The instantaneous velocity of a particle moving in a straight line is given as $\mathrm{v}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1 \mathrm{~s}$ and $2 \mathrm{~s}$ is :
Correct Option: , 2
$\mathrm{V}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$
$\frac{\mathrm{ds}}{\mathrm{dt}}=\alpha \mathrm{t}+\beta \mathrm{t}^{2}$
$\int_{s_{1}}^{s_{2}} d s=\int_{1}^{2}\left(\alpha t+\beta t^{2}\right) d t$
$\mathrm{S}_{2}-\mathrm{S}_{1}=\left[\frac{\alpha \mathrm{t}^{2}}{2}+\frac{\beta \mathrm{t}^{3}}{3}\right]_{1}^{2}$
As particle is not changing direction
So distance $=$ displacement.
Distance $=\left[\frac{\alpha[4-1]}{2}+\frac{\beta[8-1]}{3}\right]$
$=\frac{3 \alpha}{2}+\frac{7 \beta}{3}$