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Chapter 8 of Class 9 Physics is Motion. In this chapter, you will learn uniform motion and non-uniform motion. You will also understand the concept of speed and direction of motion. After understanding this chapter, you will be able to understand that all objects on the earth are in motion as the earth is constantly in motion.

NCERT Solutions Class 9 Science of Chapter 8 includes all Intext exercises 1, 2, 3, 4, 5, and an NCERT exercise for you. The NCERT exercise includes a total of 10 questions. All these questions are based on application, numerical skills, and analysis. You can practice these questions to test and enhance your application & analysis skills and numerical ability ## NCERT Solutions for Class 9 Science Chapter 8 : Motion Ex 8.1

Exercise 8.1

#### Questions 1:An object has moved through a distance. Can it have zero displacement? Ifyes, support your answer with an example.

Solution:
The shortest distance between initial and final positions of an object is called displacement of the object. Even if an object travels large distance, but eventually comes at rest at its initial position in the end, the distance travelled by the object will be non-zero but the displacement of the object would be zero. So, yes an object can move though a distance and have zero
displacement.

#### 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 2minutes 20 seconds from his initial position?

Solution:
Given:
The farmer covers the entire boundary of the square field of side 10m in 40 seconds,
So, the total distance travelled by the farmer in 40 seconds is 4 x (10) = 40 meters.
Therefore, the average distance covered by the farmer in one second is:= 1
Now, 2 minutes and 20 seconds = 140 seconds.
The total distance travelled by the farmer in 140 seconds is: 1 ×140=140
Since, the farmer is moving along the boundary of the square field,
The total number of laps completed by the farmer will be=3.5
Now, the total displacement of the farmer depends on the initial position and final position of the farmer. If the initial position of the farmer is at one corner of the field, the final position is at the opposite corner. So, the total displacement of the farmer will be equal to the length of the diagonal of the square.
Applying the Pythagoras theorem, the length of the diagonal can be obtained as follows= = 14.14 m.
This is the maximum possible displacement of the farmer.
If the initial position of the farmer is at the mid-point between two adjacent corners of the square, the net displacement of the farmer would be equal to the side of the square, which is 10 m. This is the minimum displacement.
If the farmer starts at any random point around the perimeter of the square, the total displacement of farmer after traveling 140 m will always lie between 10 m and 14.14 m.

#### 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.

Solution:
Both of the statements given in the question are false. Statement
(a) is false because the displacement of an object whose initial and final position are same is equal to zero.
(b) is false because the displacement of an object can be equal to, but can never be greater in magnitude than the total distance travelled by it.

Exercise 8.2
Questions 1:
Distinguish between speed and velocity.

#### Solution: Refer pdf.

Question 2:
Under what condition(s) is the magnitude of average velocity of an object
equal to its average speed?

Solution:
Since average speed is the total distance travelled in a time frame and velocity is the total displacement in the time frame, the magnitude of average velocity and average speed will be the same when the total distance travelled is equal to the displacement.

#### Question 3:What does the odometer of an automobile measure?

Solution:
The odometer is equipment used to measure the total distance travelled by the vehicle

#### Question 4:What does the path of an object look like when it is in uniform motion?

Solution:
The path of an object travelling in a uniform motion looks like a straight line.

#### Question 5:During an experiment, a signal from a spaceship reached the groundstation in five minutes. What was the distance of the spaceship from theground station? The signal travels at the speed of light, that is, 3 × 108 m/s.

Solution:
Given:
The signal always travels in a straight line, so, the distance between the spaceship and the ground station will be equal to the total distance travelled by the signal to reach the ground station.
5 minutes = 5 × 60 seconds = 300 seconds.
Speed of the signal = 3 × 108 m/s.
Distance travelled = Speed × Time taken
Therefore, total distance = (3 × 108 m/s) × 300s
= 9 × 1010 meters.

Exercise 8.3
Question 1:
When will you say a body is in:
(i) Uniform acceleration?
(ii) Non-uniform acceleration?

Solution:
(i) Uniform Acceleration: When the body moves along a straight line and its velocity changes at a uniform rate in a given constant time interval then this type of acceleration is called uniform acceleration.
(ii) Non-Uniform Acceleration: The acceleration in which the body moves along a straight line and its velocity changes non-uniformly in the constant time interval is called nonuniform acceleration.

#### 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.

Solution: Refer pdf.

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.

Solution: Refer pdf.

Exercise 8.4

#### Question 1:What is the nature of the distance-time graphs for uniform and nonuniform motion of an object?

Solution:
For uniform motion, the distance vs time graph will be straight line. Whereas, the distance vs time graph of an object travelling in non-uniform motion will be a curve. The first graph illustrates uniform motion and the second one describes non-uniform motion.

#### Question 2:What can you say about the motion of an object whose distance-time graphis a straight line parallel to the time axis?

Solution:
This distance-time graph can be plotted as follows.
Since distance travelled by the object (or the Y-Axis value) is not changing at any point in the X-Axis (time), therefore the object is at rest.

#### Question 3:What can you say about the motion of an object if its speed-time graph is astraight line parallel to the time axis?

Solution:
This speed vs time graph can be drawn as follows.
As there is no change in the velocity of the object (Y-Axis value) at any point of time (X-axis value), the object is said to have a uniform motion.

#### Question 4:What is the quantity which is measured by the area occupied below thevelocity-time graph?

Solution:
Considering an object in uniform motion, its velocity vs time graph can be plotted as:
Now, in the above velocity vs time graph the area of the graph is the area of the rectangle OABC, which is given by OA×OC. Where, OA is the velocity of the object and OC is time. Therefore, the shaded area can be represented as:
Area of the velocity vs time graph = velocity × time.
Substituting the value of velocity i.e.in the equation, the area of the velocity vs
time graph represents the total displacement of the object.

#### Solution: Refer pdf.

Question 3:
A trolley, while going down an inclined plane, has an acceleration of 2cms-2
.
What will be its velocity 3 s after the start?

Solution:
Given:
Initial velocity (u) = 0 (the trolley begins from the rest position)
Acceleration (a) = 0.02 ms-2
.
Time (t) = 3s
As per the first motion equation, = +
Therefore, terminal velocity of the trolley (v) = 0 + (0.02
-2
)(3 ) = 0.06 ms-2
Therefore, the velocity of the trolley after 3 seconds will be 6 cm.s-2

#### .Question 4:A racing car has a uniform acceleration of 4 m s-2. What distance will itcover in 10 s after start?

Solution: Refer pdf.

Question 5:
A stone is thrown in a vertically upward direction with a velocity of 5 ms-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?

Exercise 8

#### Solution: Refer pdf.

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?

Solution: Refer pdf.

#### Solution: Refer pdf.

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?

Solution: Refer pdf.

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?

Solution: Refer pdf.

Question 6:
Fig 8.11 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?

Solution
(a) Since the slope of line B is the greatest, B is traveling at the fastest speed.
(b) Since the three lines do not intersect at a single point, the three objects never meet at the same point on the road.
(c) Refer pdf.

#### Solution: Refer pdf.

.
Question 8:
The speed-time graph for a car is shown is Fig. 8.12
(a) Find how far the car travels 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?

Solution: Refer pdf.

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 with an acceleration but with uniform speed.
(c) an object moving in a certain direction with an acceleration in the
perpendicular direction.

Solution:
(a) It is possible; an object thrown up into the air has a constant acceleration due to gravity acting on it. However, when it reaches its maximum height, its velocity is zero.
(b) it is impossible; acceleration implies an increase or decrease in speed, and uniform speedimplies that the speed does not change over time
(c) It is possible; for an object accelerating in a circular trajectory, the acceleration is perpendicular to the direction followed by the object.

#### 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.

Solution:
Given:
Radius of the orbit = 42250 km
Therefore, circumference of the orbit = 2×π×42250km = 265571.42 km
Time taken for the orbit = 24 hours
Therefore, speed of the satellite = 11065.4 km.h-1
.
The satellite orbits the Earth at a speed of 11065.4 kilometres per hour. 