Question

The accleration of a particle is given by $a=(3t^2+2t+2)$ ${\text{ms}}^{-2}$ where $t$ is time. If the particle starts out with an initial velocity $v=2\text{m/s}$ at $t=0$, then the velocity of the particle at the end of $2\text{s}$ is

• A. $12\text{m/s}$
• B. $14\text{m/s}$
• C. $16\text{m/s}$
• D. $18\text{m/s}$

Answer 1

The correct option is D $18\text{m/s}$

As we know that
Acceleration, $a=\frac{dv}{dt}$
$\Rightarrow dv=adt$
Let the velocity of the particle at end of $2\text{sec}$ is $v$,
$\therefore {\int }_2^{v}dv={\int }_0^2(3t^2+2t+2)dt$
$\Rightarrow [v{]}_2^v=[t^3+t^2+2t{]}_0^2$
$\Rightarrow (v-2)=(16-0)$
$\Rightarrow v=18\text{m/s}$