Question

The acceleration $a$ in $m{s}^{-2}$ of a particle is given by $a=3t^2+2t+2$, where $t$ is the time. If the particle starts out with a velocity $v=2m{s}^{-1}$ at $t=0$, then find the velocity at the end of $2s$.

Answer 1

${\int }_2^{v}dv={\int }_0^{2}adt$

$v-2={[\frac{3t^3}{3}+\frac{2t^2}{2}+2t]}_0^2\Rightarrow v=18m{s}^{-1}$