AS Daily A Level Physics question
In a lab, a trolley moves along a straight air track and its displacement s from the start is recorded against time t. The s–t graph is described as follows: 0 ≤ t < 4 s: s increases linearly from 0 m to +2.0 m. 4 s ≤ t < 6 s: s is constant at +2.0 m. 6 s ≤ t ≤ 10 s: s decreases linearly from +2.0 m to −2.0 m. Which statement must be true?
Answer
The correct answer is D.
Correct: D — Over 0–10 s, the average speed is three times the magnitude of the average velocity. The total distance travelled is 2 m (out) + 4 m (back) = 6 m, while the net displacement is −2 m, so over 10 s the average speed is 0.6 m s⁻¹ and the magnitude of the average velocity is 0.2 m s⁻¹; the ratio is 3. A confuses distance and displacement: using the same total time does not make average speed equal to the magnitude of average velocity when direction changes. B is wrong because from 4–6 s the graph is flat, so at t = 5 s the instantaneous velocity is zero, not negative. C is wrong because ending farther from the start does not determine speed; the slope magnitude shows speed and is larger in 6–10 s than in 0–4 s.