Question

A point moves rectilinearly with deceleration which depends on the velocity $v$ of the particle as $a=k\sqrt{v}$, where $k$ is a positive constant. At the initial moment the velocity of the point is equal to $v_0$. What distance will it cover before it stops, and what time it will take to cover that distance? Mark this time from the given options.

• A. $\frac{2\sqrt{v_0}}{k}$
• B. $\frac{k}{2\sqrt{v_0}}$
• C. $\frac{3\sqrt{v_0}}{k}$
• D. $\frac{\sqrt{v_0}}{3k}$

Answer 1

The correct option is A $\frac{2\sqrt{v_0}}{k}$

To find the time taken to come to rest, we use $a=\frac{dv}{dt}$, as
$\frac{dv}{dt}=-k\sqrt{v}$
$\frac{dv}{\sqrt{v}}=-kdt$
On integrating ${\int }_{v_0}^v\frac{dv}{\sqrt{v}}=-{\int }_0^{t}kdt$
$2(\sqrt{v_0}-\sqrt{v})=kt$
Substituting v = 0 and t = T
$T=\frac{2\sqrt{v_0}}{k}$