WAEC 2019 · Paper 2 · Q14✱✱

(a)
A body moving at $20\text{ m s}^{-1}$ accelerates uniformly at $2\frac12\text{ m s}^{-2}$ for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest $t$ seconds after with uniform retardation. The ratio of the acceleration to the retardation is $3 : 4$ . Sketch the velocity–time graph of the motion.
Model answer
The graph starts at $20\text{ m s}^{-1}$ (not at 0), rises in a straight line to $30\text{ m s}^{-1}$ at 4 s, stays level for 8 s (to 12 s), then falls in a straight line to 0 after a further $t$ seconds. Label the speeds and the time intervals. With the $3 : 4$ ratio, the retardation is $\frac{10}{3}\text{ m s}^{-2}$ , so $t = 9$ s and the graph reaches 0 at 21 s.
(b)
Find the value of $t$ .
(c)
Find the total distance of the journey.

Velocity–time graph: the area underneath is 475 m.

Worked solution (try it first)

After 4 s: $v = 20 + 2.5 \times 4$
$= 30\text{ m s}^{-1}$ .
The graph starts at $20$ (not 0), rises to $30$ at $t = 4$ , stays level until $t = 12$ , then falls to 0.

$2.5 : r = 3 : 4$ , so the retardation is $r = \dfrac{4 \times 2.5}{3}$
$= \dfrac{10}{3}\text{ m s}^{-2}$ .
Time to stop: $t = 30 \div \frac{10}{3} = 9\text{ s}$ .