WAEC 2023 · Paper 1 · Q27

Kinematics & dynamics
Differentiation

The distance $S$ metres moved by a body in $t$ seconds is given by $S = 5t^3 - \frac{19}{2}t^2 + 6t - 4$ . Calculate the acceleration of the body after 2 seconds.

Worked solution (try it first)

Velocity is the first derivative: $v = \dfrac{dS}{dt} = 15t^2 - 19t + 6$ .
Acceleration is the second derivative: $a = \dfrac{dv}{dt} = 30t - 19$ .
At $t = 2$ : $a = 60 - 19 = 41\ \text{m s}^{-2}$ , option D.

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