Question

A particle is moving with tangential acceleration along a circle of radius $R$ in such a way that at any instant the magnitude of centripetal and tangential acceleration are equal.If at $t=0$ the speed was ${}^{\prime }{V}_o^{\prime }$, then the time to complete the first revolution will be.

Answer 1

$\frac{dV}{dt}=\frac{V^2}{R}$

${\int }_{v=0}^v\frac{dV}{V^2}={\int }_{t=0}^{t=t}\frac{dt}{R}$

$\frac{V^{-2+1}}{-2+1}=\frac{dt}{R}$

$\frac{t}{R}={\left[\frac{V^{-1}}{-1}\right]}_{V_o}^V$

$\frac{t}{R}=[-1][\frac{1}{V}-\frac{1}{V_o}]$

$-\frac{t}{R}+\frac{1}{V_o}=\frac{1}{V}$