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 h - 1 is t ( 1 – e – π 2 ) s . The value of t is ______.

Step-by-step solution

Given: Initial speed = 54 × 1000 3600 = 15   m   s - 1 a t = a c ⇒ v d v d x = v 2 R ⇒ ∫ 15 v d v v = 1 R ∫ 0 x d x ⇒ v = 15 e x R ⇒ d x d t = 15 e x R Distance covered in first quarter of revolution = 2 π R 4 = π R 2 . Therefore, ∫ 0 π R 2 e - x R d x = 15 ∫ 0 t 0 d t ⇒ t 0 = 40 1 - e - π 2 Hence, t = 40 .