A car is moving on a circular path of radius 600   m such that the magnitudes of the tangential…

Given:

Initial speed $=\frac{54\times 1000}{3600}=15 \mathrm{m} {\mathrm{s}}^{-1}$

$\left|a_t\right|=\left|a_c\right|$

$$\begin{gathered} \Rightarrow v\frac{dv}{dx}=\frac{v^2}{R} \\ \Rightarrow {\int }_{15}^v\frac{dv}{v}=\frac{1}{R}{\int }_0^{x}dx \end{gathered}$$

$\Rightarrow v=15e^{\frac{x}{R}}$

$\Rightarrow \frac{dx}{dt}=15e^{\frac{x}{R}}$

Distance covered in first quarter of revolution $=\frac{2\pi R}{4}=\frac{\pi R}{2}$.

Therefore,

${\int }_0^{\frac{\pi R}{2}}{e}^{-\frac{x}{R}}dx=15{\int }_0^{t_0}dt$

$\Rightarrow t_0=40\left(1-e^{-\frac{\pi }{2}}\right)$

Hence, $t=40$.