A particle is moving in a circle of radius 50 cm in such a way that at any instant the normal and tangential…

Given:

$\left|{\overrightarrow{a}}_{\mathrm{c}}\right|=\left|{\overrightarrow{a}}_{\mathrm{t}}\right|$

$\Rightarrow \frac{v^2}{r}=\frac{dv}{dt}$

$\Rightarrow {\int }_4^v\frac{dv}{v^2}={\int }_0^t\frac{dt}{r}$

$\Rightarrow {\left[\frac{-1}{v}\right]}_4^v=\frac{\mathit{t}}{\mathit{r}}$

$\Rightarrow \frac{-1}{v}+\frac{1}{4}=2t$

$\Rightarrow v=\frac{4}{1-8t}=\frac{\mathrm{d}s}{\mathrm{d}t}$

$\Rightarrow 4{\int }_0^t\frac{\mathrm{d}t}{1-8t}={\int }_0^s\mathrm{d}s$

$\left(r=0.5 \mathrm{m}, \text{distance} \text{covered} s=2\pi r=\pi \right)$

$\Rightarrow 4\times \frac{\left[\ln \right(1-8t){]}_0^t}{-8}=\pi$

$\Rightarrow \ell n(1-8t)=-2\pi$

$\Rightarrow 1-8t=e^{-2\pi }$

$\Rightarrow t=\left(1-{\mathrm{e}}^{-2\pi }\right)\frac{1}{8} \mathrm{s}$

So, $\alpha =8$.