Important Questions for CBSE Class 9 Science Chapter 8 - Motion
CBSE Class 9 Science Chapter-8 Important Questions - Free PDF Download
1 Marks Questions
1. Which of the following statements is correct?
(a) both speed and velocity are same
(b) speed is a scalar and velocity is a vector
(c) speed is a vector and velocity is scalar
(d) none of these
Ans. (b) speed is a scalar and velocity is a vector
2. What is the slope of the body when it moves with uniform velocity?
(a) positive
(c) negative
(b)zero
(d) may be positive or negative
Ans. (b) zero
3. Which of the following is the position time graph for a body at rest?
(i) 
(ii) 
(iii) 
(iv) 
Ans. (i)
4. What does area velocity time graph give?
(a) distance
(b) acceleration
(c) displacement
(d) none of the above
Ans. (c) displacement
5. If a body starts from rest, what can be said about the acceleration of body?
(a) Positively accelerated
(b) Negative accelerated
(c) Uniform accelerated
(d) None of the above
Ans. (a) Positively accelerated
6. What does slope of position time graph give?
(a) speed
(b) acceleration
(c) uniform speed
(d) Both (a) and (c) depending upon the type of graph.
Ans. (a) speed
7. When a body moves uniformly along the circle, then:
(a) its velocity changes but speed remains the same
(b) its speed changes but velocity remains the same
(c) both speed and velocity changes
(d) both speed and velocity remains same
Ans. (a) its velocity changes but speed remains the same
8. Which of the following statements is correct?
(a) speed distance are scalar, velocity and displacement are vector
(b) speed distance are vector, velocity and displacement are vector
(c) speed and velocity are scalar, distance and velocity are vector
(d) speed and velocity are vector, distance and displacement are scalar
Ans. (a) speed distance are scalar, velocity and displacement are vector
9. What does the slope of velocity – time graph give?
(a) Distance
(b) displacement
(c) Acceleration
(d) Change in velocity.
Ans. (c) Acceleration
10. The displacement of the body can be-
(a) Positive
(b) negative
(c) Zero
(d) All of these.
Ans. (d) All of these.
11. Which of the following gives both direction and magnitude-
(a) scalar
(b) vector
(c) Both
(d) None.
Ans. (b) vector
12. If a moving body comes to rest, then its acceleration is-
(a) Positive
(b) negative
(c) Zero
(d) All of these depending upon initial velocity.
Ans. (b) negative
2 Marks Questions
1. Distinguish between speed and velocity.
Ans. Speed of a body is the distance travelled by it per unit time while velocity is displacement per unit time of the body during movement.
2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans. If distance travelled by an object is equal to its displacement then the magnitude of average velocity of an object will be equal to its average speed.
3. What does the odometer of an auto mobile measure?
Ans. The odometer of an auto mobile measures the distance covered by that automobile.
4. What does the path of an object look like when it is in uniform motion?
Ans. Graphically the path of an object will be linear i.e. look like a straight line 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 × 108m s–1.
Ans. 
=
6. 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.
Ans. Both (a) as well as (b) are false with respect to concept of displacement.
7. When will you say a body is in
(i) uniform acceleration?
(ii) non-uniform acceleration?
Ans. (i) uniform acceleration: When an object travels in a straight line and its velocity changes by equal amount in equal intervals of time, it is said to have uniform acceleration.
(ii) non-uniform acceleration: It is also called variable acceleration. When the velocity of an object changes by unequal amounts in equal intervals of time, it is said to have non uniform acceleration.
8. A bus decreases its speed from 
to 
in 5 s. Find the acceleration of the bus.
Ans. Initial speed of bus (u) = 
= 
= 
final speed of bus (v)= 
= 
time (t) = 5 s
acceleration (a) = (v – u) /t = (16.67 – 22.22)/5 = -5.55/5 = 
9. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Ans. If an object has a uniform motion then the nature of distance time graph will be linear i.e. it would a straight line and if it has non uniform motion then the nature of distance time graph is a curved line.
10. What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans. The area occupied below the velocity-time graph measures the distance moved by any object.
11. 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.
Ans. (a) u=o, 
, t= 2min = 120 seconds.
v=u+at = 
(a) speed acquired = v= 
(b) 
= 
= 720 m
12. A trolley, while going down an inclined plane, has an acceleration of
. What will be its velocity 3 s after the start?
Ans. u = 0, 
, t= 3s
v= u +at = 
= 6 cm/s
13. A racing car has a uniform acceleration of 
. What distance will it cover in 10 s after start?
Ans. u = 0, 
, t= 10s
= 200m
14. Differentiate between distance and displacement?
Ans.
15. Derive mathematically the first equation of motion V=u + at?
Ans. Acceleration is defined as the rate of change of velocity.
Let V=final velocity; Vo= initial velocity, T= time, a =acceleration.
So by definition of acceleration
If Vo=u=initial velocity, then [V= u +at]
16. Calculate the acceleration of a body which starts from rest and travels 87.5m 5 sec?
Ans. u= 0 (starts from rest) u= initial velocity
a=? a=acceleration
T= 5 sec, t= time
S= 87.5m (S=distance)
From 2nd equation of motion –
(i)
17. Define uniform velocity and uniform acceleration?
Ans. Uniform velocity A body is said to move with uniform velocity if equal displacement takes place in equal intervals of time, however small these intervals may be.
Uniform acceleration A body is said to move with uniform acceleration if equal changes in velocity takes place in equal intervals of time, however small intervals may be.
18. A car travels at a speed of 40km/hr for two hour and then at 60km/hr for three hours. What is the average speed of the car during the entire journey?
Ans. In first case; 
= 40km/hr
= 
= 80km
In second case, 
= 3hrs
= 60km/hr
=
= 180km
The total distance = 
= 80 + 180 = 260km
Total time, 
= 2+3 = 5hrs
Average speed = 
19. The velocity time graph of two bodies A and B traveling along the +x direction are given in the figure
(a) Are the bodies moving with uniform acceleration?
(b) Which body is moving with greater acceleration A or B?
Ans. (a) Yes the bodies are moving with uniform acceleration.
(b) The body A is moving with greater acceleration.
20. Derive the second equation of motion, s = ut + 
numerically?
Ans. Let at time t = 0, body has initial velocity = 
At time ‘t’, body has final velocity = V
S = distance traveled in time ‘t’
We know, total distance traveled = Average velocity
time
Average velocity = 
= 
Total distance = s = 
(i)
Now from first equation of motion, (ii)
Use the value of (V) from (ii) in (i)
Let 
21. Calculate the acceleration and distance of the body moving with 
which comes to rest after traveling for 6sec?
Ans. Acceleration = a =?
Final velocity = V = o (body comes to rest)
Distance = s =?
Time = t = 6 sec
From, V = u + at
O = 5 + a6
-a6 = 5
a = 
a = 
Now,
-25 = 
15m=s
22. A body is moving with a velocity of 12m/s and it comes to rest in 18m, what was the acceleration?
Ans. Initial velocity =u=12m/s
Find velocity =V=0
S= distance= 18m
A= acceleration =?
From 3rd equation of motion;
=2a18
23. A body starts from rest and moves with a uniform acceleration of 
until it travels a distance of 800m, find the find velocity?
Ans. Initial velocity = u = 0
Final velocity = v = ?
Acceleration = 
Distance =s=800m
=24800
u=80 m/s
u2=6400
24. The driver of a car traveling along a straight road with a speed of 72KM ph observes a signboard which give the speed limit to be 54KM ph. The signboard is 70m ahead when the driver applies the brakes0 calculate the acceleration of the car which will cause the car to pass the signboard at the stated speed limit?
Ans. Initial speed = u=72 km/hr
Final speed = v =54 km/hr
= 
Distance = S = 70m
Now,
= 2a70
225 -400 =140a
- 175 =140a
25. Differentiate between scalars and vectors?
Ans.
3 Marks Questions
1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans. Yes, if an object has moved through a distance it can have zero displacement because displacement of an object is the actual change in its position when it moves from one position to the other. So if an object travels from point A to B and then returns back to point A again, the total displacement is zero.
2. A farmer moves along the boundary of a square field of side10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes20 seconds?
Ans. Distance covered by farmer in 40 seconds = 
Speed of the farmer = distance/time = 40m/40s = 1 m/s.
Total time given in the question = 2min20seconds = 60+60+20 =140seconds
Since he completes 1round of the field in 40seconds so in he will complete 3rounds in 120 seconds (2mins) or 120m distance is covered in 2minutes. In another 20seconds will cover another 20m so total distance covered in 2min20sec = 120 +20 =140m.
Displacement 
=
200 =
(as per diagram) 
= 14.14 m.
3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km 
in 10 minutes. Find its acceleration.
Ans. Since the train starts from rest(railway station) = u = zero
Final velocity of train =v= 
= 
= 
time (t) = 10 min = 
= 600 seconds
Since a = (v – u)/t 
4. What can you say about the motion of an object whose distance-time graph is a straight-line parallel to the time axis?
Ans. If the object’s distance time graph is a straight line parallel to the time axis indicates that with increasing time the distance of that object is not increasing hence the object is at rest i.e. not moving.
5. What can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis?
Ans. Such a graph indicates that the object is travelling with uniform velocity.
6. A train is travelling at a speed of
. Brakes are applied so as to produce a uniform acceleration of
. Find how far the train will go before itis brought to rest.
Ans. u = 
=
, v =0(train is brought to rest)
v= u+at = 25 + (-0.5)x t
0 =25 – 0.5 x
0.5t = 25, or t = 25/0.5 = 50seconds
=
= 1250 – 625 = 625m
7. A stone is thrown in a vertically upward direction with a velocity of
. If the acceleration of the stone during its motion is 
in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans. 
v = 0 (since at maximum height its velocity will be zero)
v = u + at = 
0 = 5 – 10t
10t = 5 , or, t = 5/10 =0.5second.
= 
= 2.5 – 1.25 = 1.25m
8. Derive the second equation of motion 
graphically?
Ans. let at time T=0 body moves with initial velocity u and at time ‘t’ body has final velocity ‘v’ and un time ‘t’ it covers a distance’s.
AC=v, AB=u, OA= t, DB=OA=t, BC=AC-AB =V-u
Area under a v-t curve gives displacement so,
S= Area of 
DBC + Area of rectangle OABD 
(i)
Area of DBC=
BaseHeight 
DBBC
= t (v-u) (ii)
Area of rectangle OABD= lengthBreadth
= OA BA
= tu (iii)
S= ut+t(v-u)
S= ut +tat (
use V-u=at)
S= ut+at2
9. A car moving with a certain velocity comes to a halt if the retardation was
, find the initial velocity of the car?
Ans. V=0 (comes to rest) V= final velocity
S= 62.5m
(retardation)
U=?
From 3rd equation of motion,
= 2(-5)62.5
= -1062.5
,
u= 
[u=25m/s ]
10. Two cars A and B are moving along in a straight line. Car A is moving at a speed of 80KMph while car B is moving at a speed 50KMph in the same direction, find the magnitude and direction of
(a) tive v the relative of car A with respect to B
The relative velocity of car B with respect to A.
Ans. (a) Velocity of car A = 80KMph
Velocity of Car B = - 50 kmph
(-ve sign indicates that Car B is moving in opposite direction to Car A )
Relative velocity of car A with respect to B
= velocity of car A + (- velocity of car B)
= 80 + (-(-50))
= 80+50
=+130KMph
+130 KM ph shows that for a person in car B, car A will appear to move in the same direction with speed of sum of their individual speed.
(b) Relative velocity of car B with respect to A
= velocity of car B+ (- velocity of car A)
= -50 + (-80)
= -130kmph
It shows that car B will appear to move with 130 kmph in opposite direction to car A
11. A ball starts from rest and rolls down 16m down an inclined plane in 4 s.
(a) What is the acceleration of the ball?
(b) What is the velocity of the ball at the bottom of the incline?
Ans. u= initial velocity = 0(body starts from rest)
S= distance = 16 m
T= time = 4s
(i) From, 
16=a16
(ii) From, v= u +at
v=0+24
[v= 8m/s]
12. Two boys A and B, travel along the same path. The displacement – time graph for their journey is given in the following figure.
(a) How far down the road has B travelled when A starts the journey?
(b) Without calculation, the speed, state who is traveling faster A or B?
(c) What is the speed of A?
(d) What is the speed of B?
(e) Are the speed of A and B uniform?
(f) What dose point X on the graph represent?
(g) What is the speed of approach of A towards B?
What is the speed of separation of A from B?
Ans. (a) When A starts his journey at 4 sec, B has already covered a distance of 857m
(b) A travels faster than B because A starts his journey late but crosses B and covers more distance then B in the same time as B
(c) Speed of A =
Let at t =12 min, distance covered = 3500m
(d) Speed of B = 
(e) Speed of approach of A towards B = 375 m/min- 214 m/min
= 161 m/min
(f) Speed of separation of A from B = 161 m/min.
13. A body is dropped from a height of 320m. The acceleration due to the gravity is
?
(a) How long does it take to reach the ground?
(b) What is the velocity with which it will strike the ground?
Ans. Height = h
Distance = s = 320m
Acceleration due to gravity = 
Initial velocity = u =0
(a) from s = ut + 
(b) 
14. Derive third equation of motion 
numerically?
Ans. We know;
………(i)
…….(ii)
Where, v = final velocity
u = initial velocity
a = acceleration
t = time
s = distance
From equation (i) t = 
Put the value of t in equation (ii)
15. The velocity time graph of runner is given in the graph.
(a) What is the total distance covered by the runner in 16s?
(b) What is the acceleration of the runner at t = 11s?
Ans. (a) We know that area under v-t graph gives displacement:
So, Area = distance = s = area of triangle + area of rectangle
Area of triangle = 
= 
= 30m
Area of rectangle = length breadth
= (16-6)10
=1010
= 100m
Total area = 180m
Total distance =180m
(b) Since at t = 11sec, particles travels with uniform velocity so, there is no change in velocity hence acceleration = zero.
16. A boy throws a stone upward with a velocity of 60m/s.
(a) How long will it take to reach the maximum height
?
(b) What is the maximum height reached by the ball?
(c) How long will it take to reach the ground?
Ans. u = 60 m/s; 
; v=0
(a) The time to reach maximum height is;
(b) The maximum height is;
(c) The time to reach top is equal to time taken to reach back to ground. Thus, time to reach the ground after reaching top is 6s Or the time to reach the ground after throwing is 6 + 6 =12s.
17. The displacement x of a particle in meters along the x- axis with time ‘t’ in seconds according to the equation-
(a) draw a graph if x versus t for t = 0 and t =5 sec
(b) What is the displacement come out of the particles initially?
(c) What is slope of the graph obtained?
Ans. X= 20m + (12) t
(i) At t=0
X=20+120=12 m
(ii) At t=1
X=20+12=32m
(iii) At t =2
X= 20+24= 44m
(iv) At t=5
X=20+125=72 m
(a)
(b) At T= 0 (initially)
Displacement =20m.
(c) Slope = 
18. The velocity of a body in motion is recorded every second as shown-
calculate the –
(a) Acceleration
(b) distance travelled and draw the graph.
Ans. (a) Acceleration =slope of the velocity time graph
a=
(b) Distance
= 600-300 = 300 m
(c)
19. Draw the graph for uniform retardation –
(a) position – time graph
(b) velocity – time
(c) Acceleration- time
Ans. (1) Position – time
(2) Velocity – time
(3) Acceleration- time
20. The displacement – time graph for a body is given. State whether the velocity and acceleration of the body in the region BC, CD, DE and EF are positive, negative or Zero.
Ans. (i) For AB, the curve is upward stopping i.e. slope is increasing so velocity is positive and remains same so, V= +ve but a=0
(ii) For BC, curve has still has +ve slope so, V= +ve but velocity is decreasing wrt time so, a=negative
(iii) For CD, both velocity and acceleration are Zero because slope is Zero.
(iv) For DE, velocity is the (v is increasing wrt time) and so is acceleration is +ve.
(v) For EF, velocity is +ve (positive slope of x-t graph) but acceleration is Zero because velocity remains some with time.
21. Derive the third equation of motion
as graphically?
Ans. Let at time t=0, body moves with initial velocity u and time at ‘t’ has final velocity ‘v’ and in time ‘t’ covers a distance ‘s’
Area under v-t graph gives displacement
S = Area of DBC + Area of rectangle OABD
S = 
Now, v-u = at
Put the value of ‘t’ in equation (i)
third equation of motion