Physics: Motion Problems and Solutions

A complete collection of solved conceptual and numerical problems from ICSE Class 9 Physics Chapter on Motion. Covers key topics like distance, displacement, graphs, free fall, acceleration, and velocity with detailed step-by-step solutions and reasoning. Perfect for revision and exam preparation.

1. John moving on a circular track of radius 10 m completes three-fourth of the circular track. What will be the distance travelled and displacement respectively.

Step 1: Radius of the circular track = 10 m, John completes three-fourth of the track.

Step 2: Circumference = 2πr = 2 × 3.14 × 10 = 62.8 m.

Step 3: Distance travelled = (3/4) × circumference = (3/4) × 62.8 = 47.1 m.

Step 4: Displacement is the shortest distance between the initial and final points. Since John completes three-fourth of the track, the displacement is equal to the chord of the circular arc, which can be calculated using Pythagoras theorem: displacement = √(10² + 10²) = √200 = 14.14 m (approximately).

Final answer: Distance = 47.1 m, Displacement = 14.14 m (approximately).

2. Reema throws a ball vertically upward with velocity u, find the greatest height h to which the ball will rise is?

Step 1: Initial velocity = u, acceleration = -g (since it’s thrown upward).

Step 2: Use the equation of motion to find the maximum height: v² = u² + 2as, where v = 0 at maximum height, so 0 = u² – 2gh.

Step 3: Solve for h: h = u² / (2g).

Final answer: h = u² / (2g).

3. An object moving with an initial velocity of 10 km/h comes to rest after 15 min. Then the distance covered by the object is?

Step 1: Initial velocity = 10 km/h, final velocity = 0, time = 15 min = 0.25 h.

Step 2: Convert the initial velocity to m/s: Initial velocity = 10 km/h = (10 × 1000) / 3600 = 2.78 m/s.

Step 3: Use the equation of motion to find the acceleration: v = u + at, 0 = 2.78 + a × (15 × 60), a = -2.78 / (15 × 60) = -0.0031 m/s².

Step 4: Calculate the distance covered: Using v² = u² + 2as, 0 = (2.78)² + 2 × (-0.0031) × s, s = (2.78)² / (2 × 0.0031) = 1250 m or using s = ut + (1/2)at².

Final answer: s = 20.85 m (using s = ut + (1/2)at² and converting initial velocity to m/s and time to seconds).

4. An insect is crawling on a cycle rim of radius r. What will be the distance and displacement of the insect (i) in half revolution, and (ii) in one complete revolution?

Step 1: Radius of the cycle rim = r.

Step 2: For half revolution: Distance = πr, displacement = 2r.

Step 3: For one complete revolution: Distance = 2πr, displacement = 0.

Final answer: (i) Distance = πr, displacement = 2r; (ii) Distance = 2πr, displacement = 0.

5. Can the distance travelled by an object be smaller than the magnitude of its displacement?

Step 1: Distance is the total length of the path traveled, while displacement is the shortest distance between the initial and final points.

Step 2: Distance is always greater than or equal to the magnitude of displacement.

Final answer: No.

6. In what condition is the distance covered equal to the magnitude of the displacement of a particle?

Step 1: Distance equals displacement when the particle moves in a straight line without changing direction.

Final answer: When the particle moves in a straight line without changing direction.

7. Explain with the help of two examples that motion is relative.

Step 1: Motion is relative because it depends on the observer’s frame of reference.

Step 2: Example 1: A person sitting in a train sees a tree passing by, but a person standing near the tree sees the train passing by.

Step 3: Example 2: A person in a car sees a cyclist moving, but the cyclist sees the car moving.

Final answer: Motion is relative.

8. Which of the quantity, velocity or acceleration determines the direction of motion?

Step 1: Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

Step 2: Velocity determines the direction of motion.

Final answer: Velocity.

9. Two students start from the same point and walk in opposite directions for 5 meters each. What is the displacement between them now?

Step 1: Both students walk 5 meters in opposite directions.

Step 2: Displacement = 5 + 5 = 10 meters.

Final answer: 10 meters.

10. During a given time interval the average velocity of an object is zero. What can you conclude about its displacement over the time interval?

Step 1: Average velocity = 0.

Step 2: Average velocity = displacement / time, so displacement = 0.

Final answer: Displacement is zero.

11. Under what circumstances does distance travelled equal magnitude of displacement? What is the only case in which magnitude of displacement and distance are exactly the same?

Step 1: Distance equals displacement when the object moves in a straight line without changing direction.

Step 2: When the object moves in a straight line without changing direction.

Final answer: When the object moves in a straight line without changing direction.

12. Is it possible for velocity to be constant while acceleration is not zero? Explain.

Step 1: Velocity is the rate of change of displacement, and acceleration is the rate of change of velocity.

Step 2: If velocity is constant, acceleration must be zero.

Final answer: No.

13. A car is moving in a straight line with speed 18 km h-1. It is stopped in 5 s by applying the brakes. Find: (i) the speed of car in m s-1, (ii) the retardation and (iii) the speed of car after 2 s of applying the brakes.

Step 1: Initial speed = 18 km/h, time taken to stop = 5 s.

Step 2: Convert the initial speed to m/s: Initial speed = 18 km/h = 5 m/s.

Step 3: Calculate the retardation: Using v = u + at, 0 = 5 + a × 5, a = -1 m/s².

Step 4: Calculate the speed after 2 s: v = u + at = 5 + (-1) × 2 = 3 m/s.

Final answer: (i) 5 m/s, (ii) 1 m/s², (iii) 3 m/s.

14. A car moving on a straight path covers a distance of 1 km due east in 100 s. What is (i) the speed and (ii) the velocity, of car?

Step 1: Distance = 1 km, time = 100 s, direction = due east.

Step 2: Calculate the speed: Speed = distance / time = 1000 / 100 = 10 m/s.

Step 3: Calculate the velocity: Velocity = displacement / time = 1000 / 100 = 10 m/s due east.

Final answer: (i) 10 m/s, (ii) 10 m/s due east.

15. What does the slope of a displacement-time graph represent? How does the value or sign of the slope relate to the direction of motion?

Step 1: The slope represents the velocity.

Step 2: A positive slope indicates motion in the positive direction, while a negative slope indicates motion in the negative direction.

Final answer: The slope represents velocity.

16. Is it possible for a displacement-time graph to have a zero slope for some time and then become inclined? What does this indicate about the object’s motion?

Step 1: A displacement-time graph with zero slope indicates no motion.

Step 2: A change from zero slope to a non-zero slope indicates that the object starts moving.

Final answer: Yes, it indicates that the object was at rest and then started moving.

17. A ball is thrown vertically upward and comes back down. Sketch and explain the nature of its displacement-time graph.

Step 1: The ball is thrown vertically upward.

Step 2: The displacement-time graph will be a parabola that opens downward.

Final answer: The graph will be a parabola.

18. If the displacement-time graph of a moving object becomes steeper over time, what can you say about its motion?

Step 1: The displacement-time graph becomes steeper.

Step 2: A steeper slope indicates an increase in velocity.

Final answer: The object is accelerating.

19. Compare the motion of two objects: One has a velocity-time graph that is a straight horizontal line. The other has a graph that is a straight line with positive slope. What is the difference in their motions?

Step 1: Object 1 has a constant velocity, and object 2 has a velocity with a positive slope.

Step 2: Object 1 is moving with a constant velocity, while object 2 is accelerating.

Final answer: Object 1 has a constant velocity, and object 2 is accelerating.

20. What does the slope of a velocity-time graph represent? How does the steepness of the slope affect the interpretation?

Step 1: The slope represents acceleration.

Step 2: A steeper slope indicates greater acceleration.

Final answer: The slope represents acceleration.

21. A student sees a velocity-time graph with a negative slope. What can they conclude about the motion of the object?

Step 1: The velocity-time graph has a negative slope.

Step 2: The object is decelerating or accelerating in the opposite direction.

Final answer: The object is decelerating.

22. What kind of velocity-time graph would a free-falling object (neglecting air resistance) show? What would its slope represent?

Step 1: The object is free-falling.

Step 2: The velocity-time graph will be a straight line with a positive slope.

Step 3: The slope represents acceleration due to gravity.

Final answer: A straight line with a positive slope, representing acceleration due to gravity.

23. The area under a velocity-time graph represents displacement. How is this area interpreted when: (a) The graph lies above the time axis? (b) The graph lies below the time axis?

Step 1: The area under the velocity-time graph represents displacement.

Step 2: (a) The area above the time axis represents positive displacement. (b) The area below the time axis represents negative displacement.

Final answer: (a) Positive displacement, (b) negative displacement.

24. A train starts from rest and accelerates uniformly at 100 m minute-2 for 10 minutes. Find the velocity acquired by the train. It then maintains a constant velocity for 20 minutes. The brakes are then applied and the train is uniformly retarded. It comes to rest in 5 minutes. Draw a velocity-time graph and use it to find: (i) the retardation in the last 5 minutes, (ii) total distance travelled, and iii) the average velocity of the train.

Step 1: The train accelerates uniformly at 100 m minute-2 for 10 minutes.

Step 2: Velocity acquired: v = u + at = 0 + (100/60) × 10 × 60 = 1000 m/min = 16.67 m/s.

Step 3: Distance travelled during acceleration: s1 = ut + (1/2)at² = 0 + (1/2) × (100/60) × (10 × 60)² = 5000 m.

Step 4: Distance travelled during constant velocity: s2 = v × t = 1000 × 20 × 60 = 1200000 m.

Step 5: Retardation: a = -v / t = -1000 / (5 × 60) = -3.33 m/s².

Step 6: Distance travelled during retardation: s3 = ut + (1/2)at² = 1000 × (5 × 60) – (1/2) × (1000 / (5 × 60)) × (5 × 60)² = 2500 m.

Step 7: Total distance travelled: Total distance = s1 + s2 + s3 = 5000 + 1200000 + 2500 = 1207500 m.

Step 8: Average velocity: Average velocity = total distance / total time = 1207500 / (10 + 20 + 5) × 60 = 572.37 m/min = 9.54 m/s.

Final answer: (i) -3.33 m/s², (ii) 1207500 m, (iii) 9.54 m/s.

25. A stone is thrown vertically upwards with an initial velocity of 40 m s-1. Taking g = 10 m/s2, draw the velocity-time graph of the motion of stone till it comes back on the ground. (i) Use graph to find the maximum height reached by the stone. (ii) What is the net displacement and total distance covered by the stone?

Step 1: Initial velocity = 40 m/s, g = 10 m/s².

Step 2: Time taken to reach maximum height: v = u + at, 0 = 40 – 10t, t = 4 s.

Step 3: Maximum height: h = ut + (1/2)at² = 40 × 4 – (1/2) × 10 × 4² = 80 m.

Step 4: Total distance travelled: Total distance = 2 × h = 2 × 80 = 160 m.

Step 5: Net displacement: Net displacement = 0.

Final answer: (i) 80 m, (ii) net displacement = 0, total distance = 160 m.

27. Figure shows the displacement of a body at different times. (a) Calculate the velocity of the body as it moves for time interval (i) 0 to 5 s, (ii) 5 s to 7 s and (iii) 7 s to 9 s. (b) Calculate the average velocity during the time interval 5 s to 9 s.

Step 1: Displacement-time graph.

Step 2: Calculate the velocities for each time interval:

(i) v = (20 – 0) / (5 – 0) = 4 m/s
(ii) v = (20 – 20) / (7 – 5) = 0 m/s
(iii) v = (0 – 20) / (9 – 7) = -10 m/s

Step 3: Calculate the average velocity: Average velocity = (0 – 20) / (9 – 5) = -5 m/s.

Final answer: (a) (i) 4 m/s, (ii) 0 m/s, (iii) -10 m/s; (b) -5 m/s.

28. Figure shows the velocity-time graph of a particle moving in a straight line. (i) State the nature of motion of particle. (ii) Find the displacement of particle at t = 6 s. (iii) Does the particle change its direction of motion? (iv) Compare the distance travelled by the particle from 0 to 4 s and from 4 s to 6 s. (v) Find the acceleration from 0 to 4 s and retardation from 4 s to 6 s.

Step 1: Velocity-time graph.

Step 2: (i) The particle accelerates from 0 to 4 s and decelerates from 4 s to 6 s.

Step 3: (ii) Displacement = area under the graph from 0 to 6 s.

Step 4: (iii) No, the particle does not change direction.

Step 5: (iv) Distance from 0 to 4 s is greater than from 4 s to 6 s.

Step 6: (v) Acceleration = slope from 0 to 4 s, retardation = slope from 4 s to 6 s.

Final answer: (i) Accelerating and decelerating, (ii) displacement = area under the graph, (iii) no, (iv) distance from 0 to 4 s is greater, (v) acceleration and retardation can be calculated from the slopes.

31. A ball is thrown vertically upwards with a velocity of 20 m/s from the top of a multi-storey building. The height of the point from where the ball is thrown 25 m from the ground. (i) How high will the ball rise? (ii) How long will it be before the ball hits the ground? (Take, g =10 m /s2)

Step 1: Initial velocity = 20 m/s, height = 25 m, g = 10 m/s².

Step 2: Maximum height: h = u² / (2g) = 20² / (2 × 10) = 20 m.

Step 3: Time taken to reach maximum height: t = u / g = 20 / 10 = 2 s.

Step 4: Time taken to hit the ground: Using s = ut + (1/2)at², -25 = 20t – (1/2) × 10 × t².

Step 5: Solve the quadratic equation: t² – 4t – 5 = 0, t = 5 s or t = -1 s (not valid).

Step 6: Total time: Total time = 2 + (5 – 2) = 5 s (time taken to come down from maximum height to the ground).

Final answer: (i) 20 m, (ii) 5 s.

32. A body moving with uniform acceleration travels 84 m in the first 6 s and 180 m in the next 5 s. Find: (a) the initial velocity, and (b) the acceleration of the body.

Step 1: Distance travelled in the first 6 s = 84 m, distance travelled in the next 5 s = 180 m.

Step 2: Use the equation of motion to find the initial velocity and acceleration: Using s = ut + (1/2)at², we can write two equations: 84 = 6u + (1/2)a(6)² and 264 = 11u + (1/2)a(11)² – (1/2)a(6)² (since the distance travelled in 11 s is 84 + 180 = 264 m).

Step 3: Simplify the equations: 84 = 6u + 18a and 264 = 11u + (121/2)a – 18a.

Step 4: Solve the equations: Simplifying the equations, we get: 84 = 6u + 18a and 180 = 5u + (85/2)a (by subtracting the first equation from the second equation for 11 s and 6 s).

Step 5: Calculate the initial velocity and acceleration: Solving these equations simultaneously will give the values of u and a.

Final answer: u = 2 m/s, a = 4 m/s² (after solving the equations).

33. A particle starts to move in a straight line from a point with velocity 10 m s-1 and acceleration – 2 m s-2. Find the position and velocity of the particle at (i) t = 5 s, (ii) t = 10 s.

Step 1: Initial velocity = 10 m/s, acceleration = -2 m/s².

Step 2: Use the equation of motion to find the position and velocity at t = 5 s: Using s = ut + (1/2)at², s = 10(5) + (1/2)(-2)(5)² = 25 m. v = u + at = 10 + (-2)(5) = 0 m/s.

Step 3: Use the equation of motion to find the position and velocity at t = 10 s: s = 10(10) + (1/2)(-2)(10)² = 0 m. v = u + at = 10 + (-2)(10) = -10 m/s.

Final answer: (i) s = 25 m, v = 0 m/s; (ii) s = 0 m, v = -10 m/s.

34. When brakes are applied to a bus, the retardation produced is 25 cm s-2 and the bus takes 20 s to stop. Calculate: (i) the initial velocity of bus, and (ii) the distance travelled by bus during this time.

Step 1: Retardation = 25 cm/s² = 0.25 m/s², time taken to stop = 20 s.

Step 2: Calculate the initial velocity: Using v = u + at, 0 = u + (-0.25)(20), u = 5 m/s.

Step 3: Calculate the distance travelled: Using s = ut + (1/2)at², s = 5(20) + (1/2)(-0.25)(20)² = 50 m.

Final answer: (i) 5 m/s, (ii) 50 m.

35. A body, initially at rest, starts moving with a constant acceleration 2 m s-2. Calculate: (i) the velocity acquired and (ii) the distance travelled in 5 s.

Step 1: Initial velocity = 0, acceleration = 2 m/s², time = 5 s.

Step 2: Calculate the velocity acquired: v = u + at = 0 + 2(5) = 10 m/s.

Step 3: Calculate the distance travelled: s = ut + (1/2)at² = 0 + (1/2)(2)(5)² = 25 m.

Final answer: (i) 10 m/s, (ii) 25 m.

36. A car travels a distance 100 m with a constant acceleration and average velocity of 20 m s-1. The final velocity acquired by the car is 25 m s-1. Find: (i) the initial velocity and (ii) acceleration.

Step 1: Distance = 100 m, average velocity = 20 m/s, final velocity = 25 m/s.

Step 2: Calculate the time taken: Time = distance / average velocity = 100 / 20 = 5 s.

Step 3: Use the equation of motion to find the initial velocity and acceleration: Using v = u + at, 25 = u + a(5). Using s = ut + (1/2)at² or v² = u² + 2as, we can find u and a.

Step 4: Calculate the initial velocity and acceleration: Using v² = u² + 2as, 25² = u² + 2a(100). Solving these equations simultaneously will give the values of u and a.

Final answer: (i) 15 m/s, (ii) 2 m/s² (after solving the equations).

37. A space craft flying in a straight course with a velocity of 75 km s − 1 fires its rocket motors for 6.0 s. At the end of this time, its speed is 120 km s − 1 in the same direction. Draw a velocity time graph and use it to find: (i) the space craft’s average acceleration while the motors were firing, (ii) the distance travelled by the space craft in the first 10 s after the rocket motors were started, the motors having been in action for only 6 s.

Step 1: Initial velocity = 75 km/s, final velocity = 120 km/s, time = 6 s.

Step 2: Calculate the average acceleration: a = Δv / Δt = (120 – 75) / 6 = 7.5 km/s².

Step 3: Calculate the distance travelled during the first 6 s: Using s = ut + (1/2)at² or the average velocity, s = ((75 + 120)/2) × 6 = 585 km.

Step 4: Calculate the distance travelled during the next 4 s: s = v × t = 120 × 4 = 480 km.

Step 5: Calculate the total distance travelled: Total distance = 585 + 480 = 1065 km.

Final answer: (i) 7.5 km/s², (ii) 1065 km.