particle is always positive and can never be negative or zero.
(v) Unit: In C.G.S. centimeter (cm). in S.I. system metre (m).
1.7 DISPLACEMENT
The shortest distace from the initial position to the final position of the particle is called displacement. The
displacement of a particle is measured as the change in the position of the particle in a particular direction
over a given time interval. It depends only on final and initial positions.
1.7.1 Characteristics of Displacement
(i) It is a vector quantity
(ii) The displacement of a particle between any two points is equal to the shortest distance between them.
(iii) The displacement of an object in a given time interval may be +ve, –ve or zero.
(iv) The actual distance travelled by a particle in the given interval of time is always equal to or greater
than the magnitude of the displacement and in no case, it is less than the magnitude of the displacement,
i.e. Distance  |Displacement|
(v) Unit: In C.G.S. centimeter (cm). In S.I. system metre (m).
1.8 COMPARATIVE STUDY OF DISPLACEMENT AND DISTANCE
S.No. Displacement Distance
1
It has single value between
two points
it may have more than one value
between two points
2 May be +ve, –ve or zero Distance > 0
3 It can decrease with time It can never decrease with time
4 It is a vector quantity It is a scalar quantity
Illustration 1
A boy travels a distance of 10 m towards south, then he runs towards east and travels 12 m in that
direction. Calculate (i) the total distance travelled by the boy and (ii) the displacement from his
initial position.

Solution
(i) Total distance travelled
= OA + AB = 10 + 12 = 22 m
(ii) Displacement
= OB = 2
2
12
10  = 144
00
1  = 244 m
Try yourself
1. Abody is moving along a circular path of radius r. What will be the distance and displacement of the body
when it completes half a revolution?
1.9 SPEED
Speed of an object is defined as the time rate of change of position of the object in any direction. It is
measured by the distance travelled by the object in unit time in any direction. i.e.
Speed =
taken
time
travelled
distance
(i) It is a scalar quantity
(ii) It gives no idea about the direction of motion of the objects.
(iii) It can be zero or positive but never negative
(iv) Unit: C.G.S. cm/sec., S.I. m/sec., 1 km/h =
0
6
60
1000

=
18
5
m/s
1.9.1 Types of speed:
SPEED
Uniform speed Non-Uniform speed Average speed
Uniform speed:An object is said to be moving with a uniform speed, if it covers equal distances in equal
intervals of time, however small these intervals may be. The uniform speed is shown by straight line in
distance-time graph.
For example: suppose a train travels 1000 m., in 60 sec. The train is said to be moving with uniform speed,
if it travels 500m, in 30 sec., 250 m. in 15 sec., 125 m. in 7.5 sec. and so on.
Non-uniform speed: An object is said to be moving with a variable speed if it covers equal distances in
unequal interavals of time or unequal distances in equal intervals of time, however small these intervals
may be.
For example: suppose a train travels first 1000 m. in 60 sec., next 1000 m. in 120 sec. and next 1000 m. in
50 sec., then the train is moving with variable speed.
Average speed: When an object is moving with a variable speed, then the average speed of the object is
that constant speed with which the object covers the same distance in a given time as it does while moving
with variable speed during the given time.Average speed for the given motion is defined as the ratio of the
total distance travelled by the object to the total time taken i.e..

Average speed V =
taken
time
total
travelled
distance
total
Note: If any car covers distance x1, x2, ..... in the time intervals t1, t2, ..... then.
V =
n
2
1
n
3
2
1
t
.....
t
t
x
.....
x
x
x







1.10 DISTANCE BY SPEED-TIME GRAPH
When the particle moves from time t1 to t2 with uniform speed V as shown in the graph.
Then distance covered S = V(t2 – t1) = AB × AD = Area of ABCD
Total distance travelled by particle = Area of speed-time graph
1.11 VELOCITY
The rate of change of displacement of a particle with time is called the velocity of the particle.
i.e., Velocity =
interval
Time
nt
Displaceme
(i) It is a vector quantity
(ii) The velocity of an object can be positive, zero and negative
(iii) Unit: C.G.S. cm/s.; S.I. m/s
1.11.1 Types of Velocity
VELOCITY
Uniform Velocity Non-Uniform Velocity Average Velocity
Uniform Velocity:Abody is said to move with uniform velocity, if it covers equal displacements in equal
intervals of time, howsoever, small these intervals may be. When a body is moving with uniform velocity,
then the magnitude and direction of the velocity of the body remaine same at all points of its path.
Non-uniform Velocity:The particle is said to have non-uniformmotion if it covers unequal displacements

in equal intervals of time, howsoever, small these intervals may be. In this type of motion velocity does not
remain constant.
Average Velocity : The average velocity of an object is equal to
the ratio of the displacement, to the time interval for which the
motion takes place i.e.,
Average velocity =
taken
time
nt
displaceme
If the initial and final position of a particle are 1
r

and 2
r

at time t1
and t2
respectively,
Then Displacement  r

= 2
r

– 1
r

and elapsed time t = t2
– t1
Average velocity av
V

=
Ät
r
Ä
t
–
t
r
–
r
1
2
1
2




(i) Uniform speed may be equal to uniform velocity in a linear motion because velocity = speed +
direction.
(ii) Moving body with uniform speed may have variable velocity. e.g. in uniform circular motion speed is
constant but velocity is non-uniform.
1.12 COMPARATIVE STUDY OF SPEED AND VELOCITY
S.No. Speed Velocity
1
It is the time rate of change of
distance of a body
It is the time rate of change of
displacement of a particle
2
It tells nothing about the
direction of motion of the
particle
It tells direction of motion of the
particle
3 It can be positive or zero
It can be positive or negative or
zero
4 It is a scalar quantity It is a vector quantity
Illustration 2
A train covers half of its journey with a speed of 60 m/s and other half with a speed of 40 m/s.
Calculate the average speed of the train during the whole journey.
Solution
Let the total distance = 2xm
Time taken by the train during first half =
60
x
sec.

time taken by the train during other half =
40
x
sec.
Average speed =
time
total
distance
total
=
40
x
60
x
2x

=
240
6x
4x
2x

= 2x ×
x
0
1
240
= 48 m/s
Try yourself
2. A scooterist moves from a place A to place B with a uniform speed of 30 km/h and returns from B to A
with a uniform speed of 50 km/h. Find the average speed for the complete journey.
1.13 ACCELERATION
The rate of change of velocity of an object with time is called acceleration of the object.
Let u

and v

be the velocity of the object at time t1
and t2
respecively, then acceleration of the body is
given by
Acceleration  
a

=
interval
Time
velocity
in
Change
=
2
1 t
t
u
v




(i) Acceleration is a vector quantity.
(ii) It is positive if the velocity is increasing and is negative if the velocity is decreasing.
(iii) The negative acceleration is also called retardation or deceleration.
(iv) Dimension: [M0
L1
T–2
]
1.13.1 Types of Acceleration
ACCELERATION
Uniform Acceleration Non-Uniform Acceleration
Uniform acceleration:An object is said to be moving with a uniform acceleration if its velocity changes
by equal amounts in equal intervals of time.
Non Uniform acceleration:
An object is said to be moving with a variable acceleration if its velocity changes by unequal amounts in
equal intervals of time.
Illustration 3
A car starts from rest and acquires velocity equal to 36 km h–1
after 10 second. Calculate the
acceleration of the car.
Solution
Here, u = 0 [ initially car is at rest]

v = 36 km h–1
=
36 1000
60 60


ms–1
= 10 ms–1
t = 10 sec.
Using,
v u
a
t

 or
1
10 0
10
ms
a
s


 or a = 1 ms–2
Try yourself
3. A ship is moving at a speed of 56 km/h. One second later, it is moving at 58 km/h. What is acceleration?
1.14 DISTANCE TIME GRAPH
(i) when the body is moving with uniform velocity
(ii) when the body is at rest
(iii) non uniformmotion
time (t)
Distance
(x)
(iv) motion with positive acceleration
time (t)
Distance
(x)
(v) motion with negative acceleration
Distance
(x)
time (t)
(vi) motion with zero acceleration
O
a = 0
Distance
(x
)
time (t)

• Slope of x-t graph represents the speed of object.
• For motion with uniform acceleration, x-t graph is a parabola while the v-t graph is a straight
line inclined to the time axis
1.15 VELOCITY TIME GRAPH
(i) velocity-time graph when the velocity remains constant :
(ii) velocity-time graph when velocity changes at a uniform rate
(iii) velocity-time graph when the acceleration is uniform
(iv) motion in positive direction with negative acceleration
(v) motion of an object with negative acceleration
(vi) motion of an object with negative acceleration that changes direction at time t1

(vii) motion in positive direction with positive acceleration
• Slope of v-t graph represents acceleration.
• The area under the v-t graph between time t1
and t2
is equal to the displacement of the
object during that interval of time.
1.16 EQUATION OF MOTION BY GRAPHICAL METHOD
In uniformly accelerated motion the change in velocity in equal time interval is same. Hence in uniform ly
accelerated motion the graph time between velocity and time becomes a straight line. Its slope (tan ) is
equal to the acceleration. So
Slope tan constant
dv
a
dt
   
In figure the velocity of a particle moving in uniform accelerated motion is shown by a graph in between
velocity and time.At time t = 0 (start of motion) the initial velocity of a particle is u and at time t, velocity
of the particle is v.
tan constant
BC
a
AB
   
or
v u
a
t

 or v – u = t v = u + at (1)
It is the first equation of motion.
Again by a graph in velocity-time in uniform accelerated motion on time axis (In given time interval) the
covered area is equal to the displacement of the particle. So, if the displacement in between time interval
0 to t is S then
S = Area of quadrilateral shape OACBD
or S = Area of ABC + Area of rectangle OABD
or S = ½AB × BC + OA × OD or S = ½ × t(v–u) + u × t
or S = ½t(at) + ut or S = ut + ½ at2
(2)
It is the second equation of motion.
Again from the first equation of motion
v2
= (u + at)2
= u2
+ a2
t2
+ 2uat
2 2
1
2
2
u a ut at
 
  
 
 
or v2
= u2
+ 2as (3}
It is the third equation of motion.

1.16.1 Displacement in nth second in a uniform accelerated motion
Let the initial velocity of a particle in uniform acceleration motion is u and after n second is vi if the
displacements of the particle in n and (n–1) second are sn and sn–1 respectively then from second equation
of motion
2
1
2
n
s un an
 
and
2
1
( 1) ( 1)
2
n i
s u n a n
    
In this way the displacement in nth second is
d = sn – sn–1
2 2
1 1
( 1) ( 1)
2 2
un an u n a n
     
Thus displacement in nth second with in interval of one second
(2 1)
2
a
d u n
  
Illustration 4
A stone is dropped from a height of 100 metre then find (1) time taken in fall of first 50 metre (2)
time taken in fall of last 50 metre.
Solution
(i) If time taken in the fall of first 50 metres is t1 than by the second equation of motion
2
1
2
s ut gt
 
2
1
1
50 0
2
g t
   or 1
100 100
10
10
t
g
   sec
(ii) If time taken in the fall of total 100 metre is say, t then by the second equation of motion.
2
1
2
s ut gt
 
2
1
100 0
2
gt
  
200
20 sec
10
t  
So the time taken in fall of last 50 metre.
t2 = t – t1 sec =  
20 10
 sec.
Try yourself
4. A car travels at constant acceleration of 5 m/s2
from rest. How far has it travelled by the time it reached
the velocity of 72 km/h (20 m/s)?
5. How long does it take for a car to change its velocity from 10 m/s to 25 m/s if the acceleration is 5 m/s2
?
1.17 CIRCULAR MOTION
Motion of an artificial satellite around earth, motion of a cyclist in circular path orbital motion of electron
around nucleus etc. are the examples of circular motion. For the study of circular motion, it is necessary

to understand the concepts of relevant quantities like angular displacement, angular velocity and angular
acceleration.
1.17.1 Angular Displacement
It is angle described the body at the centre of its circular part while rotating
Let a particle is moving in a circular path of radius r and centre at O. At any time t the particle is at
position. A which is at the position of angle  with reference to initial time OX. If in time t the particle
reaches point B where angular position of the particle is BOX =  +  then in the time interval t the
angular displacement gained by the particle = BOA is obtained S. I. Unit of angular displacement is
radian (rad.)
1.17.2 Angular Velocity
It is define as the rate of change of angular displacement
If in time interval velocity t angular displacement of a particle is then the ratio of angular displacement
 and time interval t is called average angular velocity, that is average angular velocity
av
t

  
 

Under the lim t  0 the average angular velocity becomes equivalent to instantaneous angular velocity.
0
lim
t
d
t dt
 

 
  
  

S. I. unit of angular velocity is rad/s. It is a vector quantity whose direction is determined by the right hand
screw rute.
If a particle is revolving in a circular path will constant angular velocity, then its angular displacement in
time t will be  = t.
Time t taken in making one revolution by a particle in a circular path is called time period and number of
revolutions n completed in unit time is called frequency. Thus, relation between magnitude of angular
velocity , time period T and frequency n is as follows
2
2 n
T

   
Illustration 5
A wheel completes 120 revolutions in one minute. Calculate its angular speed.
Solution
n = 120 rotation per minute =
120
2
60
 rotation per second
  = 2n = 2 × 2 = 4 rad s–1

Illustration 6
A scooter goes round a circular track of radius 10 m with speed of 30 ms–1
. Calculate the angular speed
of the scooter.
Solution
v = r or  =
v
r
  =
30
10
= 3 rad s–1
Solved Examples
Example 1
Let an electron moves in circular path around the nucleus. If the radius of the path is r metre find
the displacement and distance travelled by the electron in following situation
(i) For one full revolution (ii) For half revolution
(iii) For hundread revolutions
Solution
(i) In full revolution the initial and final positions are same one hence the displacement of electron will be
zero. While the distance travelled by the electron in one revolution equals to the circumference of the
circular path i.e. = 2r metre
(ii) In half revolution the initial and final positions are opposite to each other, hence the displacement of
the electron is equal to the diameter of the circular path i.e. 2r metre while the distance travelled by
the electron will be equal to r metre
(iii) For 100 revolutions the initial and final positions of the electron will remain same one hence its
displacement will be zero. While the distance travelled by the electrons in 100 full revolutions will be
equal to 100 × 2r.
Example 2*
In figure time-distance graph for two cars is given. Then find out that (i) Initially which car is
ahead and how much? (ii) Which car starts moving after and after how much time? (iii) What is
the speed of two cars (iv) After how much time and at what distance the car moving faster will
catch the second car.
Solution
(i) Initially at t = 0 the car B was at a distance 100 km and the car At t = 1 hour was at zero km. Hence
the car B is 100 km ahead of car A.
(ii) The car A starts at t = 1 hour and car B at t = 0
(iii) (a) The speed of car A = Slope of time graph

= Slope of line COA = OF/CF
= (400–0)/(4–1) km/hr = 133.33 km/hour
(b) Similarly the speed of car B = Slope of line DOB = OE/DE
= (400–100)/(4–0) km/hour = 75 km/hour
(iv) The intersection of two lines will be the point of meeting of two cars. So the two cars will meat at a
distance of 400 km. From the initial point and after 3 hours.
Example 3
A students travels with a speed of 4 km/hour from his house to his school situated at a distance of
2 km. But seeing that school is closed, return home quickly with a speed of 6 km/hour then find out
(i) Average speed of the student (ii) Average velocity of the student and
(iii) Average speed of the student for first 30 minute and first 50 minute
Solution
(i) The time taken by the student to reach the school = distance/speed
= 2 km/4 (km/hour)
Time taken in returning to home = 2 km/ (6 km/hour) = 1/2 h = 30 minute
So, the total time taken by student in whole of the journey
= 30 + 20 = 50 minute = 5/6 h
And the total distance travelled in that journey
Hence the average speed of the student = 4/(5/6) = (4×6)/5 = 4.8 km/h
(ii) After starting from home as student returns back to the home. Since the resultant displacement is
zero then velocity is also zero.
(iii) In first 30 minutes (students reaches school) the distance travelled is 2km and time taken is 1/2 hour.
So the average speed of the student is
= 2 km/30 minute = 4 km/h
In first 50 minutes students reaches school in first 30 minute and covers a distance of 2 km then in next 20
minute students returns home from the school speed with 6 km/h.
Hence the distance travelled by the student in 20 minutes when he returns to home = 6 × (1/3) = 2 km
So, the total distance travelled by the student in first 50 minutes (2 + 2) = 4 km
Hence, the average speed = 4 km/50 minute = 4/5/6 = 4.8 km/hour
Example 4
In given figure change in instantaneous speed of a particle with time is shown then find the
acceleration and distance covered by the particle in first 8 seconds.
speed
(m/s)
Solution
From figure the distance travelled by the particle = Area of speed – time graph with time axis
= 1/2 ×(8–0) × (20–0) = 80 metre
Similarly,
Acceleration = Slope of the line = (20–0)/(8–0) = 2.5 metre/sec2

Example 5*
Truck ‘A’ is going on a road with velocity 10 m/s. The second truck ‘B’ is coming from its back and
an another truck ‘C’ is coming from as front. The velocity of truck B and C is 15 m/s for each.
When the distance AB and AC is 500 metre each. The driver of truck B thinks to cross the truck A
before truck ‘C’. In this position how much minimum acceleration required for truck B.
Solution
The relative velocity of truck ‘C’ with respect to truck ‘A’
= Velocity of truck ‘C’ – Velocity of truck ‘A’ = 15 – (–10) = 25 m/s
and the distance of truck ‘C’ from truck ‘A’ = 500 metre
So the time taken by truck ‘C’ to cross truck ‘A’ = 500/25 = 20 sec
Hence the time taken by truck ‘B’ to reach up to ‘A’ should be less than 20 sec.
Again the velocity of truck ‘B’ with respect to truck ‘A’
= Velocity of truck ‘B’ – Velocity of truck ‘A’ = 15 –10 = 5m/s
If the minimum acceleration ‘a’ is required for truck ‘B’ to cover a distance of 500 metre with a relative
velocity 5 m/s in 20 seconds then by the second equation of motion.
s = ut + ½at2
500 = 5 × 20 + 1/2 × a × 20 × 20 or a = (500–100)/200 = 2 m/s
Example 6
In figure the velocity–time graph for a particle is shown. Then find out the following
(i) The average acceleration of the particle in between 6 to 8 seconds.
(ii) At what time interval the average acceleration of the particle becomes zero?
(iii) What will be the maximum acceleration?
(iv) What is the displacement of the particle in first 6 seconds?
Solution
(i) Average acceleration of the particle in between 6 to 8 seconds = Slope of line BC
= (60–30)/(8–6) = 30/2 or a = 15 metre/sec2
(ii) For zero average acceleration the velocity-time graph should be parallel to time axis. In figure portion
DE is parallel to the time axis. Hence for this portion the average acceleration will become zero. Where
the time interval is in between 10 to 14 second.
(iii) Retardation is a negative acceleration so the portion for which the velocity-time graph makes the
smallest angle with the velocity axis, the retardation is the maximum. In figure the line EF has the maximum
acceleration.
(iv) displacement of the particle in first 6 seconds will be area under the velocity time graph for first 6
seconds. So, displacement = Area OB6O = ½ × 6 × 30 = 90 m

Example 7
The velocity-time graph for a particle is according to figure. So, find the total displacement of the
particle.
Solution
According to the figure the velocity of the particle for first 4 second will be in positive direction and from
4 to 6 seconds as the velocity of particle in opposite direction so the displacement of particle will be
negative. So the total displacement.
x = Area of OAB – Areas of BED
= ½ × 4 × 10 – ½ × 2 × 10 = 10 metre
So the displacement of the particle is 10 metre and will be in positive direction.
Example 8
A runner races in straight line and his velocity time graph is shown in figure. Find the distance
covered by the runner in 20 second and also the maximum instantaneous acceleration of the
runner.
Solution
Distance covered by the runner in 20 second = Area of graph in between velocity and time
= OAM + rectangle MABN + BPC + rectangle PDQN
= ½ × (4×12) + (12×8) + ½ × (8×4) + 4 × 8 = 24 + 96 + 16 + 32 = 168 m
and the maximum acceleration = (12–0)/(4–0) = 3 metre/sec2
Example 9
If minute hand of a clock is 5 cm long. Find its angular velocity and linear velocity.
Solution
Since, minute hand of clock rotates and completes one rotation in 60 minutes.
Therefore, angled described in 60 minutes = 2 radian
Hence angular velocity of minute hand,
2ð radian 2×22radian
60 minutes 7×60×60sec
  
3 3
11 110
10 1.74 10 /
7 900 63
rad s
 
    

Now, length of minute hand, l = radius r = 5 cm
Therefore, Linear velocity of minute hand, v = r
= 5 × 1.74 × 10–3
cm/s
= 8.70 × 10–3
cm/s

Example 10*
An aircraft completes a horizontal loop of radius 1 km with a uniform speed of 900 km/h. Find the
angular velocity of the aircraft.
Solution
Given, radius of orbita of aircraft
r = 1 km
and uniform linear speed V = 900 km/h
Now, linear speed, V = radius (r) × angular velocity ()
Angular velocity,
900 /
1
v km h
r km
  
= 900 rad/h
=
900
/
60 60
rad s

= 1/4 rad/s = 0.25 rad/s
Example 11*
A body covers a distance of 20m in the 7th second and 24m in the 9th second. How much distance
shall it cover in 15th sec.
Sol.  
1
7
2
2
7 



a
u
th
S but m
th
S 20
7 
2
13
20
13
2
20 





 u
a
u ..............(i)
also s9 th = 24
2
17
24
a
u 

 ..............(ii)
from (i) equation u =
2
13
20
a
u 
 ..............(iii)
Substitute this value in (ii)
2
17
2
13
20
24
a
a



2
13
2
17
20
24
a
a



2
/
2
2
4
2
4
2
4
4 s
m
a
a
a






Use this value of (iii)
2
13
20
a
u 

13
20
2
2
13
20 





 u
u  u = 7 m/s
Now, S15th =     m
a
u 36
29
7
29
2
2
7
1
15
2
2








******

EXERCISE-I
1. What do you mean by rest?
2. What are vector quantities?
3. What do you mean by displacement?
4. What do you mean by uniform motion?
5. Which of these are vector quantities: distance, displacement, speed and velocity?
6. Convert the speed of 18 km/h into m/s.
7. Can the average velocity of a body be zero when average speed is non-zero?
8. What do you mean by uniformly accelerated motion?
9. Can the speed of a person be negative?
10. What does slope of position-time graph represent?
11. What does slope of displacement-time graph represent?
12. What is represented by slope of velocity-time graph?
13. What quantity represents area under velocity-time graph?
14. A physical quantity measured is –5 m/s, is it speed or velocity?
15. What will be direction of acceleration due to gravity at the highest point when a body is thrown upward?
16. What do you mean by uniform circular motion?
17. What do you mean by angular displacement?
18. How are linear velocity and angular velocity related?
19. What is the value of angular velocity of a (i) second’s needle (ii) minute’s needle and (iii) hour’s needle of
a clock?
EXERCISE-II
1. ‘Rest and motion’ are relative terms. Explain
2. What is the difference between scalar and vector quantities?
3. What is the nature of displacement-time graph of a body moving with constant acceleration?
4. How can you find the distance travelled by a body in uniform motion from the velocity-time graph?
5. Derive equations of uniformly accelerated motion (i) = v = u + at (ii)
2
1
2
s ut at
  (iii) v2
–u2
= 2as, from
graphs
6. Derive the relation v = r for uniform circular motion and explain the terms involved.
7. A particle moves along a circular path of radius ‘r’ and completes 3/4 part of the circle. Calculate
(i) Distance travelled by the particle
(ii) the displacement of the particle
8. A car covers a distance of 5 km in 10 minutes. Find the speed of the car in (i) km/min (ii) m/s (iii) m/min.
(iv) km/h
9. An athelete completes a circular track of radius 14 m with a uniform speed of 11 m/s. Find the time taken.
10. A non-stop bus goes from one station to another with a sped of 54 km/h. The same bus returns from second
to first station with a speed of 36 km/h. Find its average speed and average velocity for entire journey.

11. A 100 m long train crosses a 400 m bridge with a velocity of 72 km/h. Calculate the time taken to cross the
bridge.
12. A car is moving at a speed of 40 km/h. 2 seconds after its speed becomes 60 km/. Calculate the accelera-
tion of the car and the distance travelled.
13. A bus starting from rest gains a speed of 54 km/h in 10 s. What will be acceleration of the train?
14. A body starting from rest, accelerates uniformly along a straight line at the rate of 10 m/s2
for 5 sec. It
moves for 2 sec. with uniform velocity of 50 m/s. Then it travels uniformly and comes to rest in 3 sec.
Draw velocity-time graph of body and find the total distance travelled by it.
EXERCISE-III
SECTION-A
 Fill in the blanks
1. A particle moves in a circle of radius R. In half the period of revolution its displacement is ______ and
distance covered is _______.
SECTION-B
 Multiple choice question with one correct answers
1. Figures show the displacement time graph of a particle moving on the x – axis. Then match the correct
order : –
(1) (a) Object is at rest
(2) (b) Object moving with positive velocity
(3) (c) Object moving with negative velocity
(A) 1 – A, 2 – B, 3 – C (B) 1 – B, 2 – C, 3 – A (C) 1 – C, 2 – B, 3 – A (D) None of these
2. The position time (x– t) graph for two childrenAand B returning froms their school O to their homes P and
Q respectively are shown in fig. choose the correct entries in the brackets below :
(A) (B) lives closer to the school than (A).
(B) (B) starts from the school earlier than (A).
(C) (A) walks faster than (B).
(D) (A/B) overtakes (B/A) on the road (once/twice)

3. A particle has a velocity is towards east at t = o. Its accderation is towards west and is constant. Let xA
and
xB
be the magnitude of displacments in the first 10 seconds and the next 10 seconds.
(A) xA
< xB
(B) xA
= xB
(C) xA
> xB
(D) the information is in sufficient to decide the relation of xA
with xB
.
4. A person travelling on a strainght line moves with a uniform velocity v1
for some time and with uniform
velocity v2
for the next equal time. The average velocity v is given by : –
(A) v =
v v
1 2
2

(B) v = v v
1 2
(C)
2 1 1
1 2
v v v
  (D)
1 1 1
1 2
v v v
 
5. A person travelling on a straight line moves with a uniform velocity v1
for a distance x and with a uniform
velocity v2
for the next equal distance. The average velocity v is given by : –
(A) v =
v v
1 2
2

(B) v = v v
1 2
(C)
2 1 1
1 2
v v v
  (D)
1 1 1
1 2
v v v
 
6. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the
release is:
(A) a upward (B) (g – a) upward (C) (g – a) downward (D) g downward
7. A person standing near the edge of the top of a building throws two balls A and B. The ball A is throwns
vertically upward and B is thrown vertically downward with the same speed The ballAhits the ground with
a speed vA
and the ball B hits the ground with a speed vB
. we have : –
(A) VA
> VB
(B) VA
< VB
(C) VA
= VB
(D) the relation between VA
and VB
depends on height of the building above the ground
8. The acceleration of a car started at t = o, varies with times as shown in figure. the distance travelled in 15
seconds is : –
(A) 100 m (B) 25 m (C) 150 m (D) 0 m
9. A player throws a ball upwards with an initial speed of 29.4 m s–1
. The time taken by the ball to return to the
player’s hands is. (Take g = 9.8 m s–2
and neglect air resistance).
(A) 10 sec (B) 8 sec (C) 12 sec (D) 6 sec
10. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h–1
. Finding
the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1
. the average speed
of the man over the interval of time 0 to 40 min.
(A) 5 km h–1
(B) 7.5 km h–1
(C) 5.625 km h–1
(D) None of these
11. What is the speed with which a stone leaves a string, if it is let go when the length of the string holding the
stone is 0.7 m and is revolving at 3 rev/s?
(A) 15.6 m/s (B) 10.2 m/s (C) 13.2 m/s (D) none of these
12. Ratio of displacment to distance is:
(A) always < 1 (B) always = 1 (C) always > 1 (D) = or < 1

SECTION-C
 Assertion & Reason
Instructions: In the following questions as Assertion (A) is given followed by a Reason (R). Mark your
responses from the following options.
(A) Both Assertion and Reason are true and Reason is the correct explanation of ‘Assertion’
(B) Both Assertion and Reason are true and Reason is not the correct explanation of ‘Assertion’
(C) Assertion is true but Reason is false
(D) Assertion is false but Reason is true
1. Assertion: The v–t graph perpendicular to the time axis is not possible.
Reason: If v–t graph is perpendicular to the time axis, then acceleration of the particle should be infinite.
2. Assertion: Retardation is directed opposite to the velocity
Reason: Retardation is equal to the time rate of decrease of speed.
3. Assertion: Relative velocity when particles are moving on the same straight line path can be greater in
magnitude than velocity of either particle.
Reason: When the particles are moving with velocities v1
and v2
in opposite directions, then relative
velocity = v1
+ v2
4. Assertion: A body can have acceleration even if its velocity is zero at a given instant of time.
Reason: A body is momentarily at rest when it reverses its direction of motion.
5. Assertion: The velocity of a particle may vary even when its speed is constant.
Reason: Such a body may move along a circular path
6. Assertion: The x–t graph for a body at rest is a straight line parallel to time axis.
Reason: A body at rest does not change its position with the lapse of time.
SECTION-D
 Match the following (one to one)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. Only One entries of column-I may have the matching with the some entries of column-
II and one entry of column-II Only one matching with entries of column-I
1. Column I Column II
(Distance time graph) (Type of motion)
(A) (P) Body moves with constant speed
(B) (Q) Non-uniform motion

(C) (R) Body is stationary
(D) (S) Not possible
(Type of motion) (Nature of Acceleration)
(A) Circular Motion (P)Acceleration is along the direction of motion
(B) UniformMotion (Q)Acceleration is along perpendicular
direction of the motion.
(C) Free Fall (R)Acceleration is in the opposite direction of motion
(D) A stone thrown upward with (S) Acceleration is Zero
initial velocity‘u’
EXERCISE-IV
SECTION-A
 Multiple choice question with one correct answers
1. The following shows the time-velocity graph for a moving object.
The maximum acceleration will be-
(A) 1 m/sec2
(B) 2 m/sec2
(C) 3 m/sec2
(D) 4 m/sec2
2. The adjoining curve represents the velocity time graph of a
particle, its acceleration values along OA, AB and BC in m,/sec2
.
are respectively-
(A) 1, 0, –0.5 (B) 1 , 0, 0.5
(C) 1, 1, 0.5 (D) 1, 0.5, 0
3. The v-t graph of a linear motion is shown in adjoining figure.
The distance from origin after 8 sec. is-
(A) 18 meters (B) 16 meters
(C) 8 meters (D) 6 meters
4. Amotor car covers 1/3 part of total distance with v1
= 10 km/hr, second 1/3 part with v2
= 20 km/hr and rest
1/3 part with v3
= 60 km/hr. What is the average speed the car-
(A) 18 km/hr (B) 45 km/hr (C) 6 km/hr (D) 22.5 km/hr

5. A car covers a distance of 2 km. in 2.5 min. if it covers half of the distance with speed 40 km/hr the rest
distane it will cover with speed-
(A) 56 km/hr (B) 60 km/hr (C) 50 km/hr (D) 48 km/hr
6. A cyclist moving on a circular track of radius 40 m completes half a revolution in 40 sec. Its average
velocity is-
(A) Zero (B) 5 m/sec. (C) 2 m/sec. (D) 3.5 m/sec.
7. A truck starts from rest with an acceleration of 1.5 m/sec2
while a car 150 metre behind starts from rest
with an acceleration of 2 m/sec2
. How long will it take before both the truck and car are side by side-
(A) 5.24 sec. (B) 24.5 sec. (C) 2.45 sec. (D) 52.4 sec.
8. Atrain of length 100 m is crossing a bridge 200 m in length at the speed of 72 km per hour. The time taken
by the train to cross the bridge is-
(A) 24 sec (B) 15 sec (C) 10 min (D) 10 sec
9. Which of the following statements is wrong about a ball thrown vertically up-
(a) It is moving with constant acceleration
(b) It may have different velocities at the same position
(c) It may have two positions at the same time
(d) The angular momentum of the particle about origin remains conserved
(A) c only (B) c, d (C) b, c, d (D) a, b, c and d
10. A body travelling with uniform acceleration crosses two points A and B with velocities 20 m/sec and 30
m/sec respectively. Then the speed of the body at mid-point of A and B is-
(A) 25 m/sec (B) 25.5 m/sec (C) 24 m/sec (D) 10 6 m/sec
11. A bus accelerates uniformly from rest and acquires a speed of 36 km/hour in 10 seconds. The acceleration
is
(A) 1000 m/sec2
(B) 1 m/sec2
(C) 100 m/sec2
(D) 10 m/sec2
12. Abody thrown up with a velocity reaches a maximum height of 50 meters. Another body with double the
mass thrown up, with double the initial velocity will reach a maximum height of
(A) 100 metres (B) 200 metres (C) 10 metres (D) 400 metres
13. One 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
(A) 40 km/hr (B) 80 km/hr (C) 46
3
2
km/hr (D) 36 km/hr
14. 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.
(A) 8.3 m (B) 9.3 m (C) 10.3 m (D) none of these
15. Acceleration of a particle changes when
(A) direcion of velocity changes (B) magnitude of velocity changes
(C) both of above (D) speed changes
16. If a body of mass 3 kg is dropped from the top of a tower of height 50 m, then its K.E. after 3 sec will be
(A) 1296J (B) 1048J (C) 735J (D) zero
17. A body released from a great height falls freely towards earth. Another body is released from the same
height exactly one second later. The separation between the two bodies two second after the release of the
second body is
(A) 9.8 m (B) 49 m (C) 24.5 m (D) 19.6 m

SECTION-B
 Multiple choice question with one or more than one correct answers
1. Which of the following statements regarding the freely falling body is/are correct?
(A) The body is uniformly accelerated
(B) The body is non-uniformly accelerated
(C) The distance travelled by the body in the first second, second second and third second are in the ratio
of 1 : 3 : 5
(D) The distance travelled by the body in the first second, first two seconds and first three seconds are in
the ratio of 1 : 4 : 9
2. The velocity time graph of an object is shown in figure.
Which of the following statement is/are correct ?
(A) The slanted portion of the v-t graph represents constant acceleration.
(B) The horizontal portion represents constant velocity
(C) Area under v-t graph gives the momentum change.
(D) Area under v-t graph is equal to acceleration
3. In an examination, the students were asked to draw distance -time graph of a school boy going from home
straight to school and starting back home immediately (without any time loss). The following graphs were
drawn by four students. Which of these graphs do not depict the real situation ?
(A) (B) (C) (D)
SECTION-C
 Comprehension
A “sun yacht” is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in
every day circumstances, it can be large enough to send the space craft outward from the Sun on a cost-
free but slow trip. Suppose that the spacecraft has a mass of 900 kg and received a push of 20 N.
1. The magnitude of the resulting acceleration is :
(A) 45 ms2
(B) 18000 ms2
(C)
45
1
ms2
(D) None of these
2. If the craft starts from rest how far will it travel in 1 min
(A) 40 m (B) 30 m (C) 100 m (D) None of these
3. How fast will it then be moving
(A) 3/4 m/s (B) 4/3 m/s (C) 5/4 m/s (D) None of these
SECTION-D
 Match the following (one to many)
Column-I and column-II contains four entries each. Entries of column-I are to be matched with some
entries of column-II. One or more than one entries of column-I may have the matching with the some entries
of column-II and one entry of column-II may have one or more than one matching with entries of column-I
Displacement time(x - t)/velocity-time graph (v - t) Type of motion
(A) (P) Uniform motion

(B) (Q) Non-uniform motion
(C) (R) constant velocity
(D) (S) uniformly accelerated motion
EXERCISE-V
1. An object travels 16m in 4 s and then another 16 m in 2 s. What is the average speed of the object?
2. The odometer of a car reads 2000 km at the start of a trip and 2400 km at the end of the trip. If the trip
took 8 h, calculate the average speed of the car in km h–1
and m s–1
.
3. Ushaswimsina90mlongpool.Shecovers180minoneminutebyswimmingfromoneendtotheother
and backalongthe same straight path. Find the average speed and average velocityof Usha.
4. Starting froma stationaryposition, Rahupaddles his bicycle toattain a velocityof6 m s–1
in30s. Then
he applies brakes such that the velocityof the bicycle comes down to4 m s–1
in the next 5 s. Calculate
the accelerationof the bicycle inboth the cases
5. The brakes appliedto a car produce an accelerationof 6m s2
inthe oppositedirection to the motion. If
the cartakes 2 s tostop after the application of brakes, calculate thedistance it travels duringthis time.
EXERCISE-VI
1. An object has moved through a distance. Can it have zero displacement? If yes. support your answer
withanexample.
2. Afarmermoves alongtheboundaryof asquare fieldofside10m in40s.What will bethemagnitudeof
displacement ofthefarmerat theendof2mintues20seconds from hisinitial position?
3. Whichofthefollowingistruefordisplacement?
(a) It cannot be zero
(b) its magnitudeis greater thanthe distance travelled bythe object.
4. Distinguishbetweenspeedandvelocity.
5. Under what condition(s) is the magnitudeof average velocityofan object equal toits average speed?
6. What does the odometer of anautomobile measure?
7. What does thepath of anobject look like when it is in uniform motion?
8. Duringanexperiment, asignal from a spaceship reached thegroundstation in fiveminutes. What was
the distance of the spaceship from the ground station? The signal travels at the speed of light, that is
3×108
m s–1
.

9. When willyou saya bodyis in
(i)Uniformacceleration?
(ii)Non-uniformacceleration?
10. A bus decreases its speed from 80 km h–1
to 60 km h–1
in 5 s. Find the acceleration of the bus.
11. Atrainstartingfromarailwaystationandmoveingwithuniformaccelerationattainsaspeed40kmh–1
in
10 minutes.Finditsacceleration.
12. What is thenatureofthedistance-time graphs foruniformandnon-uniform motionofan object?
13. What can yousayabout the motionof an object whose distance-timegraph is a straight line parallel to
thetimeaxis?
14. What can you sayabout the motion of an object if its speed-time graph is a straight line parallel to the
timeaxis?
15. What is thequantitywhich is measuredbythe area occupied belowthe velocity-timegraph?
16. Abus startingfrom rest moves withauniform accelerationof0.1m s2
for2minutes.Find (a)thespeed
acquired, (b)thedistance travelled.
17. A train is travellingat a speed of90 km h–1
. Brakes areapplied so as to producea uniform acceleration
of –0.5 m s–2
. Findhow far thetrain will go before it is brought to rest.
18. Atrolley,whilegoingdownaninclinedplane,hasanaccelerationof2cms–2
.Whatwill beitsvelocity3
s after the start?
19. A racingcarhas a uniform acceleration of4 m s–2
. What distancewill it cover in 10s after start.
20. A stone is thrown in a verticallyupward direction with a velocityof 5 m s–1
. If the acceleration of the
stone during its motion is 10 m s–2
in the downward direction, what will be the height attained bythe
stoneand howmuch time will it taketoreach there?
******
Answers
Knowledge base questions
1. (B) 2. (A) 3. (C) 4. (B) 5. (B)
6. (C) 7. (B) 8. (A) 9. (C) 10. (C)
Try Yourself
1. Distance = r, Displacement = 2r
2. 37.5 km/h 3. 0.56 ms–2
4. 40 m 5. 3 sec
Exercise-I
1. If an object do not change its position with respect to time it is said to be at rest.
2. Those quantities which are represented completely by magnitude and direction both. Examples are: dis-
placement, velocity, acceleration, force, momentum, electric field, angular velocity, angular-acceleration.
3. Displacement is the change in position of a body along the shortest path from its initial position to final
position.
4. Motion is said to be uniform if the body covers equal distances or/displacements in equal intevals of time,
however small, the interval may be

5. Displacement and velocity are vector quantities
6. V = 18 km/h =
18 1000
5 /
(60 60)
m
m s
s



7. Yes. If a body moves fromA to B along straight path AB, and if it comes back to A, then net displacement
= 0 and hence its average velocity =
( )
0
Net displacement AB BA
Total time t t

 


 

But average speed
2
0
2
AB BA AB AB
t t t t

   

8. Uniformly accelerated motion is that in which velocity of the body increases continuously and at constant
rate w.r.t time.
9. No, speed can never be negative, therefore distance covered is always positive.
10. Slope = tan =
distance
=speed
time
11. Slope = tan =
displacement
=velocity
time interval
12. Slope of velocity-time graph,
change in velocity
tan =acceleration
time interval
 
13. It represents, displacement S


= Velocity × time = Area under V–t graph
14. Velocity, Speed =
distance
time
, can never be negative
15. Vertically downward directed a g
 

, at the highest point
16. Motion in which an object revolve in a circle with constant speed is called uniform circular motion.
17. Angle subtended by an object at the centre of circle in given time.
18. Linear velocity, v = (radius ‘r’) × (angular velocity ‘’)
19. Angular velocity
Angle described
=
time taken
(i) For second’s needle,
2
= /
60sec
s rad s

 (1 complete rotation in 60 sec, angled described = 2)
(ii) For minute needle,
2 2
= /
60min 60 60
m rad s
 
 

(iii) For hour’s needle,
2 2
= /
12 12 60 60 sec.
h rad s
hour
 
 
 
Exercise-I
7. (i) r

2
3
, (ii) r
2
8. (i) 0.5 km/h, (ii) 8.33 m/sec, (iii) 500 m/min.
Exercise-III
Section-A
1. 2r, 2

Section-B
1. (B) 2. (D) 3. (D) 4. (A) 5. (C)
6. (D) 7. (C) 8. (A) 9. (D) 10. (C)
11. (C) 12. (D)
Section-C
1. (B) 2. (A) 3. (A) 4. (A) 5. (B)
6. (A)
Section-D
1. (A)-(R), (B)-(S), (C)-(P), (D)-(Q) 2. (A)-(Q), (B)-(S), (C)-(P), (D)-(R)
Exercise-IV
Section-A
1. (D) 2. (A) 3. (A) 4. (A) 5. (B)
6. (C) 7. (B) 8. (B) 9. (A) 10. (B)
11. (B) 12. (B) 13. (D) 14. (A) 16. (C)
16. (A) 17. (C)
Section-B
1. (A,C,D) 2. (A,B) 3. (B,C)
Section-C
1. (C) 2. (A) 3. (B)
Section-D
1. (A)-(Q,S), (B)-(P,R), (C)-(Q), (D)-(P,R)
Exercise-V
1. 5.33 m/sec 2. 50 km/hr., 13.9 m/sec 3. 3 m/sec, 0
4. 0.2 m/sec2
, –0.4 m/sec2
5. 12 m
Exercise-VI
2. 14.1 m

class 9 science chapter 7 motion pdf download

More Related Content

What's hot (20)

Similar to class 9 science chapter 7 motion pdf download (20)

More from Vivekanand Anglo Vedic Academy (20)

Recently uploaded (20)

class 9 science chapter 7 motion pdf download