Class 9 Chapter Motion NCERT Solution

 

1. An object has moved through a distance. Can it have zero displacement? If yes, support your answer with an example.

  1. Yes, an object can have zero displacement even if it has moved through a distance. This is possible when the object returns to its initial position after completing its motion. For example, suppose a person walks 10 meters to the east and then walks back 10 meters to the west. The total distance traveled by the person is 20 meters, but the displacement is zero because the person ends up at the same position where they 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?

The farmer moves along the boundary of a square field of side 10 m, so the total distance traveled by the farmer would be the perimeter of the square, which is 4 times the length of the side, or 40 meters. We don’t know the path that the farmer takes, so we can’t directly calculate the displacement. However, we can use the fact that displacement is the straight-line distance between the initial and final positions.

After 2 minutes 20 seconds, the farmer has been walking for a total of 160 seconds (40 seconds initially + 2 minutes 20 seconds = 40 + 120 = 160 seconds). We can use the formula:

displacement = speed x time

where speed is the average speed of the farmer during the entire 160 seconds. The average speed is the total distance traveled divided by the total time taken:

average speed = total distance ÷ total time = 40 m ÷ 160 s = 0.25 m/s

Now we can calculate the displacement:

displacement = average speed x time = 0.25 m/s x 160 s = 40 m

So the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds from his initial position is 40 meters.

3. Which of the following is true for displacement?
(a) It cannot be zero.
(b) Its magnitude is greater than the distance traveled by the object.

The correct statement about displacement is (b) Its magnitude is greater than the distance traveled by the object. Displacement is a vector quantity that represents the straight-line distance between an object’s initial and final positions, along with the direction from the initial position to the final position. It can be positive, negative, or zero depending on the direction of the motion. The magnitude of displacement is always less than or equal to the distance traveled by the object, except in cases where the object returns to its initial position, in which case the displacement is zero and the distance traveled is greater than zero.

 

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Question 1: Explain the difference between speed and velocity.

 

Answer: Speed is the rate of change of distance traveled by an object per unit time, while velocity is the rate of change of displacement per unit time. In other words, speed is a scalar quantity that only describes how fast an object is moving, while velocity is a vector quantity that describes both how fast an object is moving and the direction in which it is moving.

Question2 : When is the average velocity of an object equal to its average speed?

 

Answer: The average velocity of an object is equal to its average speed only when the object moves in a straight line. This is because the direction of motion and displacement are the same in this case. When an object moves in a curved path, its displacement and distance traveled are different, and the average velocity and average speed are not equal.

Question 3 : What is the purpose of an odometer in an automobile?

 

Answer: An odometer in an automobile measures the distance traveled by the vehicle. It does this by counting the number of revolutions of the vehicle’s wheels and converting it into distance using the wheel circumference. This information is useful for vehicle maintenance and for determining fuel efficiency.

Question 4 : How does the path of an object appear when it is moving with uniform motion?

Answer: When an object is moving with uniform motion, its path is a straight line. This is because uniform motion means that the object is moving with a constant speed in a constant direction. If the object changes speed or direction, then its path is not a straight line.

Question 5. : Calculate the distance of a spaceship from a ground station if the signal from the spaceship was received by the ground station in 5 minutes, given that the signal travels at the speed of light which is 3 x 10^8 m/s.

 

Answer: The distance of the spaceship from the ground station can be calculated using the formula:

distance = speed x time

where speed is the speed of light, which is 3 x 10^8 m/s, and time is the time taken for the signal to travel from the spaceship to the ground station, which is 5 minutes or 300 seconds.

Converting the time to seconds, we have:

distance = speed x time = 3 x 10^8 m/s x 300 s = 9 x 10^10 m

Therefore, the distance of the spaceship from the ground station is 9 x 10^10 meters.

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

Answer: (i) A body is said to be in uniform acceleration if it is moving in a straight line and its velocity is changing at a constant rate. In other words, if the body is moving with a constant acceleration, it is said to be in uniform acceleration. (ii) A body is said to be in non-uniform acceleration if its velocity is changing at a non-constant rate. In other words, if the body is moving with a varying acceleration, it is said to be in 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. Answer: The initial speed of the bus, u = 80 km/h The final speed of the bus, v = 60 km/h The time taken, t = 5 s

To find the acceleration of the bus, we use the formula:

a = (v – u) / t

Substituting the values, we get:

a = (60 – 80) / 5 a = -20 / 5 a = -4 km/h^2

Therefore, the acceleration of the bus is -4 km/h^2.

Note: The negative sign indicates that the acceleration is in the opposite direction to the motion of the bus.

 

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.

To find the acceleration of the train, we can use the formula:

acceleration = (final velocity – initial velocity) / time

Here, the initial velocity of the train is 0 km/h (since it starts from rest), the final velocity is 40 km/h, and the time taken for the change in velocity is 10 minutes.

First, we convert the velocities to meters per second (m/s) as follows:

Initial velocity = 0 km/h = 0 m/s Final velocity = 40 km/h = 11.11 m/s

Now, we need to convert the time to seconds:

Time = 10 minutes = 600 seconds

Now, we can substitute these values into the formula:

acceleration = (final velocity – initial velocity) / time = (11.11 m/s – 0 m/s) / 600 s = 0.0185 m/s^2

Therefore, the acceleration of the train is 0.0185 m/s^2.

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

The distance-time graph for uniform motion of an object is a straight line with a constant slope, indicating that the object is covering equal distances in equal intervals of time. The distance-time graph for non-uniform motion of an object is a curve, indicating that the object is covering unequal distances in equal intervals of time.

2. What can you say about the motion of an object whose distance-time graph is a straight
line parallel to the time axis?

If the distance-time graph of an object is a straight line parallel to the time axis, it means that the object is at rest and not moving.

3. What can you say about the motion of an object if its speedtime graph is a straight line
parallel to the time axis?

If the speed-time graph of an object is a straight line parallel to the time axis, it means that the speed of the object is constant and it is moving with uniform motion.

4. What is the quantity which is measured by the area occupied below the velocity-time graph?

 

The area occupied below the velocity-time graph represents the displacement of an object. The quantity measured by the area below the velocity-time graph is the displacement of the object. The displacement is a vector quantity that describes the change in position of an object with respect to its initial position.

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  1. A bus starting from rest moves with a uniform acceleration of 0.1 m/s² for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

(a) The initial velocity of the bus, u = 0 m/s Acceleration of the bus, a = 0.1 m/s² Time taken, t = 2 minutes = 120 seconds

Using the formula, v = u + at, we get: v = 0 + (0.1 × 120) = 12 m/s

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

(b) Using the formula, s = ut + 1/2 at², we get: s = 0 × 120 + 1/2 × 0.1 × (120)² = 720 m

Therefore, the distance travelled by the bus is 720 m.

  1. A train is travelling at a speed of 90 km/h. Brakes are applied so as to produce a uniform acceleration of –0.5 m/s². Find how far the train will go before it is brought to rest.

Initial velocity of the train, u = 90 km/h = 25 m/s Acceleration of the train, a = –0.5 m/s² Using the formula, v² = u² + 2as, we get: 0 = (25)² + 2(-0.5)s Solving for s, we get: s = 625 m

Therefore, the train will go 625 m before it is brought to rest.

  1. A trolley, while going down an inclined plane, has an acceleration of 2 cm/s². What will be its velocity 3 s after the start?

Acceleration of the trolley, a = 2 cm/s² = 0.02 m/s² Time taken, t = 3 s Using the formula, v = u + at, we get: v = 0 + (0.02 × 3) = 0.06 m/s

Therefore, the velocity of the trolley 3 s after the start is 0.06 m/s.

  1. A racing car has a uniform acceleration of 4 m/s². What distance will it cover in 10 s after start?

Initial velocity of the car, u = 0 m/s Acceleration of the car, a = 4 m/s² Time taken, t = 10 s Using the formula, s = ut + 1/2 at², we get: s = 0 × 10 + 1/2 × 4 × (10)² = 200 m

Therefore, the distance covered by the racing car in 10 s after start is 200 m.

  1. 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 m/s² in the downward direction, what will be the height attained by the stone and how much time will it take to reach there?

Initial velocity of the stone, u = 5 m/s Acceleration of the stone, a = –10 m/s² Using the formula, v² = u² + 2as, we get: 0 = (5)² + 2(-10)s Solving for s, we get: s = 1.25 m

Therefore, the height attained by the stone is 1.25 m.

Using the formula, v = u + at, we get: 0 = 5 + (-10)t Solving for t, we get: t = 0.5 s

Exercise
  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 = Circumference of the circular track = πd = π x 200 m = 628.32 m

Time taken to complete 2 minutes 20 seconds = 140 seconds

Number of rounds completed in 140 seconds = 140/40 = 3.5

Total distance covered = 3.5 x 628.32 m = 2199.12 m

Displacement = 0 (since the athlete ends up at the same point where he started)

  1. 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?

(a) Average speed = Total distance covered/Total time taken = (300 m + 300 m)/(2 min 30 s) = 4.8 km/h

Average velocity = Displacement/Total time taken = (300 m – 0)/(2 min 30 s) = 2.0 km/h towards B

(b) Total distance covered = 300 m + 100 m = 400 m

Total time taken = 2 min 30 s + 1 min = 3 min 30 s

Average speed = Total distance covered/Total time taken = 400 m/3.5 min = 6.86 km/h

Average velocity = Displacement/Total time taken = (100 m – 0)/(3 min 30 s) = 1.43 km/h towards C

  1. 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?

Let the distance between Abdul’s home and school be d.

Total time taken = Time taken to reach school + Time taken to return home

= d/20 + d/30 = (3d + 2d)/60 = 5d/60 = d/12

Average speed = Total distance covered/Total time taken = 2d/(d/12) = 24 km/h

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