Question

The instantaneous velocity of a particle moving in a straight line is given as $v=\alpha t+\beta t^2$, where $\alpha$ and $\beta$ are constants. The distance travelled by the particle between $1\text{s}$ and $2\text{s}$ is

• A. $3\alpha +7\beta$
• B. $\frac{3}{2}\alpha +\frac{7}{3}\beta$
• C. $\frac{\alpha }{2}+\frac{\beta }{3}$
• D. $\frac{3}{2}\alpha +\frac{7}{2}\beta$

Answer 1

The correct option is B $\frac{3}{2}\alpha +\frac{7}{3}\beta$

Given that,

$v=\alpha t+\beta t^2$

$\Rightarrow \frac{ds}{dt}=\alpha t+\beta t^2$

Integrating both side within suitable limit

${\int }_{S_1}^{S_2}ds={\int }_1^2(\alpha t+\beta t^2)dt$

$S_2-S_1={[\frac{\alpha t^2}{2}+\frac{\beta t^3}{3}]}_1^2$

As the velocity is always positive, the particle does not change the direction.

So, distance = displacement.

$\therefore \text{Distance},D=[\frac{\alpha [4-1]}{2}+\frac{\beta [8-1]}{3}]$

$D=\frac{3\alpha }{2}+\frac{7\beta }{3}$

Hence, option $\left(\text{B}\right)$ is correct.