JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Kinematics

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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

t (1 - e^{- π /2}) s

. The value of t is ______.

Answer: 40

Solution

With the tangential and centripetal accelerations equal,

\frac{d v}{d t} = \frac{v ^{2}}{R}

, and writing

\frac{d v}{d t} = v \frac{d v}{d s}

gives

\frac{d v}{d s} = \frac{v}{R}

, so

v = v_{0} e^{s / R}

. Since

v = \frac{d s}{d t}

, separating and integrating from

s = 0

to a quarter circle

s = \frac{π R}{2}

gives

\int_{0}^{π R /2} e^{- s / R} d s = v_{0} t_{0}

, that is

R (1 - e^{- π /2}) = v_{0} t_{0}

. With

v_{0} = 54

km/h

= 15

m/s and

R = 600

t_{0} = \frac{600}{15} (1 - e^{- π /2}) = 40 (1 - e^{- π /2})

s, so

t = 40

JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 29 Jan 2023 · Shift 2 · Kinematics

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