An electric bus has these (time in s, speed in m/s) readings: (0, 0), (4, 12), (10, 12), (14, 4), (16, 4), (18, 0). Join successive points with straight lines. (a) Describe the motion. (b) Calculate acceleration from 0–4 s and 10–14 s. (c) Calculate total distance. (d) Calculate average speed. (e) Explain why speed data alone may not give displacement.

Question

Accepted Answer

Accelerations +3.0 m/s² and −2.0 m/s²; distance 140 m; average speed 7.78 m/s; displacement also needs direction. The rising line from 0–4 s is uniform acceleration from rest, 4–10 s is constant speed, 10–14 s is uniform slowing, 14–16 s is constant slower motion and 16–18 s is uniform slowing to rest. Acceleration is graph gradient: (12 − 0)/4 = +3.0 m/s² initially and (4 − 12)/(14 − 10) = −2.0 m/s² during the first slowing interval. The sign describes the chosen direction. Distance is area below the graph: ½(4)(12) + (6)(12) + ½(12 + 4)(4) + (2)(4) + ½(2)(4) = 24 + 72 + 32 + 8 + 4 = 140 m. Average speed is total distance divided by total time: 140/18 = 7.78 m/s. Averaging the highest and lowest speeds would ignore the unequal durations spent at each speed. Speed has no sign. If the bus reverses, distance still accumulates positively but displacement must subtract travel in the opposite direction; a velocity–time graph or explicit directions are required.

Electric-bus speed–time evidence

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Electric-bus speed

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Electric-bus speed — supplied values
Time / s	Speed / m s⁻¹
0	0
4	12
10	12
14	4
16	4
18	0