9th Standard Chapter 7 Motion Ncert Question Answers with Solutions

Class 9 Science Chapter: Motion - Exercises with Easy Answers

Here are the detailed and simple explanations for your exercise questions from the chapter Motion to help you revise effectively.

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?

Answer: Diameter = 200 m, so the circumference = π × 200 = 628 m (approx).
Total time = 140 s. Number of rounds = 140 / 40 = 3.5 rounds.
Distance covered = 3.5 × 628 = 2198 m (approx).
Displacement after 3.5 rounds will be the diameter of the circle (since 0.5 round means opposite side) = 200 m.

2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 min 30 s and then turns around and jogs 100 m back to point C in another 1 min. What are Joseph's average speeds and velocities in jogging (a) from A to B and (b) from A to C?

Answer:
(a) From A to B: Distance = 300 m, Time = 150 s. Average speed = 300/150 = 2 m/s. Average velocity = 2 m/s (since it is a straight path).
(b) From A to C: Distance = 300 + 100 = 400 m, Time = 150 + 60 = 210 s. Average speed = 400/210 = 1.9 m/s. Displacement = 200 m (since C is 100 m back from B, so 300 - 100). Average velocity = 200/210 = 0.95 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?

Answer: Let the distance be x km. Time taken during onward trip = x/20, during return = x/30.
Total time = x/20 + x/30 = (3x + 2x)/60 = 5x/60 = x/12.
Total distance = 2x.
Average speed = Total distance / Total time = 2x / (x/12) = 2 × 12 = 24 km/h.

4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s² for 8.0 s. How far does the boat travel during this time?

Answer: Initial speed u = 0, acceleration a = 3 m/s², time t = 8 s.
Distance travelled, s = ut + (1/2)at² = 0 + 0.5 × 3 × 64 = 96 m.

5. A driver of a car travelling at 52 km/h applies the brakes. (a) 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?

Answer: (a) The area under the speed-time graph shows the distance travelled by the car.
(b) The horizontal part of the speed-time graph before braking represents uniform motion of the car.

6. Fig 7.10 shows the distance-time graph of three objects A, B, and C. Study the graph and answer the following questions:

Answer: (a) The steepest slope on the distance-time graph shows the fastest speed.
(b) If the lines intersect, they are at the same position at the same time.
(c) By checking where B crosses A, check the distance of C at that time.
(d) Similarly, check B's position when it crosses C on the graph.

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?

Answer: Using v² = u² + 2as, u = 0, a = 10 m/s², s = 20 m.
v² = 0 + 2 × 10 × 20 = 400, so v = 20 m/s.
Using v = u + at: 20 = 0 + 10 × t, so t = 2 s.

8. The speed-time graph for a car is shown as Fig 7.11. (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?

Answer: (a) The area under the speed-time graph for the first 4 seconds gives the distance.
(b) The horizontal part of the graph represents uniform speed of the car.

9. State which of the following situations are possible and give an example for each:

Answer: (a) An object with constant acceleration but zero velocity (Example: An object thrown up at the highest point, it has zero velocity but constant downward acceleration due to gravity).
(b) An object moving with acceleration but uniform speed (Not possible, as acceleration changes velocity).
(c) An object moving in a certain direction with acceleration in the perpendicular direction (Example: Circular motion, like a satellite around Earth, where acceleration is towards the center).

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.

Answer: Radius r = 42250 km = 4.225 × 10⁷ m, Time T = 24 × 3600 s = 86400 s.
Speed = Distance / Time = 2πr / T = 2 × 3.14 × 4.225 × 10⁷ / 86400 ≈ 3.07 km/s.

This webpage is designed for Class 9 students to help them revise the Motion chapter easily with detailed yet simple answers. Keep practicing these questions to strengthen your understanding of motion, speed, and graphs.