WAEC Further Mathematics 2015 Paper

Concept:Distance is found by integrating acceleration twice with respect to time.The initial conditions, starting from rest, give zero constants of integration.Explanation:Given acceleration: $a = 3t - 2 \,\mathrm{m\,s^{-2}}$.Velocity is the first integral of acceleration:$v = \int (3t - 2)\,dt = \frac{3}{2}t^2 - 2t + C_1$.Since the particle starts from rest, $v(0) = 0$, so $C_1 = 0$.Thus $v = \frac{3}{2}t^2 - 2t \,\mathrm{m\,s^{-1}}$.Distance is the integral of velocity:$s = \int \left(\frac{3}{2}t^2 - 2t\right)dt = \frac{t^3}{2} - t^2 + C_2$.Assuming displacement is measured from the starting point, $s(0) = 0$, so $C_2 = 0$.At $t = 3$ seconds:$s(3) = \frac{3^3}{2} - 3^2 = \frac{27}{2} - 9 = \frac{27 - 18}{2} = \frac{9}{2}$.Therefore, the distance covered is $\frac{9}{2}\,\mathrm{m}$.Answer:$s = \frac{9}{2}\,\mathrm{m}$.Correct option: D. $\frac{9}{2}\,\mathrm{m}$