AS Daily A Level Physics question
A student walks along a straight corridor. A displacement–time graph for their motion has two straight sections: from t = 0 to 20 s the displacement increases uniformly from 0 m to +16 m; from t = 20 s to 50 s the displacement decreases uniformly to +4 m. Which statement must be true for the whole 0–50 s interval?
Answer
The correct answer is D.
Correct: D — The average speed is 0.56 m/s and seven times the magnitude of the average velocity; the average velocity is positive. Distance = 16 m + 12 m = 28 m so average speed = 28/50 = 0.56 m/s; net displacement = +4 m so average velocity = +4/50 = +0.08 m/s, and 0.56 is 7×0.08 with a positive direction. A confuses distance with displacement and wrongly asserts equality; reversals make average speed larger than the magnitude of average velocity. B infers sign from the longer-time segment, but the sign of average velocity depends on net displacement, which is positive here. C uses displacement/time to compute speed; that gives the magnitude of the average velocity, not the average speed, which must use total distance 28 m. D correctly compares distance (28 m) with displacement (4 m) and gives the correct sign and ratio.