WAEC 2020 · Paper 1 · Q35

Kinematics & dynamics

The distance(s) in metres covered by a particle in motion at any time, $t$ seconds, is given by $S = 120t - 16t^2$ . Find, in metres, the distance covered by the body before coming to rest.

Worked solution (try it first)

Differentiate to get the velocity: $v = \dfrac{dS}{dt} = 120 - 32t$ .
At rest $v = 0$ , so $32t = 120$ and $t = 3.75$ s.
Then $S = 120(3.75) - 16(3.75)^2 = 450 - 225$ .
So the distance is 225 m, option D.

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