Question

A particle moves in a circular path such that its speed v varies with distance $s$ as $v=\sqrt{s}$, where $\alpha$ is $\alpha$ positive constant. Find the acceleration of the particle after traversing a distance $s$.

Answer 1

Total acceleration, $a=\sqrt{a_t^2+a_r^2}=\sqrt{{\left(\frac{dv}{dt}\right)}^2+{\left(\frac{v^2}{R}\right)}^2}$
where $v=\alpha \sqrt{s}$
Differentiating $v=\alpha \sqrt{s}$ with respect to time, we have
$\frac{dv}{dt}=\alpha \frac{s^{-1/2}}{2}\frac{ds}{dt}$
Substituting $\frac{ds}{dt}=\alpha \sqrt{s}$, we have $\frac{dv}{dt}=\frac{{\alpha }_2}{2}$
Now substituting $dv/dt$ and $v$ in the expression of $a$, we have
$a=\sqrt{{\left(\frac{{\alpha }^2}{2}\right)}^2+{\left[\frac{\alpha \sqrt{s}}{R}\right]}^2}$
This gives $a={\alpha }^2\sqrt{\frac{1}{4}+\frac{s^2}{R^2}}$