# NCERT Solutions Class 9 Science Chapter 8 Motion

### In-Text Questions Solutions

#### In Text-Questions Page 100

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.
Ans:- Yes, an object moving through a distance can have zero displacement. This happens when final position of the object coincides with its initial position. For example, if a person moves around a park and returns to the same point from where he had started then the displacement will be zero.
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?
Ans:- Figure ABCD is a square field of side 10 m.  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
.
Ans:-

(a) Not true
Displacement can become zero when the initial and final positions of the object are the same.
(b) Not true Displacement is the shortest distance between the initial and final positions of an object.
It cannot be greater than the magnitude of the distance travelled by an object. However, sometimes, it may be equal to the distance travelled by the object.

#### In Text-Questions Page 102

1. Distinguish between speed and velocity.
Ans:-

2. Under what condition(s) is the magnitude of average velocity of an object equal to its average speed?
Ans:- If the total distance covered by an object is the same as its displacement, then its average speed would be equal to its average velocity, i.e. when the object moves along a straight line path.

3. What does the odometer of an automobile measure?
Ans:- Odometer of an automobile measures the distance covered by automobile.

4. What does the path of an object look like when it is in uniform motion?
Ans:- An object having uniform motion has a straight line path.

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 × 108 m s–1.
Ans:- Time taken = 5 mins = 5 x 60 s = 300 s
Speed of signal = 3 x 108 m/s
Distance = speed x time = 3 x 108 x 300 = 9 x 1010 m

#### In Text-Questions Page 103

1. When will you say a body is in
(i) uniform acceleration? (ii) non uniform acceleration?

Ans:- (i) Uniform acceleration:- A body is said to be in uniform acceleration if it moves in a straight line and its velocity changes by equal amounts in equal intervals of time.
(ii) Non-uniform acceleraiion:- A body is said to be in non-uniform acceleration if its velocity does not change by equal amounts in equal intervals of time.
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.
Ans:– Initial velocity, u = 80 km/h = 80 x 5/18 = 22.22 m/s
Final velocity, v = 60 km/h = 60 x 5/18 = 16.66 m/s
Time taken, t = 5 s
Acceleration, a = = = -1.11 m/s2
Negative sign shows retardation.
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.
Ans:- Here, initial speed, u = 0
Final speed, v = 40 km/h = 40 x 5/18 = 11.11 m/s
Time taken, t = 10 mins = 10 x 60 s = 600 s
Acceleration, a = = = 1.85 x 10 -2 m/s2

#### In Text-Questions Page 107

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?
Ans:- When the motion is uniform, the distance-time graph is a straight line with some slope.
When the motion is non-uniform, the distance time graph is not a straight line.
2. What can you say about the motion of an object whose distance-time graph is a straight line parallel to the time axis?
Ans:- It means that distance of the object does not change with time and the object is in a state of rest.
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 ?
Ans:- It means that the speed of the object is uniform.
4. What is the quantity which is measured by the area occupied below the velocity-time graph?
Ans:- Area occupied below the velocity-time graph is a measure of the displacement of the body.

#### In Text-Questions Page 109-110

1. A bus starting from rest moves with a uniform acceleration of 0.1 ms-2 for 2 minutes. Find
(a) the speed acquired,

(b) the distance travelled.
Ans:- Here, u = 0, a = 0.1 m/s2,
t = 2 min = 2 x 60 s = 120 s
(a) Using v = u + at = 0 + 0.1 x 120 = 12 m/s
(b) Using S = ut + ½ at2 = 0 x 120 + ½ 0.1 (120)2 = 7.2 x 102 m
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.
Ans:- Here, initial speed, u = 90 km/h = 90 x 5/18 m/s = 25 m/s
Acceleration, a = -0.5 m/s2
Final velocity, v = 0
Using, 2aS = v2 – u2
Distance travelled, S = = = 625 m
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?
Ans:- Here, acceleration, a = 2 cm/s2
Time taken, t = 3 s
Initial velocity, u = 0
Final velocity, v = u + at = 0 + 2 x 3 = 6 cm/s
4. A racing car has a uniformm acceleration of 4 m s-2. What distance will it cover in 10 s after start?
Ans:- Here, a = 4 m s-2, t = 10 s, u = 0
Using, S = ut + ½ at2
S = 0 x 10 + ½ x 4 x 102
S = 0 + ½ x 4 x 100 = 200 m
The distance covered in 10 s by the car is 200 m.
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-1 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?
Ans:- Here, u = 5 m/s, v = 0, a = -10 m/s2
(i) Using, v = u + at
t = = = = 0.5 s
(ii) Using 2aS = v2 – u2
⇒ 2 x (-10) x S = 02 – 52
⇒ -20 x S = -25
⇒ S = 1.25 m

### Chapter End 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?
Ans:- Displacement after 3.5 rounds = diameter of the track = 200 m
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 50 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? Ans:- (a) From A to B  (b) From A to C,
Total time taken = 2 min 50 s + 1 min = 150 s + 60 s = 210 s
Total distance = 300 + 100 = 400 m
Displacement = 300 – 100 = 200 m 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 40 km h-1. What is the average speed for Abdul’s trip?
Ans:- Let the school be at a distance of x km.
If t1 is time taken to reach the school, then 4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m s-1 for 8.0 s. How far does the boat travel during this time?
Ans:- Here, u = 0, a = 3 m/s2, t = 8 s
Using, S = ut + ½ at2
⇒ S = 0 x 8 + ½ x 3 x 82
⇒ S = ½ x 3 x 64 = 96 m
So, boat travels a distance of 96 m.
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?
Ans:- For first car:
u = 52 km/h = 52 x 5/18 = 14.4 m/s,
v = 0 and t = 5 s
For second car:
u = 3 km/.h = 3 x 5/18 m/s = 0.8 m/s,
v = 0 and t = 10 s The distance traveled by a moving body is given by the area under its speed-time graph.
So, distance traveled by the first car
= Area of the triangle AOB
= ½ x OB x AO
= ½ x 14.4 x 5 = 36 m
Similarly, Distance travelled by the second car
= Area of triangle COD
= ½ x OD x CO
= ½ x 0.83 x 10 = 4.1 m
So, the second car traveled farther after the brakes were applied.
6. Figure given below shows the distance-time 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?
Ans:-
(a) B is travelling fastest.
(b) As three lines do not meet at any point, the three objects never meet on the road.
(c) When B crosses A, then distance traveled by C = 9 – 2 = 7 km.
(d) By the time B passes C, it has travelled 5.5 km.
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?
Ans:- Here, s = 20 m, u= 0, a = 10 m/s2
Using, s = ut + ½ at2
⇒ 20 = 0 x t + ½ x 10 x t2
⇒ 20 = 5t2
⇒ 4 = t2
⇒ t = 2 s
Usingt, v = u + at
⇒ v = 0 + 10 x 2 = 20 m/s
The ball strike the ground after 2 s with the velocity of 20 m s-1.
8. The speed-time graph for a car is shown in the figure: (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?
Ans:- (a) The motion during first 4 seconds is not uniformly accelerated. So, distance travelled by car in first 4 seconds is calculated by graphical method. Number of squares in shaded part of the graph = 61.5
One small square in x-axis represents, t = ⅖ sec
One small square on y-axis represents, v = ⅔ m/s
So, area of each square = v x t = ⅔ x ⅖ = 4/15 m
Total area = 61.5 x 4/15 = 16.4 m
Hence, distance traveled by the car = 16.4 m

(b) Therefore, portion of the graph almost after time t = 6 s describes the 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 in a certain direction with an acceleration in the perpendicular direction.
Ans:- (a) It is possible. For example, when an object is thrown vertically upwards then at the maximum height, velocity becomes zero but acceleration due to gravity is constant
(b) It is possible. For ex, when a car is moving on a circular track, its acceleration is perpendicular to its 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.
Ans:- Radius of the orbit, r = 42250 km = 42250 x 1000 m
Time taken for one revolution = 24 hours = 24 × 60 × 60 s 