Kinematics

JEE Main level, numerical answer, medium

292 of 308

Question 292 of 308

A car is moving on a circular path of radius 600m600m 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 54kmh−154km{h}^{-1} is t(1–e–π2)st(1–{e}^{–\dfrac{\pi }{2}})s. The value of tt is ______.

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Answer: 40

Given:
Initial speed =54×10003600=15ms−1=\dfrac{54\times 1000}{3600}=15m{s}^{-1}
∣at∣=∣ac∣\left|{a}_{t}\right|=\left|{a}_{c}\right|
⇒vdvdx=v2R⇒∫15vdvv=1R∫0xdx\Rightarrow v\dfrac{dv}{dx}=\dfrac{{v}^{2}}{R} \Rightarrow {\int }_{15}^{v}\dfrac{dv}{v}=\dfrac{1}{R}{\int }_{0}^{x}dx
⇒v=15exR\Rightarrow v=15{e}^{\dfrac{x}{R}}
⇒dxdt=15exR\Rightarrow \dfrac{dx}{dt}=15{e}^{\dfrac{x}{R}}
Distance covered in first quarter of revolution =2πR4=πR2=\dfrac{2\pi R}{4}=\dfrac{\pi R}{2}.
Therefore,
∫0πR2e−xRdx=15∫0t0dt{\int }_{0}^{\dfrac{\pi R}{2}}{e}^{-\dfrac{x}{R}}dx=15{\int }_{0}^{{t}_{0}}dt
⇒t0=40(1−e−π2)\Rightarrow {t}_{0}=40\left(1-{e}^{-\dfrac{\pi }{2}}\right)
Hence, t=40t=40.

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