WAEC 2008 · Paper 2 · Q7

(a)
A car travelling at a velocity of $50\text{ km h}^{-1}$ covers a distance of $20\text{ km}$ . If it was accelerating at $6\text{ km h}^{-2}$ , calculate, correct to one decimal place, the time the car took to cover the distance.

Worked solution (try it first)

Use $s = ut + \frac12 at^2$ with $u = 50$ , $a = 6$ and $s = 20$ .
Substitute: $20 = 50t + 3t^2$ .
Rearrange into a quadratic: $3t^2 + 50t - 20 = 0$ .
Use the formula: $t = \dfrac{-50 \pm \sqrt{50^2 + 4(3)(20)}}{2(3)}$
$= \dfrac{-50 \pm \sqrt{2740}}{6}$ .
$\sqrt{2740} \approx 52.345$ .
Time cannot be negative, so $t = \dfrac{2.345}{6} \approx 0.391$ .
The car took about $0.4$ hours (about 23 minutes).