If the velocity of a particle is $\mathrm{v} = \mathrm{At} + {\mathrm{Bt}}^2$, where A and B are constants, then the distance travelled by it between Is and 2s is     [NEET 2016]

Velocity of the particle is given as $\mathrm{v} = \mathrm{At} + {\mathrm{Bt}}^2$

where, A and B are constants.

$\Rightarrow \frac{dx}{dt}=At+B{t}^2$                                                $\left(\because v=\frac{dx}{dt}\right)$

$\Rightarrow dx=\left(At+B{t}^2\right)dt$

Integrating both sides, we get

               ${\int }_{x_1}^{x_2}dx={\int }_1^2\left(At+B{t}^2\right)dt$

$\Rightarrow \Delta x=x_2-x_1=A{\int }_1^{2}tdt+B{\int }_1^2{t}^{2}dt$

                        $=A{\left[\frac{t^2}{2}\right]}_1^2+B{\left[\frac{t^3}{3}\right]}_1^2$

                         $=\frac{A}{2}\left(2^2-1^2\right)+\frac{B}{3}\left(2^3-1^3\right)$

$\therefore$   Distance travelled by the particle between Is and 2s is

                    $\Delta x=\frac{A}{2}\times \left(3\right)+\frac{B}{3}\left(7\right)=\frac{3A}{2}+\frac{7B}{3}$