A car moves on a circular path of radius 600 m with tangential acceleration equal in magnitude to centripetal acceleration. Starting at 54 km/h, the time for the first quarter revolution is t(1- e − π 2 e^{-\frac{\pi}{2}} e − 2 π ) s. Find t.
Physics · Circular Motion · kinematics of circular motion
JEE Mains 2023

Question

A car moves on a circular path of radius 600 m with tangential acceleration equal in magnitude to centripetal acceleration. Starting at 54 km/h, the time for the first quarter revolution is t(1- e − π 2 e^{-\frac{\pi}{2}} e − 2 π ) s. Find t.

Physics · Circular Motion · kinematics of circular motion

JEE Mains 2023

G1 Physics · Accepted Answer

aₜ=a(c) gives dv/dt=v²/R. Since ds/dt=v, dv/ds=v/R, so v=v₀ e s / R e^{s/R} e s / R . For s=πR/2, time t(q)=∫₀^{πR/2} ds/[v₀ e s / R e^{s/R} e s / R ]=(R/v₀)(1- e − π / 2 e^{-π/2} e − π /2 ). With R=600 m and v₀=15 m/s, R/v₀=40 s. Thus t=40.

A car moves on a circular path of radius 600 m with tangential acceleration equal in magnitude to centripetal acceleration. Starting at 54 km/h, the time for the first quarter revolution is t(1- e − π 2 e^{-\frac{\pi}{2}} e − 2 π ​ ) s. Find t.

A car moves on a circular path of radius 600 m with tangential acceleration equal in magnitude to centripetal acceleration. Starting at 54 km/h, the time for the first quarter revolution is t(1- e − π 2 e^{-\frac{\pi}{2}} e − 2 π ) s. Find t.