Question 7
On the same journey, during the last 3 seconds the object goes from 15 m s-1 to rest.
| t (s) | 0 | 6 | 12 | 15 |
|---|---|---|---|---|
| v (m s⁻¹) | 0 | 15 | 15 | 0 |
What is the object's retardation during those last 3 seconds?
The right answer is the one marked above.
5 m s-2. Work out the gradient of the final, downward-sloping part of the graph:
acceleration = (0 − 15) m s-1 ÷ 3 s = −5 m s-2
The value is negative, so this is a retardation of magnitude 5 m s-2. That is, the velocity drops by 5 m s-1 every second: 15, 10, 5, 0.
2.5 m s-2 is the acceleration over the first 6 s. The object takes half as long to stop as it took to speed up, so the retardation is twice the acceleration.