Explanation

(b) Given, $\ \ \ \ \ \ \ \ a = \frac{dv}{dt} = 6t + 5$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ dv = (6t + 5)\ dt$
Integrating, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{v}{dv} = \int_{0}^{t}(6t + 5)\ dt$
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ v = \left( \frac{6t^{2}}{2} + 5t \right)$
Again $\ \ \ \ \ \ \ \ \ \ v = \frac{ds}{dt}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ds = \left( \frac{6t^{2}}{2} + 5t \right)\ dt$
Integrating again, we get
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \int_{0}^{s}{ds} = \int_{0}^{t}{\left( \frac{6t^{2}}{2} + 5t \right)dt}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = \frac{3t^{3}}{3} + \frac{5t^{2}}{2}$
When, $t = 2\ s,\ s = 3 \times \frac{2^{3}}{3} + \frac{5\ \times 2^{2}}{2} = 3 \times \frac{8}{3} + \frac{5\ \times 4}{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 8 + 10 = 18\ m$