WAEC 2013 · Paper 2 · Q18

(a)
The displacement, $S$ metres, of a particle from a fixed point $O$ at time $t$ seconds is given by $S = t^2 - 6t + 5$ . On a graph sheet, draw a displacement–time graph for the interval $0 \le t \le 6$ .
Model answer
Plot $(0, 5), (1, 0), (2, -3), (3, -4), (4, -3), (5, 0), (6, 5)$ and join them with a smooth U-shaped curve. It crosses $S = 0$ at $t = 1$ and $t = 5$ and is lowest, $S = -4$ , at $t = 3$ .
(b)
From the graph, find the: (i) time at which the velocity is zero; (ii) average velocity over the interval $0 \le t \le 4$ ; (iii) total distance covered in the interval $0 \le t \le 5$ .
Separate values with commas, e.g. 3, −2

Displacement–time graph: S = t² − 6t + 5.

Worked solution (try it first)

Work out $S = t^2 - 6t + 5$ at each whole second: $S = 5, 0, -3, -4, -3, 0, 5$ at $t = 0, 1, 2, 3, 4, 5, 6$ .
Plot them and join with a smooth U-shaped curve.

The velocity is the gradient of the displacement–time graph.
It is zero at the lowest point, $t = 3$ s (also $\dfrac{dS}{dt} = 2t - 6 = 0$ ).

Average velocity $= \dfrac{\text{change in displacement}}{\text{time}}$
$= \dfrac{S(4) - S(0)}{4}$
$= \dfrac{-3 - 5}{4}$
$= -2\text{ m s}^{-1}$ , that is $2\text{ m s}^{-1}$ back towards $O$ .