Question

A particle retards from a velocity $v_0$ while moving in a straight line. If the magnitude of deceleration is directly proportional to the square root of the speed of the particle, find its average velocity for the total time of its motion.

Answer 1

The average velocity is given as:
$v_{av}=\frac{s}{t}=\frac{Totaldisplacement}{Totaltime}$
We know a = v$\frac{dv}{ds}\Rightarrow$ v dv = a ds
Let us calculate the displacement by substituting a = $-\alpha \sqrt{v}$ (for retardation) in vdv = a ds.
We have v dv = ($-\alpha \sqrt{v}ds$ $\Rightarrow$ $\sqrt{v}dv$ = $-\alpha ds$
When the particle slows down from $v_0$ to 0, it covers a distance s.
Hence, ${\int }_{v_0}^0\sqrt{v}dv=-{\int }_0^s\alpha ds$
This gives s = $\frac{2v_0^{3/2}}{3\alpha }$
Now we will find the time of motion by substituting a = $-\alpha \sqrt{v}$ in a = $\frac{dv}{dt}$, we get
$-\alpha \sqrt{v}=\frac{dv}{dt}\Rightarrow \frac{dv}{\sqrt{v}}=-\alpha dt$
If the particle takes time t to stop, we have
${\int }_{v_0}^0\frac{dv}{\sqrt{v}}=-\alpha {\int }_0^{t}dt$
This yields t = $\frac{\sqrt{v_0}}{\alpha }$
Finally, substituting s and t in $v_{av}$ = s/t, we have
$v_{av}=\frac{2v_0}{3}$