A2 Daily A Level Physics question
In a lab, a trolley moves along a straight track and its displacement x from a marker is plotted against time t. The x–t graph is three straight segments: from t = 0.0 s to 3.0 s, x increases uniformly from 0.00 m to 0.90 m; from 3.0 s to 7.0 s, x stays constant at 0.90 m; from 7.0 s to 10.0 s, x decreases uniformly to 0.30 m. Which option gives the average speed over 0–10.0 s and the sign of the instantaneous velocity at t = 8.0 s?
Answer
The correct answer is B.
Correct: B — 0.15 m s^-1; negative. Total distance = 0.90 m out + 0.60 m back = 1.50 m, so average speed over 10.0 s is 1.50/10.0 = 0.15 m s^-1; at 8.0 s the graph is sloping down, so velocity is negative. A uses net displacement (0.30 m) divided by time to get 0.03 m s^-1, confusing average speed with average velocity, and also takes the velocity at 8 s as positive. B is correct because it uses total distance for speed and recognises the negative gradient (moving back) between 7–10 s. C ignores the return motion (effectively using 0.90 m/10.0 s) and wrongly infers zero velocity at 8 s from the earlier flat section. D gets the average speed right but wrongly assigns a positive sign to the instantaneous velocity despite the decreasing displacement at 8 s.