Question

A car move with a speed of 126km/h and breaks are appled and the car comes to a stop within 200m .calculate the retardation and time taken for the car to stop

Answer 1

Speed = Initial velocity = u =  126 km/h = 126 * (5/18) = 35 m/s
Distance = S = 200 m
Final velocity = v = 0 m/s
We know, 
v^2 = u^2 - 2aS (Since, retardation)
0 = (35)^2 - 2*a*200
- 1225 = -400a
a = 1225/400
= 3.06 m/s^2
Therefore, time taken = (u-v)/t = 3.06
= 35/3.06 = time taken
= 11.44 seconds

Answer 2

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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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