A point initially at rest moves along X-axis. Its acceleration varies with time as $\mathrm{a} =(6\mathrm{t} + 5) {\mathrm{ms}}^{-2}$. If it starts from origin, then the distance covered in 2 s is

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 = 3t² + 5t + C, where C is a constant of integration.

When t = 0, V = 0, so C = 0

$\therefore$             $v=\frac{ds}{dt}=3t^2+5t\text{ or } ds=\left(3t^2+5t\right)dt$

Integrating it within the conditions of motion, i. e. as t changes from 0 to 2s, s changes from 0 to s, we have

                 ${\int }_0^{s}ds={\int }_0^2\left(3t^2+5t\right)dt$

$\therefore$                      $s={\left[t^3+\frac{5}{2}{t}^2\right]}_0^2=8+10=18 \mathrm{m}$