Question 3
The table shows how the displacement of a child cycling along a straight path varied second by second.
| t (s) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|---|
| s (m) | 0 | 2 | 4 | 6 | 8 | 8 | 8 | 8 | 8 | 4 | 0 |
What kind of motion does the child have during the first 4 seconds?
The right answer is the one marked above.
Forwards at a uniform velocity of 2 m s-1. Over the first 4 s the displacement grows by 2 m every second — 0, 2, 4, 6, 8.
velocity = (8 − 0) m ÷ 4 s = 2 m s-1
Since the change in displacement is the same in each second, the velocity is uniform. 8 m s-1 is a slip — 8 m is the total displacement over the 4 s, not the displacement per second.