A car is moving on a circular path of radius $$600$$ m such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete first quarter of revolution, if it is moving with an initial speed of $$54$$ km h$$^{-1}$$ is $$t(1-e^{-\frac{\pi}{2}})$$ s. The value of $$t$$ is ______.

Solution :

Given :

Radius of circular path,

$$r = 600\text{ m}$$

Initial speed,

$$u = 54\text{ km h}^{-1}$$

$$= \frac{54 \times 1000}{3600}$$

$$= 15\text{ m s}^{-1}$$

Given tangential acceleration equals centripetal acceleration :

$$a_t = a_c$$

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

Separating variables :

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

Integrating :

$$\int_{u}^{v}\frac{dv}{v^2} = \int_{0}^{t}\frac{dt}{r}$$

$$\left[-\frac{1}{v}\right]_{u}^{v} = \frac{t}{r}$$

$$\frac{1}{u} - \frac{1}{v} = \frac{t}{r}$$

Hence,

$$v = \frac{ur}{r-ut}$$

Angular speed :

$$\omega = \frac{v}{r}$$

Angular displacement for quarter revolution :

$$\theta = \frac{\pi}{2}$$

Therefore,

$$\int_0^t \omega dt = \frac{\pi}{2}$$

$$\int_0^t \frac{u}{r-ut}dt = \frac{\pi}{2}$$

$$\left[-\ln(r-ut)\right]_0^t = \frac{\pi}{2}$$

$$\ln\left(\frac{r}{r-ut}\right) = \frac{\pi}{2}$$

$$\frac{r}{r-ut} = e^{\pi/2}$$

$$r-ut = re^{-\pi/2}$$

$$ut = r(1-e^{-\pi/2})$$

$$t = \frac{r}{u}(1-e^{-\pi/2})$$

Substituting values :

$$t = \frac{600}{15}(1-e^{-\pi/2})$$

$$t = 40(1-e^{-\pi/2})$$

Comparing with given form :

$$t(1-e^{-\pi/2})$$

Hence,

$$t = 40$$

Final Answer :

$$40$$