WAEC 2020 · Paper 2 · Q15

(a)
A particle starts from rest at a point $O$ and accelerates uniformly for 3 minutes until it attains a velocity of $133\text{ m s}^{-1}$ . It moves with this velocity for 3 minutes and then retards uniformly for another 2 minutes before coming to rest at $P$ . (i) Sketch the velocity–time graph for the motion. (ii) Calculate the: (α) total distance covered by the particle; (β) average velocity of the particle.
Separate values with commas, e.g. 3, −2
(b)
A particle is projected from a point $P$ with an initial velocity of $60\text{ m s}^{-1}$ at an angle of $60^\circ$ to the horizontal. Find its vertical displacement when its horizontal displacement is $75$ metres. $[\text{Take } g = 10\text{ m s}^{-2}]$

Velocity–time graph: the area under it is the distance.

Worked solution (try it first)

Times: $3\text{ min} = 180\text{ s}$ and $2\text{ min} = 120\text{ s}$ .
The graph rises from 0 to $133$ over $0$ – $180\text{ s}$ , is level to $360\text{ s}$ , then falls to 0 at $480\text{ s}$ .

(α)** The area is a trapezium with parallel sides $180$ and $480$ : $\frac12(180 + 480) \times 133 = 43\,890\text{ m}$ .

Horizontally: $x = (60\cos60^\circ)t = 30t$ , so $t = 2.5\text{ s}$ when $x = 75$ .
Vertically: $y = (60\sin60^\circ)(2.5) - 5(2.5)^2$
$= 75\sqrt3 - 31.25$
$\approx 98.65\text{ m}$ .