NECO 2024 · Paper 1 · Q58

Calculus (JAMB bridge)

A particle moves a distance of $S$ metres in $t$ seconds, where $S = 5t^3 - 12t^2 + 7$ . At what time is its acceleration zero?

Worked solution (try it first)

Velocity is $v = \frac{dS}{dt} = 15t^2 - 24t$ .
Acceleration is $a = \frac{dv}{dt} = 30t - 24$ .
Set $a = 0$ : $30t = 24$ , so $t = 0.8$ s, option A.

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