Question

A particle is moving in a circular path of radius 1 $m$. Under the action of a centripetal force, the speed $\left(\sqrt{2}\right)\pi {ms}^{-1}$ of the particle is constant. Find the average velocity (in ${ms}^{-1}$) between $A$ and $B$.

Answer 1

So the displacement of the particle between A and B will be :

$d=\sqrt{r^2+r^2}=r\sqrt{2}$

and the time taken to travel that distance=$\frac{T}{4}=\frac{2\pi r}{\sqrt{2}\pi \times 4}$

So the average velocity=$\frac{d}{\frac{T}{4}}$

Or, $v_{avg}=\frac{r\sqrt{2}}{\frac{2\pi r}{\sqrt{2}\pi \times 4}}=4m{s}^{-1}$