8 . Motion

CBSE Chapter 8 . Motion

Chapter 8 . Motion

Class 9 Science Chapter 8 Motion internal / Example Questions and Answers :

Internal Questions :

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 can have zero displacement even if it has moved through a distance.

Example: If an object moves forward 5 meters and then returns back 5 meters to its starting position, the total distance traveled by the object is 10 meters (= 5 meters forward + 5 meters backward). However, the displacement of the object is zero because the net change in position is zero – it ended up at the same position from where it started.

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?

Solution: Given, The side of the square field m

Time taken to move along the boundary of the square field second

Perimeter of the square field m = 40 m

The total time taken by the farmer, which is 2 minutes and 20 seconds.

Total time = 120 seconds + 20 seconds = 140 seconds

The farmer will complete rounds around the square.

The magnitude of the displacement of the farmer field

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.

Answer: (a) false , because the initial and final positions are the same.

(b) false , because the displacement takes into account the shortest straight-line distance between the initial and final positions, which can be greater than the actual distance traveled.

Example 8.1 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

Therefore, the average speed of the object is 5.33 m/s .

Internal Questions :

1. Distinguish between speed and velocity .

Answer: Difference between speed and velocity :

Speed	Velocity
The speed of an object is the distance covered per unit time.	The velocity is the displacement per unit time.
The speed of an object need not be constant.	The velocity remains constant with time.
The speed of an objects will be in non-uniform motion.	The velocity of an object can be uniform or variable
The speed of an objects has only magnitude. (Scalar quantity)	The velocity of an objects has magnitude and direction . (Vector quantity)

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

Answer: When the object travels in a straight line, the magnitude of average velocity is equal to its average speed.
3. What does the odometer of an automobile measure?

Answer: The odometer of an automobile measures the total distance traveled by the vehicle .
4. What does the path of an object look like when it is in uniform motion?

Answer: When an object is in uniform motion, its path is a straight line .
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, m/s .

Solution: Given, m/s , min = 5 × 60 = 300 seconds

We know that,

Therefore, the distance of the spaceship from the ground station is m .

Example 8.2 : 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 and m/s .

Solution: Distance covered by the car

Time

Average speed of the car km/h

Again, Average speed of the car km/h

Example 8.3 : 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 = 180 m.

Displacement of Usha in 1 min m

Total time taken = 1 minute = 60 seconds

Average speed m/s

Average velocity m/s

The average speed of Usha is 3 m/s and her average velocity is 0 m/s .

Example 8.4 : Starting from a stationary position, Rahul paddles his bicycle to attain a velocity of 6 m/s in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m/s in the next 5 s. Calculate the acceleration of the bicycle in both the cases.

Solution: In first case :

Given, Initial velocity, m/s

Final velocity, m/s

Time s

We have,

m/s²

In second case : Initial velocity, m/s

Final velocity, m/s

Time s

We have,

m/s²

The acceleration of the bicycle in the first case is 0.2 m/s² and in the second case, it is – 0.4 m s² .

Internal Questions :

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

Answer: (i) If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the object is said to be uniform acceleration .

(ii) If an object travels in a straight line and its velocity increases or decreases by unequal amounts in equal intervals of time, then the object is said to be non- uniform acceleration .

2. A bus decreases its speed from 80 km/h to 60 km/h in 5 s. Find the acceleration of the bus.

Solution: Given, Initial speed km/h

m/s

Final speed 60 km/h

m/s

Time 5 seconds

We can use the equation of motion:

The acceleration

Therefore, the acceleration of the bus is – 1.11 m/s².

3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km/h in 10 minutes. Find its acceleration.

Solution: Given, Initial velocity m/s

Final velocity 40 km/h

m/s

Time 10 minutes = 10 × 60 seconds = 600 seconds

We have,

m/s²

Therefore, the acceleration of the train is 0.0185 m/s² .

Internal Questions :

1. What is the nature of the distance-time graphs for uniform and non-uniform motion of an object?

Answer: The distance-time graph for uniform motion is a straight line, while for non-uniform motion, it is a curved 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?

Answer: If the distance-time graph is a straight line parallel to the time axis, it indicates that the object is at rest or not moving.
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?

Answer: If the speed-time graph is a straight line parallel to the time axis, it indicates that the object is at a constant speed.
4. What is the quantity which is measured by the area occupied below the velocity-time graph ?

Answer: The quantity measured by the area below the velocity-time graph is the distance traveled by the object.

Example 8.5 : A train starting from rest attains a velocity of 72 km/h 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: Given, m/s

km/h m/s

And Time minutes = 5 × 60 s = 300 s.

(i) We know that ,

m/s²

(ii) We have ,

The acceleration of the train is 0.067 m/s² and the distance travelled is 3 km.

Example 8.6 : A car accelerates uniformly from 18 km/h to 36 km/h in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car in that time.

Solution: Given, km/h 5 m/s

km/h m/s and s

(i) We know that,

m/s²

(ii) We have,

The acceleration of the car is 1 m/s² and the distance covered is 37.5 m.

Example 8.7 : The brakes applied to a car produce an acceleration of 6 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: Given, m/s² s and m/s

We know that,

m/s

We get,

Thus, the car will move 12 m before it stops after the application of brakes.

Internal Questions :

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

Solution: Given, m/s² , min = 2 × 60 = 120 seconds , m/s

(a) We have,

m/s

Therefore, the speed acquired by the bus is 12 m/s.

(b) We have,

Therefore, the distance traveled by the bus is 720 meters.

2. A train is travelling at a speed of 90 km/h . Brakes are applied so as to produce a uniform acceleration of – 0.5 . Find how far the train will go before it is brought to rest.

Solution: Given, Initial velocity km/h m/s

Acceleration m/s²

Final velocity m/s (train comes to a rest)

Using the equation of motion, we have,

Therefore, the train will travel 625 meters before coming to a rest when the brakes are applied with a uniform acceleration of – 0.5 m/s².

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

Solution: Given, Initial velocity cm/s (since the trolley starts from rest)

Acceleration cm/s²

Time seconds

Using the equation of motion, we have

cm/s

Therefore, the velocity of the trolley 3 seconds after the start is 6 cm/s.

4. A racing car has a uniform acceleration of 4 . What distance will it cover in 10 s after start?

Solution: Given: Initial velocity m/s (Since the car starts from rest)

Acceleration m/s²

Time seconds

Using the equation of motion, we have

Therefore, the racing car will cover a distance of 200 meters in 10 seconds after the start

5. A stone is thrown in a vertically upward direction with a velocity of 5 m/s . If the acceleration of the stone during its motion is 10 in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Solution: Given, Initial velocity m/s (upward)

Final velocity m/s

Acceleration m/s² (downward)

Using the equations of motion, we have

second

Again,

Therefore, the stone attains a height of 1.25 meters, and it takes 0.5 seconds to reach that height.

Class 9 Science Chapter 8 Motion Exercise Questions and Answers :

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?

Solution: The diameter of the track = 200 m

The radius m

The circumference of the circle

meters

Total Time = 2 minutes 20 seconds

= 2 × 60 + 20 = 140 seconds

The number of rounds completed is given by:

Number of Rounds

Number of Rounds rounds

Total Distance Covered m

Since the athlete completes an integral number of rounds, the starting and ending positions coincide, resulting in a displacement of 0.

Therefore, the distance covered by the athlete at the end of 2 minutes 20 seconds is m, and the displacement is 0.

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: We know that,

Average Speed

Average Velocity

(a) From A to B:

Total Distance = 300 m

Total Time = 2 minutes 30 seconds = 2 minutes + 30 seconds

= (2 × 60 + 30) seconds = 150 seconds

Average Speed (A to B) m/s

Average Velocity (A to B)

The displacement from A to B is 300 m in the positive direction (forward).

Average Velocity (A to B) m/s

(b) From A to C:

Total Distance = 300 m + 100 m = 400 m

Total Time = 2 minutes 30 seconds + 1 minute

= (2 × 60 + 30 + 60) seconds = (120+ 90) seconds = 210 seconds

Average Speed (A to C) m/s

Average Velocity (A to C)

The displacement from A to C is 100 m in the positive direction (forward).

Average Velocity (A to C) m/s

Therefore, Joseph's average speed from A to B is 2 m/s, with an average velocity of 2 m/s. His average speed from A to C is approximately 1.9 m/s, with an average velocity of approximately 0.48 m/s.

3. Abdul, while driving to school, computes the average speed for his trip to be 20 km/h . On his return trip along the same route, there is less traffic and the average speed is 30 km/h . What is the average speed for Abdul’s trip?

Solution: We know that,

Speed

Let, the distance from Abdul's home to school is kilometers.

On the outbound trip: Speed = 20 km/h

Time taken

On the return trip: Speed = 30 km/h

Time taken

The total distance for the round trip is 2d kilometers (since he travels the same distance both ways).

The total time taken for the round trip

The average speed is given by:

Average Speed

Therefore, the average speed for Abdul's trip is 24 km/h.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 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?

Solution: Given, initial velocity m/s (Since the boat starts from rest)

The acceleration m/s²

and the time taken s

Using the equation of motion, we have

Therefore, the motorboat travels a distance of 96.0 meters during the 8.0 second time interval.

5. A driver of a car travelling at 52 km/h applies the brakes and accelerates uniformly in the opposite direction. The car stops in 5 s. Another driver going at 3 km/h 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: For the first car:

Initial speed km/h m/s

Time taken s

Final speed m/s (as the car stops)

Using the equation of motion, we have

The graph for the first car will be a straight line with a negative slope, representing uniform deceleration.

We have,

For the second car:

Initial speed km/h m/s

Time taken s

Final speed m/s (as the car stops)

Using the equation of motion, we have

We have,

The graph for the second car will also be a straight line with a negative slope, representing a slower uniform deceleration compared to the first car.

Therefore, the second car traveled farther after the brakes were applied, covering a distance of 7.885 meters compared to the first car, which traveled a distance of 0 meters (came to a stop).

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?
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?

Solution: Given, height m

Final velocity ,

The initial velocity m/s (Since the ball is dropped) ,

The acceleration m/s² and t is the time taken.
We know that,

We have,

m/s

And seconds

seconds

Therefore, the ball will strike the ground with a velocity of 20 m/s, and it will take 2 seconds to reach the ground.

8. The speed-time 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?
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.

Answer: (a) It is possible for an object to have a constant acceleration but with zero velocity. One example is an object thrown vertically upward at its maximum height. At the highest point, the object momentarily stops and its velocity becomes zero while still experiencing the acceleration due to gravity.

(b) It is not possible for an object to move with an acceleration but with uniform speed. Acceleration is the rate of change of velocity, so if there is an acceleration, the speed of the object cannot remain constant.

(c) It is possible for an object to move in a certain direction with an acceleration in the perpendicular direction. One example is a car moving on a curved road. The car's velocity is tangential to the curve, while the acceleration acts towards the center of the curve, which is perpendicular to the velocity vector.

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 km = 42250 × 1000 m = 42250000 m

Time hours = 24 × 60 × 60 seconds

We know that,

The speed of the artificial satellite

m/s

Therefore, the speed of the artificial satellite in its circular orbit is 3,073.75 m/s