A car is moving on a circular path of radius 600 m such that the magnitudes of 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 is t(1 − e⁻π/2) s. The value of t is ______.
Physics · Circular Motion
JEE Mains 29 January 2023 Shift 2 Evening Shift

Question

A car is moving on a circular path of radius 600 m such that the magnitudes of 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 is t(1 − e⁻π/2) s. The value of t is ______.

Physics · Circular Motion

JEE Mains 29 January 2023 Shift 2 Evening Shift

G1 Physics · Accepted Answer

Equality of tangential and centripetal accelerations gives dv/dt = v²/R. Using v = dx/dt, dv/dx = v/R. Integrating from v₀ = 15 m s⁻¹ at x = 0 gives v = 15e^(x/R). Therefore dt = dx/[15e^(x/R)]. For a quarter revolution x = πR/2. Hence time = (R/15)[1 − e⁻π/2] = (600/15)[1 − e⁻π/2] = 40(1 − e⁻π/2) s. Thus t = 40.