Given:
Initial speed =360054×1000=15ms−1
∣at∣=∣ac∣
⇒vdxdv=Rv2⇒∫15vvdv=R1∫0xdx
⇒v=15eRx
⇒dtdx=15eRx
Distance covered in first quarter of revolution =42πR=2πR.
Therefore,
∫02πRe−Rxdx=15∫0t0dt
⇒t0=40(1−e−2π)
Hence, t=40.
A car is moving on a circular path of radius 600m 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−1 is t(1–e–2π)s. The value of t is ______.
Held on 29 Jan 2023 · Verified 6 Jul 2026.
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