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JEE Main 2023 (29 Jan, Shift 2) · Non-Uniform Circular Motion

Non-Uniform Circular Motion question from JEE Main 2023 (29 Jan, Shift 2)

Q14JEE Main 2023 · 29 JAN · SHIFT 2Difficult

Non-Uniform Circular Motion

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−π/2) st(1 - e^{-\pi/2})\text{ s}. The value of t is ____________.

Numerical answer type. Work it out, then open the answer.

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Answer: 40

  1. at=aca_t = a_c: dvdt=v2R\frac{dv}{dt} = \frac{v^2}{R} gives s=−Rln⁡(1−v0tR)s = -R\ln\left(1 - \frac{v_0 t}{R}\right)
  2. Quarter revolution, s=πR2s = \frac{\pi R}{2}: t=Rv0(1−e−π/2)t = \frac{R}{v_0}\left(1 - e^{-\pi/2}\right)
  3. v0=54v_0 = 54 km/h =15= 15 m/s, Rv0=60015=40\frac{R}{v_0} = \frac{600}{15} = 40
  4. So t=40t = 40 s
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