Question

The magnitude of radius vector of a point varies with time as $r=\left(\beta t\right(1-\alpha t)$ where $\alpha$ and $\beta$ are positive constant. The distance travelled by this body over a closed path must be then :

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

Answer 1

The correct option is A $\frac{\beta }{2\alpha }$

The velocity is given as,

$\overrightarrow{v}=\frac{d\overrightarrow{r}}{dt}$

$\overrightarrow{v}=\frac{d}{dt}\left(\beta t\left(1-\alpha t\right)\right)$

$\overrightarrow{v}=\beta \left(1-2\alpha t\right)$

The distance travelled by the body is given as,

$s=\underset{0}{\overset{\frac{1}{2\alpha }}{\int }}\beta \left(1-2\alpha t\right)dt+\underset{\frac{1}{2\alpha }}{\overset{\frac{1}{\alpha }}{\int }}\beta \left(2\alpha t-1\right)dt$

$s=\frac{\beta }{2\alpha }$