A car moving along a straight highway with a speed of 72 km/h is brought to rest with in a distance of 100m. What is the retardation of the car and how long does it takes to stop.
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Initial velocity of the car, u = 126 km/h = 35 m/s
Final velocity of the car, v = 0
Distance covered by the car before coming to rest, s = 200 m
Retardation produced in the car = a
From third equation of motion, a can be calculated as:
V^2 - U^2 = 2aS
(0)^2 - (35)^2 = 2 . a . 200
a = ( 35×35)/(2× 200)
a= 3.06 m/ s^2
From first equation of motion, time (t) taken by the car to stop can be obtained as:
V = U + at
t = (V- U )/a = (-35)/(-3.06) = 11.44 sec
I hope, this will help you
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