Rotational Motion Question 10
Question 10 - 29 January - Shift 2
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 $54 km / hr$ is $t(1-e^{-\pi / 2}) s$. The value of $t$ is
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Answer: (40)
Solution:
Formula: Instantaneous Velocity, Instantaneous Acceleration
$v \frac{dv}{dx}=\frac{v^{2}}{R} \Rightarrow \int _15^{v} \frac{dv}{v}=\frac{1}{R} \int_0^{x} dx$
$v=15 e^{x / R}$
$\frac{dx}{dt}=15 e^{x / R}$
$\pi R$
$\int_0^{2} e^{-x / R} d x=15 \int_0^{t_0} d t$
$t_0=40(1-e^{-\pi / 2})$
