Question 8
Consider the journey as a whole: accelerating to 15 m s-1 in 6 s, 6 s at uniform velocity, then stopping in 3 s — 15 s altogether.
| t (s) | 0 | 6 | 12 | 15 |
|---|---|---|---|---|
| v (m s⁻¹) | 0 | 15 | 15 | 0 |
What total distance does the object travel over the whole time?
The right answer is the one marked above.
157.5 m. The total distance is the area of the shaded trapezium. Its two parallel sides are the total time (15 s) and the time at uniform velocity (6 s), and its height is 15 m s-1:
total distance = (15 + 6) ÷ 2 × 15 = 21 ÷ 2 × 15 = 157.5 m
Working out the three sections separately and adding them gives the same answer: 45 m (triangle) + 90 m (rectangle) + 22.5 m (triangle) = 157.5 m.
225 m comes from treating the whole 15 s as though it were spent at 15 m s-1 (15 × 15) — but the object is slower than that at both the start and the end.