In everyday life, we see some objects at rest
and others in motion. Birds fly, fish swim,
blood flows through veins and arteries, and
cars move. Atoms, molecules, planets, stars
and galaxies are all in motion. We often
perceive an object to be in motion when its
position changes with time. However, there
are situations where the motion is inferred
through indir
ect evidences. For example, we
infer the motion of air by observing the
movement of dust and the movement of leaves
and branches of trees. What causes the
phenomena of sunrise, sunset and changing
of seasons? Is it due to the motion of the
earth? If it is true, why don’t we directly
perceive the motion of the earth?
An object may appear to be moving for
one person and stationary for some other. For
the passengers in a moving bus, the roadside
trees appear to be moving backwards. A
person standing on the road–side perceives
the bus alongwith the passengers as moving.
However, a passenger inside the bus sees his
fellow passengers to be at rest. What do these
observations indicate?
Most motions are complex. Some objects
may move in a straight line, others may take
a circular path. Some may rotate and a few
others may vibrate. There may be situations
involving a combination of these. In this
chapter, we shall first learn to describe the
motion of objects along a straight line. We
shall also learn to express such motions
through simple equations and graphs. Later,
we shall discuss ways of describing
circular motion.
Activity ______________ 7.1
• Discuss whether the walls of your
classr
oom are at rest or in motion.
Activity ______________ 7.2
• Have you ever experienced that the
train in which you are sitting appears
to move while it is at rest?
• Discuss and share your experience.
Think and Act
We sometimes are endangered by the
motion of objects around us, especially
if that motion is erratic and
uncontrolled as observed in a flooded
river, a hurricane or a tsunami. On the
other hand, controlled motion can be a
service to human beings such as in the
generation of hydro-electric power. Do
you feel the necessity to study the
erratic motion of some objects and
learn to control them?
7.1 Describing Motion
We describe the location of an object by
specifying a reference point. Let us
understand this by an example. Let us
assume that a school in a village is 2 km north
of the railway station. We have specified the
position of the school with respect to the
railway station. In this example, the railway
station is the reference point. We could have
also chosen other reference points according
to our convenience. Therefore, to describe the
position of an object we need to specify a
reference point called the origin.
7
MM
M
M
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OTIONOTION
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OTION
C
hapter
Rationalised 2023-24
7.1.1 MOTION ALONG A STRAIGHT LINE
The simplest type of motion is the motion
along a straight line. We shall first learn to
describe this by an example. Consider the
motion of an object moving along a straight
path. The object starts its journey from O
which is treated as its reference point
(Fig. 7.1). Let A, B and C represent the position
of the object at different instants. At first, the
object moves through C and B and reaches A.
Then it moves back along the same path and
reaches C through B.
are used to describe the overall motion of an
object and to locate its final position with
reference to its initial position at a given time.
Activity ______________ 7.3
• Take a metre scale and a long rope.
• Walk from one corner of a basket-ball
court to its oppposite corner along its
sides.
• Measure the distance covered by you
and magnitude of the displacement.
• What difference would you notice
between the two in this case?
Activity ______________ 7.4
• Automobiles are fitted with a device
that shows the distance travelled. Such
a device is known as an odometer. A
car is driven from Bhubaneshwar to
New Delhi. The difference between the
final reading and the initial reading of
the odometer is 1850 km.
• Find the magnitude of the displacement
between Bhubaneshwar and New Delhi
by using the Road Map of India.
The total path length covered by the object
is OA + AC, that is 60 km + 35 km = 95 km.
This is the distance covered by the object. To
describe distance we need to specify only the
numerical value and not the direction of
motion. There are certain quantities which
are described by specifying only their
numerical values. The numerical value of a
physical quantity is its magnitude. From this
example, can you find out the distance of the
final position C of the object from the initial
position O? This difference will give you the
numerical value of the displacement of the
object from O to C through A. The shortest
distance measured from the initial to the final
position of an object is known as
the displacement.
Can the magnitude of the displacement be
equal to the distance travelled by an object?
Consider the example given in (Fig. 7.1). For
motion of the object from O to A, the distance
covered is 60 km and the magnitude of
displacement is also 60 km. During its motion
from O to A and back to B, the distance covered
Fig. 7.1: Positions of an object on a straight line path
= 60 km + 25 km = 85 km while the magnitude
of displacement = 35 km. Thus, the magnitude
of displacement (35 km) is not equal to the path
length (85 km). Further, we will notice that the
magnitude of the displacement for a course of
motion may be zero but the corresponding
distance covered is not zero. If we consider the
object to travel back to O, the final position
concides with the initial position, and therefore,
the displacement is zero. However, the distance
covered in this journey is OA + AO = 60 km +
60 km = 120 km. Thus, two different physical
quantities—the distance and the displacement,
MOTION 73
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SCIENCE74
uestions
1. An object has moved through a
distance. Can it have zero
displacement? If yes, support
your answer with an example.
2. A farmer moves along the
boundary of a square field of side
10 m in 40 s. What will be the
magnitude of displacement of the
farmer at the end of 2 minutes 20
seconds from his initial position?
3. Which of the following is true for
displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than
the distance travelled by the
object.
7.1.2 UNIFORM MOTION AND NON-
UNIFORM MOTION
Consider an object moving along a straight
line. Let it travel 5 m in the first second,
5 m more in the next second, 5 m in the
third second and 5 m in the fourth second.
In this case, the object covers 5 m in each
second. As the object covers equal distances
in equal intervals of time, it is said to be in
uniform motion. The time interval in this
motion should be small. In our day-to-day
life, we come across motions where objects
cover unequal distances in equal intervals
of time, for example, when a car is moving
on a crowded street or a person is jogging
in a park. These are some instances of
non-uniform motion.
Activity ______________ 7.5
• The data regarding the motion of two
different objects A and B are given in
Table 7.1.
• Examine them carefully and state
whether the motion of the objects is
uniform or non-uniform.
Q
(a)
(b)
Fig. 7.2
Table 7.1
Time Distance Distance
travelled by
travelled by
object A in m object B in m
9:30 am 10 12
9:45 am 20 19
10:00 am 30 23
10:15 am 40 35
10:30 am 50 37
10:45 am 60 41
11:00 am 70 44
7.2 Measuring the Rate of Motion
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MOTION 75
Look at the situations given in Fig. 7.2. If
the bowling speed is 143 km h
–1
in Fig. 7.2(a)
what does it mean? What do you understand
from the signboard in Fig. 7.2(b)?
Different objects may take different
amounts of time to cover a given distance.
Some of them move fast and some move
slowly. The rate at which objects move can
be different. Also, differ
ent objects can move
at the same rate. One of the ways of
measuring the rate of motion of an object is
to find out the distance travelled by the object
in unit time. This quantity is referred to as
speed. The SI unit of speed is metre per
second. This is represented by the symbol
m s
–1
or m/s.
The other units of speed include
centimetre per second (cm s
–1
) and kilometre
per hour (km h
–1
). To specify the speed of an
object, we require only its magnitude. The
speed of an object need not be constant. In
most cases, objects will be in non-uniform
motion. Therefore, we describe the rate of
motion of such objects in terms of their
average speed. The average speed of an object
is obtained by dividing the total distance
travelled by the total time taken. That is,
average speed =
Total distance travelled
Total time taken
If an object travels a distance s in time t then
its speed v is,
v =
s
t
(7.1)
Let us understand this by an example. A
car travels a distance of 100 km in 2 h. Its
average speed is 50 km h
–1
. The car might
not have travelled at 50 km h
–1
all the time.
Sometimes it might have travelled faster and
sometimes slower than this.
Example 7.1 An object travels 16 m in 4 s
and then another 16 m in 2 s. What is
the average speed of the object?
Solution:
Total distance travelled by the object =
16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed =
Total distance travelled
Total time taken
=
32 m
6 s
= 5.33 m s
–1
Therefore, the average speed of the object
is 5.33 m s
–1
.
7.2.1 SPEED WITH DIRECTION
The rate of motion of an object can be more
comprehensive if we specify its direction of
motion along with its speed. The quantity that
specifies both these aspects is called velocity.
Velocity is the speed of an object moving in a
definite direction. The velocity of an object
can be uniform or variable. It can be changed
by changing the object’s speed, direction of
motion or both. When an object is moving
along a straight line at a variable speed, we
can express the magnitude of its rate of
motion in terms of average velocity. It is
calculated in the same way as we calculate
average speed.
In case the velocity of the object is
changing at a uniform rate, then average
velocity is given by the arithmetic mean of
initial velocity and final velocity for a given
period of time. That is,
average velocity =
initial velocity + final velocity
2
Mathematically, v
av
=
u + v
2
(7.2)
where v
av
is the average velocity, u is the initial
velocity and v is the final velocity of the object.
Speed and velocity have the same units,
that is, m s
–1
or m/s.
Activity ______________ 7.6
• Measure the time it takes you to walk
from your house to your bus stop or
the school. If you consider that your
average walking speed is 4 km h
–1
,
estimate the distance of the bus stop
or school from your house.
Rationalised 2023-24
SCIENCE76
=
50
km 1000 m 1h
× ×
h 1km 3600s
= 13.9 m s
–1
The average speed of the car is
50 km h
–1
or 13.9 m s
–1
.
Example 7.3 Usha swims in a 90 m long
pool. She covers 180 m in one minute
by swimming from one end to the other
and back along the same straight path.
Find the average speed and average
velocity of Usha.
Solution:
Total distance covered by Usha in 1 min
is 180 m.
Displacement of Usha in 1 min = 0 m
Average speed =
Total distance covered
Total timetaken
=
180 m 180 m 1 min
= ×
1min 1min 60 s
= 3 m s
-1
Average velocity =
Displacement
Total timetaken
=
0 m
60 s
= 0 m s
–1
The average speed of Usha is 3 m s
–1
and her average velocity is 0 m s
–1
.
7.3 Rate of Change of Velocity
During uniform motion of an object along a
straight line, the velocity remains constant
with time. In this case, the change in velocity
of the object for any time interval is zero.
However, in non-uniform motion, velocity
varies with time. It has different values at
different instants and at differ
ent points of
the path. Thus, the change in velocity of the
object during any time interval is not zero.
Can we now express the change in velocity of
an object?
Activity ______________ 7.7
• At a time when it is cloudy, there may
be frequent thunder and lightning. The
sound of thunder takes some time to
reach you after you see the lightning.
• Can you answer why this happens?
• Measure this time interval using a
digital wrist watch or a stop watch.
• Calculate the distance of the nearest
point of lightning. (Speed of sound in
air = 346 m s
-1
.)
uestions
1. Distinguish between speed and
velocity.
2. Under what condition(s) is the
magnitude of average velocity of
an object equal to its average
speed?
3. What does the odometer of an
automobile measure?
4. What does the path of an object
look like when it is in uniform
motion?
5. During an experiment, a signal
from a spaceship reached the
ground station in five minutes.
What was the distance of the
spaceship from the ground
station? The signal travels at the
speed of light, that is, 3 × 10
8
m s
–1
.
Example 7.2 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
.
Solution:
Distance covered by the car,
s = 2400 km – 2000 km = 400 km
Time elapsed, t = 8 h
Average speed of the car is,
v
av
=
400 km
8 h
=
s
t
= 50 km h
–1
Q
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MOTION 77
To answer such a question, we have to
introduce another physical quantity called
acceleration, which is a measure of the
change in the velocity of an object per unit
time. That is,
acceleration =
change in velocity
time taken
If the velocity of an object changes from
an initial value u to the final value v in time t,
the acceleration a is,
v – u
a =
t
(7.3)
This kind of motion is known as
accelerated motion. The acceleration is taken
to be positive if it is in the direction of velocity
and negative when it is opposite to the
direction of velocity. The SI unit of
acceleration is m s
–2
.
If an object travels in a straight line and
its velocity increases or decreases by equal
amounts in equal intervals of time, then the
acceleration of the object is said to be
uniform. The motion of a freely falling body
is an example of uniformly accelerated
motion. On the other hand, an object can
travel with non-uniform acceleration if its
velocity changes at a non-uniform rate. For
example, if a car travelling along a straight
road increases its speed by unequal amounts
in equal intervals of time, then the car is said
to be moving with non-uniform acceleration.
Activity ______________ 7.8
• In your everyday life you come across
a range of motions in which
(a) acceleration is in the direction of
motion,
(b) acceleration is against the
direction of motion,
(c) acceleration is uniform,
(d) acceleration is non-uniform.
• Can you identify one example each
for the above type of motion?
Example 7.4 Starting from a stationary
position, Rahul paddles his bicycle to
attain a velocity of 6 m s
–1
in 30 s. Then
he applies brakes such that the velocity
of the bicycle comes down to 4 m s
-1
in
the next 5 s. Calculate the acceleration
of the bicycle in both the cases.
Solution:
In the first case:
initial velocity, u = 0 ;
final velocity, v = 6 m s
–1
;
time, t = 30 s .
From Eq. (8.3), we have
v – u
a =
t
Substituting the given values of u,v and
t in the above equation, we get
(
)
–1
–1
6 m s – 0 m s
=
30 s
a
= 0.2 m s
–2
In the second case:
initial velocity, u = 6 m s
–1
;
final velocity, v = 4 m s
–1
;
time, t = 5 s.
Then,
(
)
–1
–1
4 m s – 6 m s
=
5 s
a
= –0.4 m s
–2
.
The acceleration of the bicycle in the
first case is 0.2 m s
–2
and in the second
case, it is –0.4 m s
–2
.
uestions
1. When will you say a body is in
(i) uniform acceleration? (ii) non-
uniform acceleration?
2. A bus decreases its speed from
80 km h
–1
to 60 km h
–1
in 5 s.
Find the acceleration of the bus.
3. A train starting from a railway
station and moving with uniform
acceleration attains a speed
40 km h
–1
in 10 minutes. Find its
acceleration.
Q
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SCIENCE78
7.4 Graphical Representation of
Motion
Graphs provide a convenient method to
present basic information about a variety of
events. For example, in the telecast of a
one-day cricket match, vertical bar graphs
show the run rate of a team in each over. As
you have studied in mathematics, a straight
line graph helps in solving a linear equation
having two variables.
To describe the motion of an object, we
can use line graphs. In this case, line graphs
show dependence of one physical quantity,
such as distance or velocity, on another
quantity, such as time.
7.4.1 DISTANCE–TIME GRAPHS
The change in the position of an object with
time can be represented on the distance-time
graph adopting a convenient scale of choice.
In this graph, time is taken along the x–axis
and distance is taken along the y-axis.
Distance-time graphs can be employed under
various conditions where objects move with
uniform speed, non-uniform speed, remain
at rest etc.
Fig. 7.3: Distance-time graph of an object moving
with uniform speed
We know that when an object travels equal
distances in equal intervals of time, it moves
with uniform speed. This shows that the
distance travelled by the object is directly
proportional to time taken. Thus, for uniform
speed, a graph of distance travelled against
time is a straight line, as shown in Fig. 7.3.
The portion OB of the graph shows that the
distance is increasing at a uniform rate. Note
that, you can also use the term uniform
velocity in place of uniform speed if you take
the magnitude of displacement equal to the
distance travelled by the object along the
y-axis.
We can use the distance-time graph to
determine the speed of an object. To do so,
consider a small part AB of the distance-time
graph shown in Fig 7.3. Draw a line parallel
to the x-
axis from point A and another line
parallel to the y-axis from point B. These two
lines meet each other at point C to form a
triangle ABC. Now, on the graph, AC denotes
the time interval (t
2
– t
1
) while BC corresponds
to the distance (s
2
– s
1
). We can see from the
graph that as the object moves from the point
A to B, it covers a distance (s
2
– s
1
) in time
(t
2
– t
1
). The speed, v of the object, therefore
can be represented as
v =
2 1
2 1
–
–
s s
t t
(7.4)
We can also plot the distance-time graph
for accelerated motion. Table 7.2 shows the
distance travelled by a car in a time interval
of two seconds.
Table 7.2: Distance travelled by a
car at regular time intervals
Time in seconds Distance in metres
0 0
2 1
4 4
6 9
8 16
10 25
12 36
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MOTION 79
The distance-time graph for the motion
of the car is shown in Fig. 7.4. Note that the
shape of this graph is different from the earlier
distance-time graph (Fig. 7.3) for uniform
motion. The nature of this graph shows non-
linear variation of the distance travelled by
the car with time. Thus, the graph shown in
Fig 7.4 represents motion with non-uniform
speed.
7.4.2 VELOCITY-TIME GRAPHS
The variation in velocity with time for an object
moving in a straight line can be represented
by a velocity-time graph. In this graph, time is
represented along the x-
axis and the velocity
Fig. 7.4: Distance-time graph for a car moving with
non-uniform speed
Fig. 7.5: Velocity-time graph for uniform motion
of a car
is represented along the y-axis. If the object
moves at uniform velocity, the height of its
velocity-time graph will not change with time
(Fig. 7.5). It will be a straight line parallel to
the x-axis. Fig. 7.5 shows the velocity-time
graph for a car moving with uniform velocity
of 40 km h
–1
.
We know that the product of velocity and
time give displacement of an object moving
with uniform velocity. The ar
ea enclosed by
velocity-time graph and the time axis will be
equal to the magnitude of the displacement.
To know the distance moved by the car
between time t
1
and t
2
using Fig. 7.5, draw
perpendiculars from the points corresponding
to the time t
1
and t
2
on the graph. The velocity
of 40 km h
–1
is represented by the height AC
or BD and the time (t
2
– t
1
) is represented by
the length AB.
So, the distance s moved by the car in
time (t
2
– t
1
) can be expressed as
s = AC × CD
= [(40 km h
–1
) × (t
2
– t
1
) h]
= 40 (t
2
– t
1
) km
= area of the rectangle ABDC (shaded
in Fig. 7.5).
We can also study about uniformly
accelerated motion by plotting its velocity–
time graph. Consider a car being driven along
a straight road for testing its engine. Suppose
a person sitting next to the driver records its
velocity after every 5 seconds by noting the
reading of the speedometer of the car. The
velocity of the car, in km h
–1
as well as in
m s
–1
,
at different instants of time is shown
in table 7.3.
Table 7.3: Velocity of a car at
regular instants of time
Time Velocity of the car
(s) (m s
–1
) (km h
–1
)
0 0 0
5 2.5 9
10 5.0 18
15 7.5 27
20 10.0 36
25 12.5 45
30 15.0 54
Rationalised 2023-24
SCIENCE80
In this case, the velocity-time graph for the
motion of the car is shown in Fig. 7.6. The
nature of the graph shows that velocity
changes by equal amounts in equal intervals
of time. Thus, for all uniformly accelerated
motion, the velocity-time graph is a
straight line.
Fig. 7.6: Velocity-time graph for a car moving with
uniform accelerations.
You can also deter
mine the distance
moved by the car from its velocity-time graph.
The area under the velocity-time graph gives
the distance (magnitude of displacement)
moved by the car in a given interval of time.
If the car would have been moving with
uniform velocity, the distance travelled by it
would be represented by the area ABCD
under the graph (Fig. 7.6). Since the
magnitude of the velocity of the car is
changing due to acceleration, the distance s
travelled by the car will be given by the area
ABCDE under the velocity-time graph
(Fig. 7.6).
That is,
s = area ABCDE
= area of the rectangle ABCD + area of
the triangle ADE
= AB × BC +
1
2
(AD × DE)
In the case of non-uniformly accelerated
motion, velocity-time graphs can have any
shape.
Fig. 7.7: Velocity-time graphs of an object in non-
uniformly accelerated motion.
Fig. 7.7(a) shows a velocity-time graph that
represents the motion of an object whose
velocity is decreasing with time while
Fig. 7.7 (b) shows the velocity-time graph
representing the non-uniform variation of
velocity of the object with time. Try to interpret
these graphs.
Activity ______________ 7.9
• The times of arrival and departure of
a train at three stations A, B and C
and the distance of stations B and C
from station A are given in Table 7.4.
Table 7.4: Distances of stations B
and C from A and times of arrival
and departure of the train
Station Distance Time of Time of
from A arrival departure
(km) (hours) (hours)
A 0 08:00
08:15
B 120 11:15 11:30
C 180 13:00
13:15
• Plot and interpret the distance-time
graph for the train assuming that its
motion between any two stations is
uniform.
Velocity (km h
–1
)
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MOTION 81
4. What is the quantity which is
measured by the area occupied
below the velocity-time graph?
7.5 Equations of Motion
When an object moves along a straight line
with uniform acceleration, it is possible to
relate its velocity, acceleration during motion
and the distance covered by it in a certain
time interval by a set of equations known as
the equations of motion. For convenience, a
set of three such equations are given below:
v = u + at (7.5)
s = ut + ½ at
2
(7.6)
2 a s = v
2
– u
2
(7.7)
where u is the initial velocity of the object which
moves with uniform acceleration a for time t,
v is the final velocity, and s is the distance
travelled by the object in time t. Eq. (7.5)
describes the velocity-time relation and Eq.
(7.6) represents the position-time relation. Eq.
(7.7), which represents the relation between the
position and the velocity, can be obtained from
Eqs. (7.5) and (7.6) by eliminating t. These
three equations can be derived by graphical
method.
Example 7.5 A train starting from rest
attains a velocity of 72 km h
–1
in 5
minutes. Assuming that the acceleration
is uniform, find (i) the acceleration and
(ii) the distance travelled by the train for
attaining this velocity.
Solution:
We have been given
u = 0 ; v = 72 km h
–1
= 20 m s
-1
and
t = 5 minutes = 300 s.
(i) From Eq. (7.5) we know that
(
)
v – u
a =
t
–1
–1
–2
20 m s – 0 m s
=
300 s
1
= m s
15
Activity _____________ 7.10
• Feroz and his sister Sania go to school
on their bicycles. Both of them start at
the same time from their home but take
different times to reach the school
although they follow the same route.
Table 7.5 shows the distance travelled
by them in different times
Table 7.5: Distance covered by
Feroz and Sania at different
times on their bicycles
Time Distance Distance
travelled
travelled
by Feroz by Sania
(km)
(km)
8:00 am 0 0
8:05 am 1.0 0.8
8:10 am 1.9 1.6
8:15 am 2.8 2.3
8:20 am 3.6 3.0
8:25 am – 3.6
Q
• Plot the distance-time graph for their
motions on the same scale and
interpret.
uestions
1. What is the nature of the
distance-time graphs for uniform
and non-uniform motion of an
object?
2. What can you say about the
motion of an object whose
distance-time graph is a straight
line parallel to the time axis?
3. What can you say about the
motion of an object if its speed-
time graph is a straight line
parallel to the time axis?
Rationalised 2023-24
SCIENCE82
Example 7.7 The brakes applied to a car
produce an acceleration of 6 m s
-2
in the
opposite direction to the motion. If the
car takes 2 s to stop after the application
of brakes, calculate the distance it
travels during this time.
Solution:
We have been given
a = – 6 m s
–2
; t = 2 s and v = 0 m s
–1
.
From Eq. (7.5) we know that
v = u + at
0 = u + (– 6 m s
–2
) × 2 s
or u = 12 m s
–1
.
From Eq. (7.6) we get
s = u t +
1
2
a t
2
= (12 m s
–1
) × (2 s) +
1
2
(–6 m s
–2
) (2 s)
2
= 24 m – 12 m
= 12 m
Thus, the car will move 12 m before it
stops after the application of brakes. Can
you now appreciate why drivers are
cautioned to maintain some distance
between vehicles while travelling on the
road?
uestions
1. A bus starting from rest moves
with a uniform acceleration of
0.1 m s
-2
for 2 minutes. Find (a)
the speed acquired, (b) the
distance travelled.
2. A train is travelling at a speed
of 90 km h
–1
. Brakes are applied
so as to produce a uniform
acceleration of – 0.5 m s
-2
. Find
how far the train will go before it
is brought to rest.
3. A trolley, while going down an
inclined plane, has an
acceleration of 2 cm s
-2
. What will
be its velocity 3 s after the start?
(ii) From Eq. (7.7) we have
2 a s = v
2
– u
2
= v
2
– 0
Thus,
2
–1 2
–2
=
2
(20 m s )
=
2×(1/15) m s
v
s
a
= 3000 m
= 3 km
The acceleration of the train is
1
15
m s
– 2
and the distance travelled is 3 km.
Example 7.6 A car accelerates uniformly
from 18 km h
–1
to 36 km h
–1
in 5 s.
Calculate (i) the acceleration and (ii) the
distance covered by the car in that time.
Solution:
We are given that
u = 18 km h
–1
= 5 m s
–1
v = 36 km h
–1
= 10 m s
–1
and
t = 5 s .
(i) From Eq. (7.5) we have
v – u
a =
t
=
-1
-1
10 m s – 5 m s
5s
= 1 m s
–2
(ii) From Eq. (7.6) we have
s = u t +
1
2
a t
2
= 5 m s
–1
× 5 s +
1
2
× 1 m s
–2
× (5 s)
2
= 25 m + 12.5 m
= 37.5 m
The acceleration of the car is 1 m s
–2
and
the distance covered is 37.5 m.
Q
Rationalised 2023-24
MOTION 83
4. A racing car has a uniform
acceleration of 4 m s
-2
. What
distance will it cover in 10 s after
start?
5. A stone is thrown in a vertically
upward direction with a velocity
of 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
by the stone and how much time
will it take to reach there?
7.6 Uniform Circular Motion
When the velocity of an object changes, we say
that the object is accelerating. The change in
the velocity could be due to change in its
magnitude or the direction of the motion or
both. Can you think of an example when an
object does not change its magnitude of
velocity but only its direction of motion?
straight parts AB, BC, CD and DA of the track.
In order to keep himself on track, he quickly
changes his speed at the corners. How many
times will the athlete have to change his
direction of motion, while he completes one
round? It is clear that to move in a rectangular
track once, he has to change his direction of
motion four times.
Now, suppose instead of a rectangular
track, the athlete is running along a
hexagonal shaped path ABCDEF, as shown
in Fig. 7.8(b). In this situation, the athlete will
have to change his direction six times while
he completes one round. What if the track
was not a hexagon but a regular octagon,
with eight equal sides as shown by
ABCDEFGH in Fig. 7.8(c)? It is observed that
as the number of sides of the track increases
the athelete has to take turns more and more
often. What would happen to the shape of the
track as we go on increasing the number of
sides indefinitely? If you do this you will
notice that the shape of the track approaches
the shape of a circle and the length of each of
the sides will decrease to a point. If the athlete
moves with a velocity of constant magnitude
along the circular path, the only change in
his velocity is due to the change in the
direction of motion. The motion of the athlete
moving along a circular path is, therefore, an
example of an accelerated motion.
We know that the circumference of a circle
of radius r is given by
π
2 r
. If the athlete takes
t seconds to go once around the circular path
of radius r
, the speed v is given by
π
2 r
v =
t
(7.8)
When an object moves in a circular path
with uniform speed, its motion is called
uniform circular motion.
(a) Rectangular track
(b) Hexagonal track
(d) A circular track(c) Octagonal shaped track
Fig. 7.8: The motion of an athlete along closed tracks
of different shapes.
Let us consider an example of the motion
of a body along a closed path. Fig 8.9 (a)
shows the path of an athlete along a
rectangular track ABCD. Let us assume that
the athlete runs at a uniform speed on the
Rationalised 2023-24
SCIENCE84
Activity _____________ 7.11
• Take a piece of thread and tie a small
piece of stone at one of its ends. Move
the stone to describe a circular path
with constant speed by holding the
thread at the other end, as shown in
Fig. 7.9.
If you carefully note, on being released
the stone moves along a straight line
tangential to the circular path. This is
because once the stone is released, it
continues to move along the direction it has
been moving at that instant. This shows that
the direction of motion changed at every point
when the stone was moving along the circular
path.
When an athlete throws a hammer or a
discus in a sports meet, he/she holds the
hammer or the discus in his/her hand and
gives it a circular motion by rotating his/
her own body. Once released in the desired
direction, the hammer or discus moves in
the direction in which it was moving at the
time it was released, just like the piece of
stone in the activity described above. There
are many more familiar examples of objects
moving under uniform circular motion,
such as the motion of the moon and the
earth, a satellite in a circular orbit around
the earth, a cyclist on a circular track at
constant speed and so on.
What
you have
learnt
• Motion is a change of position; it can be described in terms
of the distance moved or the displacement.
• The motion of an object could be uniform or non-uniform
depending on whether its velocity is constant or changing.
• The speed of an object is the distance covered per unit time,
and velocity is the displacement per unit time.
• The acceleration of an object is the change in velocity per
unit time.
• Uniform and non-uniform motions of objects can be shown
through graphs.
• The motion of an object moving at uniform acceleration can
be described with the help of the following equations, namely
v = u + at
s = ut + ½ at
2
2as = v
2
– u
2
Fig. 7.9: A stone describing a circular path with
a velocity of constant magnitude.
• Now, let the stone go by releasing the
thread.
• Can you tell the direction in which
the stone moves after it is released?
• By repeating the activity for a few
times and releasing the stone at
different positions of the circular
path, check whether the direction in
which the stone moves remains the
same or not.
Rationalised 2023-24
MOTION 85
where u is initial velocity of the object, which moves with
uniform acceleration a for time t, v is its final velocity and
s
is the distance it travelled in time t.
• If an object moves in a circular path with uniform speed, its
motion is called uniform circular motion.
Exercises
1. An athlete completes one round of a circular track of diameter
200 m in 40 s. What will be the distance covered and the
displacement at the end of 2 minutes 20 s?
2. Joseph jogs from one end A to the other end B of a straight
300 m road in 2 minutes 30 seconds and then turns around
and jogs 100 m back to point C in another 1 minute. What
are Joseph’s average speeds and velocities in jogging (a)
from A to B and (b) from A to C?
3. Abdul, while driving to school, computes the average speed
for his trip to be 20 km h
–1
. On his return trip along the
same route, there is less traffic and the average speed is
30 km h
–1
. What is the average speed for Abdul’s trip?
4. A motorboat starting from rest on a lake accelerates in a
straight line at a constant rate of 3.0 m s
–2
for 8.0 s. How
far does the boat travel during this time?
5. A driver of a car travelling at 52 km h
–1
applies the brakes
Shade the area on the graph that represents the distance
travelled by the car during the period.
(b) Which part of the graph represents uniform motion of
the car?
6. Fig 7.10 shows the distance-time graph of three objects A,B
and C. Study the graph and answer the following questions:
Fig. 7.10
Rationalised 2023-24
SCIENCE86
(a) Which of the three is travelling the fastest?
(b) Are all three ever at the same point on the road?
(c) How far has C travelled when B passes A?
(d) How far has B travelled by the time it passes C?
7. A ball is gently dropped from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m s
-2
, with what velocity
will it strike the ground? After what time will it strike the
ground?
8. The speed-time graph for a car is shown is Fig. 7.11.
(a) Find how far does the car travel in the first 4 seconds.
Shade the area on the graph that represents the distance
travelled by the car during the period.
(b) Which part of the graph represents uniform motion of
the car?
9. State which of the following situations are possible and give
an example for each of these:
(a) an object with a constant acceleration but with zero
velocity
(b) an object moving with an acceleration but with uniform
speed.
(c) an object moving in a certain direction with an
acceleration in the perpendicular direction.
10. An artificial satellite is moving in a circular orbit of radius
42250 km. Calculate its speed if it takes 24 hours to revolve
around the earth.
Fig. 7.11
Rationalised 2023-24
