WAEC 2023 · Paper 2 · Q8

Kinematics & dynamics
Integration

The acceleration, $a$ , of a particle starting from rest and moving at any time $t$ seconds is given by $a = (20t - 3t^2)\text{ m s}^{-2}$ . Find the:

(a)
time taken for the particle to come to rest again;
(b)
distance covered by the particle after 5 seconds.

Try it on a graph

Velocity v = 10t² − t³ (x-axis is time). The area under it from 0 to 5 is the distance.

Worked solution (try it first)

(a)

$v = \displaystyle\int (20t - 3t^2)\,dt$
$= 10t^2 - t^3 + c$ .
It starts from rest, so $c = 0$ .
At rest again when $v = 0$ : $t^2(10 - t) = 0$ , so $t = 10$ s.

(b)

$v$ is positive from $t = 0$ to $10$ , so the distance in 5 s is $\displaystyle\int_0^5 (10t^2 - t^3)\,dt$ .
$= \left[\dfrac{10t^3}{3} - \dfrac{t^4}{4}\right]_0^5$
$= \dfrac{1250}{3} - \dfrac{625}{4}$
$= \dfrac{3125}{12}$
$\approx 260.42$ m.

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