Explanation

(b) Given acceleration $a = 6t + 5$
$\therefore a = \frac{dv}{dt} = 6t + 5,\ dv = (6t + 5)dt$
Integrating it, we have $\int_{0}^{v}{dv} = \int_{0}^{t}{(6t + 5)dt}$
$v = 3t^{2} + 5t + C,$ where $C$ is a constant of integration
When $t = 0,\ v = 0$ so $C = 0$
$\therefore v = \frac{ds}{dt} = 3t^{2} + 5t$ or, $ds = (3t^{2} + 5t)dt$
Integrating it within the conditions of motion, $i.e.,$ as $t$ changes from $0$ to $2s,\ s$ changes from $0$ to $s$, we have
$\int_{0}^{s}{ds} = \int_{0}^{2}{\left( 3t^{2} + 5t \right)dt}$
$\therefore s = \left\lbrack t^{3} + \frac{5}{2}t^{2} \right\rbrack_{0}^{2} = 8 + 10 = 18\ m$