A2 Daily A Level Physics question
A cyclist travels along a straight cycle path. Displacement from the start is defined positive in the forward direction. From 0–20 s the displacement increases uniformly at 2.0 m/s; from 20–40 s it is constant; from 40–60 s it decreases uniformly at 1.0 m/s. Which statement must be true about the motion over the entire 60 s?
Answer
The correct answer is C.
Correct: C — The average speed is three times the magnitude of the average velocity; the average velocity is directed away from the start. Total distance = 40 + 0 + 20 = 60 m, so average speed = 60/60 = 1.0 m/s; net displacement = +40 − 20 = +20 m, so average velocity = +20/60 ≈ +0.33 m/s in the forward direction; the ratio is about 3:1. A The claim treats distance and displacement as equal and also assigns 1.0 m/s to the average velocity, but the magnitude of the average velocity is about 0.33 m/s, not 1.0 m/s. B This assumes the cyclist returns to the start; the net displacement is +20 m, not zero, so the average velocity is not zero. C Matches the calculated 3:1 ratio and the forward (positive) direction for the average velocity. D By definition, average speed (total distance/time) cannot be less than the magnitude of average velocity (|net displacement|/time), so this reverses the inequality.