NECO 2024 · Paper 1 · Q60

A particle moves a distance of $S$ metres in $t$ seconds, where $S = 5t^3 - 12t^2 + 7$ . Find the acceleration after 12 seconds.

Worked solution (try it first)

Velocity is $v = 15t^2 - 24t$ , so acceleration is $a = \frac{dv}{dt} = 30t - 24$ .
At $t = 12$ : $30 \times 12 - 24 = 360 - 24$
$= 336\text{ m/s}^2$ , option C.