Problems Based on Average Speed for Class 9 Science

Let the one-way distance be ‘d’. The time taken for the forward journey is t₁ = d/30. The time taken for the return journey is t₂ = d/45. Total distance = 2d. Total time = (d/30) + (d/45) = (3d + 2d)/90 = 5d/90.

(a) To pass a pole, the train must travel a distance equal to its own length. Speed = Distance / Time = 100m / 5s = 20 m/s.
(b) To cross a bridge, the train must travel a distance equal to the bridge’s length plus its own length. Total distance = 500m + 100m = 600m.

Distance covered in one revolution is equal to the circumference of the circle (π × d). For 3 revolutions, Distance = 3 × 3.14 × 5m = 47.1m. Because the particle returns exactly to its starting point after full revolutions, the shortest path between initial and final position is zero.

The total path covered by the body is ‘h’ upwards and ‘h’ downwards, making the total distance h + h = 2h. Since the body falls back to the exact point it was thrown from, its initial and final positions are identical, making displacement zero.

Total distance is simply the sum of the actual path lengths: 15m + 20m = 35m. Since the second movement is at a right angle, the initial, turning, and final points form a right-angled triangle. Displacement is the hypotenuse.

(i) For half a circle, distance is half the circumference (πr), and displacement is the straight line connecting opposite ends, which is the diameter (2r).
(ii) For a full circle, distance is the full circumference (2πr), and displacement is zero as it returns to the start.

First, find the total distance travelled: 16m + 16m = 32m. Then, find the total time taken: 4s + 2s = 6s. Finally, divide total distance by total time.

Case 1: Initial velocity (u) = 0 m/s, Final velocity (v) = 6 m/s, Time (t) = 30s. Acceleration = (v – u)/t = (6 – 0)/30 = 0.2 m/s².
Case 2: Initial velocity (u) = 6 m/s, Final velocity (v) = 4 m/s, Time (t) = 5s. Acceleration = (v – u)/t = (4 – 6)/5 = -0.4 m/s².

First convert velocity to m/s: 45 km/hr = 45 × (5/18) = 12.5 m/s.
(a) 1 minute = 60 seconds. Distance = Speed × Time = 12.5 m/s × 60s = 750 m.
(b) 1 second. Distance = 12.5 m/s × 1s = 12.5 m.

Total distance covered = 2400 km – 2000 km = 400 km. Total time = 8 hrs. Average speed in km/hr = 400 / 8 = 50 km/hr. To convert to m/s, multiply by 5/18.

First, convert velocity to m/s: 120 km/hr = 120 × (5/18) = 33.33 m/s. Time is 30s.

Uniform motion implies that the velocity of the body remains constant over time. There is no acceleration. Therefore, the velocity after 10 seconds will still be 15 m/s.

When a journey is split into equal time intervals, the average speed is simply the arithmetic mean of the speeds. Average Speed = (v₁ + v₂) / 2.

Initial velocity (u) = 0.5 m/s. Final velocity (v) = 0 m/s (since it stops). Acceleration (a) = -0.05 m/s² (negative because velocity decreases). Using the first equation of motion: v = u + at.

Convert speeds to m/s: u = 36 km/hr = 10 m/s; v = 70 km/hr ≈ 19.44 m/s.
Acceleration = (19.44 – 10) / 5 = 1.888 m/s².
For stopping: u = 19.44 m/s, v = 0, t = 20s. a = (0 – 19.44) / 20 ≈ -0.97 m/s². Retardation is the positive magnitude of negative acceleration.

Convert max speed: 36 km/hr = 10 m/s.
Case 1 (starting): u = 0, v = 10 m/s, t = 10s. a = (10 – 0) / 10 = 1 m/s².
Case 2 (stopping): u = 10 m/s, v = 0, t = 20s. a = (0 – 10) / 20 = -0.5 m/s².

Total target time for the 120km trip at an average of 60 km/hr is 120/60 = 2 hours. The time taken for the first 30km is 30/30 = 1 hour. Time remaining for the next 90km is 2 – 1 = 1 hour.

Calculate total distance (d = v×t): (60×0.52) + (30×0.24) + (70×0.71) = 31.2 + 7.2 + 49.7 = 88.1 km. Total time = 0.52 + 0.24 + 0.71 = 1.47 hr.

Total distance = 180m. Total time = 1 min = 60s. Average speed = 180/60 = 3 m/s. Since he returns to his starting point, displacement is 0m. Average velocity = Displacement / Time = 0 / 60 = 0 m/s.

(a) Total distance = 4 rounds × 200m/round = 800m.
(b) Displacement is 0 because the start and finish lines are the exact same point.
(c) Non-uniform. Even if their speed is constant, their direction of motion is constantly changing on a closed track, meaning velocity changes.
(d) No, displacement (0) is not equal to distance (800m).