
Question 1. An object has moved through
a distance. Can it have zero
displacement? If yes, support your
answer with an example.
Answer : Yes,
an object even after it has moved
through a distance, it can have zero
displacement. As we know distance is
just length of the path an object has
covered irrespective of its direction or
position with reference to certain
point, where as the shortest distance
measured from the initial to the final
position of an object is known as the
displacement.
For example, an object starts from point
A and after covering a distance of say
50 meters, reaches at point B. Here
after, it again moves back to point A.
Here the distance covered by object is =
AB + BA = 50 m + 50 m = 100 m
where as displacement of object is = AB
 BA = 50 m  50 m = 0 m
As initial position of object is same as
that of its final position hence its
displacement, which is distance measured
from the initial to the final position,
is zero.
A 
>50 m>
<50 m< 
B 
Question 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?
Answer : Suppose,
a farmer moves along the boundary of a
square field of side 10 m in 40 s as
shown in the figure given below.
Distance cover by the farmer as
he moves from A to B to
C to D to A, along the boundary
wall of square field 
= Perimeter of Square field 

= 4 x side of square field 

= 4 × 10 m 

= 40 m 
∴ speed of farmer 
= 40 m/40 s 

= 1 m/s 
Distance covered by farmer in 2
minutes 20 seconds 
= Speed × Time 

= 1 m/s × [(2×60) s + 20 s] 

= 140 m 
Number of round in covering 40
m of distance
along the boundary wall 
= 1 round 
∴ Number of round in covering
140 m of distance
along the boundary wall 
= 1×140 /40 rounds 

= 3.5 round 

= 3 ^{1}/_{2} rounds 
Which means the farmer will be at point
C just diagonally opposite of point A
∴ Relative Displacement of farmer from
point A at the end of 3 ^{1}/_{2} round
will be = length of AC
which can be determined by the
mathematical theorem as given below :
AC 
= √AB^{2} +
√BC^{2} 

= √10^{2} +
√10^{2} 

=10 √2^{2} 

= 10 × 1.414 m 

= 14.14m 
Question 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.
Answer : Both
of the statements are not true as
(a) Displacement can be zero
(b) Its magnitude is either less or
equal to the distance travelled by the
object

Page 102 (CBSE
Class IX ( 9th) Science Textbook 
Chapter 8. Motion )
Question 1. Distinguish between speed
and velocity.
Answer : The
speed of an object is the distance
covered per unit time,and velocity is
the displacement per unit time. The
speed is a scalar quantity as it has
just magnitude where as velocity is a
vector quantity as it has both direction
as well as magnitude.The
speed can be changed by the distance
travelled by a body in a particular time
where as the velocity
can be changed by changing the object's
speed, direction of motion or both.
Question 2. Under what condition(s) is
the magnitude of average velocity of an
object equal to its average speed?
Answer : The
magnitude of average velocity of an
object is equal to its average speed,
only when it is moving in a straight
line.
Question 3. What does the odometer of an
automobile measure?
Answer : Odometer
of an automobile measures the distance
covered by an automobile. All
Automobiles are fitted with Odometer.
Earlier Odometer used to be mechanical
device, now a days we have electronic
odometer.
Question 4. What does the path of an
object look like when it is in uniform
motion?
Answer : The
path of an object looks like a straight
line when it is in uniform motion.j
Question 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}
Answer :
Speed of Signal (v) 
= Speed of light 

=3 × 10^{8} ms^{1} 
Time taken by Signal to reach
the ground station (t) 
= 5 minutes 

= 5 × 60 seconds 

= 300 seconds 
Distance between the spaceship
and the ground station (S) 
= vt 

= 3 × 10^{8} m
s^{1} ×
300 m 

=9×10^{10} m 

Page 103 (CBSE
Class IX ( 9th) Science Textbook 
Chapter 8. Motion )
Question 1. When will you say a body is
in (i) uniform acceleration? (ii)
nonuniform acceleration?
Answer : (i)
A body is said to be in uniform
acceleration if it travels in a straight
line and
its velocity increases or decreases by
equal amounts in equal intervals of time
(ii) A body is said to be in nonuniform
acceleration if the rate of change of
its velocity is not constant .
Question 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.
Answer :
Initial Speed of the Bus (u) 
= 80 km h^{1} 

= (80 × 1000)/ (60 × 60) ms ^{1} 

= 800/36 ms ^{1 } 
Final Speed of the Bus (v) 
= 60 km h^{1} 

= (60 × 1000)/ (60 × 60) ms ^{1} 

= 600/36 ms ^{1} 
Time in transition (t) 
= 5 s 
The acceleration of the Bus (a) 
= (vu)
/ t
= [(800/36)  (600/36)]
/ 5 ms ^{2 }
= (200/36) / 5 ms ^{2 }
= 5.55 / 5 ms ^{2 }
= 1.11 ms ^{2} 
Question 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.
Answer :
Initial Speed of the Train (u) 
= 0 ms ^{1 } 
Final Speed of the Train (v) 
= 40 km h^{1} 

= (40 × 1000)/ (60 × 60) ms ^{1} 

= 400/36 ms ^{1} 
Time in transition (t) 
= 10 minutes 

= 10 × 60 s 

= 600 s 

= 600 s 

= 600 s 
The acceleration of the Train (a) 
= (vu)
/ t
= [(400/36)  0] / 600
ms ^{2 }
= (11.11) / 600 ms ^{2 }
= 0.0185 ms ^{2 }
^{
} 

Page 107 (CBSE
Class IX ( 9th) Science Textbook 
Chapter 8. Motion )
Question 1. What is the nature of the
distancetime graphs for uniform and
nonuniform motion of an object?
Answer :
(i) For uniform speed, a graph of
distance travelled against time is a
straight line and not inclined along the
time axis, as shown in the figure below
(iI) For uniform speed, a graph of
distance travelled against time is a
curve and as shown in the figure below
Question 2. What can you say about the
motion of an object whose distancetime
graph is a straight line parallel to the
time axis?
Answer :
Motion of an object whose distancetime
graph is a straight line parallel to the
time axis is not moving at all and is in
state of rest as shown in the figure
below :
Question 3. What can you say about the
motion of an object if its speedtime
graph is a straight line parallel to the
time axis?
Answer :The
motion of an object if its speedtime
graph is a straight line parallel to the
time axis indicates that the object is
moving with uniform speed.
Question 4. What is the quantity which
is measured by the area occupied below
the velocitytime graph?
Answer :The
area occupied below the velocitytime
graph measures the distance covered by
the object.

Page 109110 (CBSE
Class IX ( 9th) Science Textbook 
Chapter 8. Motion )
Question 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.
Answer :
Here as given, Initial speed (u) 
= 0 
Acceleration (a) 
=0.1 m s^{2} 
Time in transition (t) 
=2 minutes 

=2 × 60 seconds= 120 s 
We know that Final speed 
= u + at 
∴ (a) the speed acquired 
= 0 + 0.1 m s^{2} ×
120 m s^{1 } 

= (1/10)120 ms^{1 } 

= 12 ms^{1} 

= ut +
(1/2)at^{2} 
∴ (b) the distance travelled. 
= 0 ×120
+ (1/2)
× 0.1×(120)^{2} 

= 0 + (120 × 120)
/2 ×
10 

= 14400/20 = 720 m 

=720 m 
Question 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.
Answer :
Given Initial speed of train (u) 
=90 km h^{1} 

= (90 1000) / (60×60) m s^{1} 

= 25 m s^{1} 
Final speed of train (v) 
= 0 ms^{1} 
Braking acceleration (a) 
= 0.5 m s^{2} 
We know 2as 
= v^{2}
u^{2} 
Or distance (s) 
=(v^{2}u^{2})/2a 
∴ Distance covered by the train
before it came to rest 
=(0^{2}25^{2})/(2
×0.5 )m 

=  (25 × 25)×10/1 m 

=625 m 


Question 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?
Answer :
Initial Velocity of trolley (u) 
=0 cms^{1} 
Acceleration (a) 
= 2 cm s^{2} 
Time (t) 
= 3 s 
We know that final velocity (v) 
= u
+ at 

= 0 + 2 x 3 cms^{1} 
∴ The velocity of train after 3
seconds 
= 6 cms^{1} 
Question 4. A racing car has a uniform
acceleration of 4 m s2. What distance
will it cover in 10 s after start?
Answer :
Initial Velocity of the car (u) 
=0 ms^{1} 
Acceleration (a) 
= 4 m s^{2} 
Time (t) 
= 10 s 
We know Distance (s) 
= ut + (1/2)at^{2} 
∴ Distance covered by car in 10
second 
= 0 × 10 + (1/2) × 4 × 10^{2} 

= 0 + (1/2) × 4× 10 × 10 m 

= (1/2)× 400 m 

= 200 m 
Question 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?
Answer:
Given Initial velocity of stone
(u) 
=5 m s^{1} 
Downward of negative
Acceleration (a) 
= 10 m s^{2} 
We know that 2 (a)(s) 
= v^{2} u^{2} 
∴Height attained by the stone (s) 
= (0^{2}5^{2})/
2 × (10)m 

= 25/20 m 

= 1.25 m 
Also we know that final velocity
(v) 
= u + at 
or Time (t) 
= (vu)/a 
∴ Time
(t) taken by stone to
attain the height(s) 
= (05)/ 10 s 

= 0.5 s 

Exercises (CBSE
Class IX ( 9th) Science Chapter 8.
Motion)
Question 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? Distance covered by
the athlete
Answer :
Diameter of circular track (D) 
= 200 m 
Radius of circular track (r) 
= 200/2=100 m 
Time taken by the athlete for
one round (t) 
= 40 s 
Distance covered by athlete in
one round (s) 
= 2π r 

= 2 × ( 22 / 7 )×100 
Speed of the athlete (v) 
= Distance / Time 

= (2 × 2200)/(7 × 40) 

= 4400 / 7 × 40 
∴ Distance
covered in 2 minutes 20
seconds (s) or 140 s 
= Speed (s) × Time(t) 

= 4400 / (7 × 40) × (2×60 + 20) 

= 4400 /( 7 × 40) × 140 

= 4400 × 140 /7 × 40 

=2200 m 
Number of round in 40 s 
=1 round 
Number of round in 140 s 
=140/40 

=3 ^{1}/_{2} 
After taking start from position
X,
the athlete will be at postion Y
after
3 ^{1}/_{2} rounds
as shown in figure 

∴ Hence
Displacement of the athlete
with respect to initial position
at X 
=XY 

= Diameter of circular track 

= 200 m 
Question 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?
Answer :
Total Distance covered from AB 
= 300 m 
Total time taken 
= 2 × 60 + 30 s 

=150 s 


∴ Average Speed from AB 
= Total Distance / Total Time 

=300/150 m s^{1} 

=2 m s^{1} 
∴ Velocity from AB 
=Displacement AB / Time =
300/150 m s^{1} 

=2 m s^{1} 
Total Distance covered from AC 
=AB + BC 

=300 + 200 m 
Total time taken from A to C 
= Time taken for AB + Time taken
for BC 

= (2×60+30)+60 s 

= 210 s 
∴Average Speed from AC 
= Total Distance /Total Time 

= 400 /210 m s^{1} 

= 1.904 m s^{1} 
Displacement (S) from A to C 
= AB  BC 

= 300100 m 

= 200 m 
Time (t) taken for displacement
from AC 
= 210 s 
∴Velocity from AC 
= Displacement (s) / Time(t) 

= 200 / 210 m s^{1} 

= 0.952 m s^{1} 
Question 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?
Answer : Let
us assume:
The distance Abdul commutes
while driving from Home to
School 
= S 
Let us assume time taken by
Abdul to commutes this distance 
= t_{1} 
Distance Abdul commutes while
driving from School to Home 
= S 
Let us assume time taken by
Abdul to commutes this distance 
= t_{2} 
Average speed from home to
school v_{1av} 
= 20 km h^{1} 
Average speed from school to
home v_{2av} 
= 30 km h^{1} 
Also we know Time taken form
Home to School t_{1} 
=S/v_{1av} 
Similarly Time taken form School
to Home t2 
=S/v_{2av} 
Total distance from home to
school and backward 
= 2 S 
Total time taken from home to
school and backward (T) 
= S/20+ S/30 
∴ Average speed (V_{av})
for covering total distance
(2S) 
= Total Dostance/Total Time 

= 2S/(S/20 +S/30) 

= 2S/[(30S+20S)/600] 

= 1200S/50S 
_{
} 
= 24 kmh^{1} 
Question 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?
Answer :
Given Initial velocity of
motorboat u 
= 0 
Acceleration of motorboat a 
= 3.0 m s^{2} 
Time under consideration t 
= 8.0 s 
We know that Distance s 
= ut + (1/2)at^{2} 
∴ The distance travel by
motorboat 
= 0 × 8 + (1/2)3.0 × 8 ^{2} 

= (1/2) × 3 × 8 × 8 m 

= 96 m 
Question 5. A driver of a car travelling
at 52 km h^{1} applies
the brakes and accelerates uniformly in
the opposite direction. The car stops in
5 s. Another driver going at 3 km h^{1} in
another car applies his brakes slowly
and stops in 10 s. On the same graph
paper, plot the speed versus time graphs
for the two cars. Which of the two cars
travelled farther after the brakes were
applied?
Answer : As
given in the figure below AB (in red
line) and CD(in red line) are the
Speedtime graph for given two cars with
initial speeds 52 kmh^{1} and
3 kmh^{1} respectively.
Distance Travelled by first car
before coming to rest 
=Area of △ OAB 

= (1/2) × OB× OA 

= (1/2) × 5 s × 52 kmh^{1} 

= (1/2) × 5× (52×1000)/3600) m 

= (1/2) × 5× (130/9) m 

= 325/9 m 

= 36.11 m 
Distance Travelled by second car
before coming to rest 
=Area of △ OCD 

= (1/2) × OD× OA 

= (1/2) × 10 s× 3 kmh^{1} 

= (1/2) × 10× (3×1000)/3600) m 

= (1/2) × 10× (5/6) m 

= 5× (5/6) m 

= 25/6 m 

= 4.16 m 
∴Clearly
the first car will travel farther (36.11
m) than
the first car (4.16
m).
Question 6. Fig 8.11 shows the
distancetime graph of three objects A,B
and C. Study the graph and answer the
following questions:
(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?
Answer :Drawing
x and y coordinates from points A, B and
C we get :
Speed of object A 
= Slope of MA 

= AP / MP 

= 4/2.2 kmh^{1} 

= 1.81 kmh^{1} 
Speed of object B 
= Slope of OB 

= BQ / OQ 

= 12/1.4 kmh^{1} 

= 8.57 kmh^{1} 
Speed of object C 
= Slope of LC 

= CR / LR 

= 8/1.4 kmh^{1} 

= 5.71 kmh^{1} 
(a) Object B is travelling the fastest
(b) None of three are ever at the same
point on the road as their graph LM, OB
and LC do not coincided simultaneously
at any point.
(c) C has travelled a distance JK =
IMKM = 9.141.14 km =8 km when B
passes A.
(d) B has travelled a distance XY = 6 km
when B passes C.
Question 7. A ball is gently dropped
from a height of 20 m. If its velocity
increases uniformly at the rate of 10 m
s2, with what velocity will it strike
the ground? After what time will it
strike the ground?
Answer :
Let us assume, the final velocity with
which ball will strike the ground be 'v'
and time it takes to strike the ground
be 't'
Initial Velocity of ball u 
=0 
Distance or height of fall s 
=20 m 
Downward acceleration a 
=10 m s^{2} 
As we know, 2as 
=v^{2}u^{2} 
v^{2} 
= 2as+ u^{2} 

= 2×10×20 + 0 

= 400 
∴ Final
velocity of ball, v 
= 20 ms^{1} 
t 
=
(vu)/a 
∴Time
taken by the ball to strike 
=
(200)/10 

= 20/10 

= 2 seconds 
Question 8. The speedtime graph for a
car is shown is Fig. 8.12.
(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?
Answer :
(a)The shaded area with blue color under
Speedtime graph represents the
distance which the car will travel in
first 4 second.
(b) The straight line part of graph,
from point A to point B represents a
uniform motion.
Question 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 in a certain
direction with an acceleration in the
perpendicular direction.
Answer : Both
the situations are possible.
(a) An
object with a constant acceleration can
still have the zero velocity. For
example an object which is at rest on
the surface of earth will have zero
velocity but still being acted upon by
the gravitational force of earth with an
acceleration of 9.81 ms^{2} towards
the center of earth.
(b) When
an 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. Here, the motion of the athlete
moving along a circular path is,
therefore, an example of an accelerated
motion where acceleration is always
perpendicular to direction of motion of
an object at a given instance. .
Question 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.
Answer : Let
us assume An artificial satellite, which
is moving in a circular orbit of radius
42250 km covers a distance 's' as it
revolve around earth with speed 'v' in
given time 't' of 24 hours.
= 42250 km
Radius of circular orbit r 


= 42250 × 1000 m 
Time taken by artificial
satellite t 
= 24 hours 

= 24 × 60 × 60 s 
Distance covered by satellites 
=circumference of circular orbit 

=2π
r 
∴ Speed
of sattellite v 
=(2π
r)/t 

=[2× (22/7)×42250 × 1000] / (24
× 60 × 60) 

=(2×22×42250×1000) / (7 ×24 × 60
× 60) m s ^{1} 

=3073.74 m s ^{1} 

