A car is moving on a circular path of radius 600m such that the...

Given:

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

\left|{a}_{t}\right| = \left|{a}_{c}\right|$</p><p>$⇒ v\dfrac{dv}{dx} = \dfrac{{v}^{2}}{R} ⇒ {∫}_{15}^{v}\dfrac{dv}{v} = \dfrac{1}{R}{∫}_{0}^{x}dx$</p><p>$⇒ v = 15e\dfrac{x}{R}$</p><p>$⇒ \dfrac{dx}{dt} = 15e\dfrac{x}{R}$</p><p>Distance covered in first quarter of revolution$= \dfrac{2\piR}{4} = \dfrac{\piR}{2}$.</p><p>Therefore,</p><p>${∫}_{0}^{\dfrac{\piR}{2}}{e}^{- \dfrac{x}{R}}dx = 15{∫}_{0}^{{t}_{0}}dt$</p><p>$⇒ {t}_{0} = 40\left(1 - {e}^{- \dfrac{\pi}{2}}\right)$</p><p>Hence,$t = 40$.