Question

A car is moving along a straight pathwith speed of 126km/hr is brought tostop within a distance of 200m . what is the retardation of the car

Answer 1

u = 126 km/hr = 35 m/s
v =0
s = 200 m
as, v² - u² = 2as
⇒(0)² - (35)² = 2×a×200
⇒400a = -1225
⇒a = - 3.0625 m/s²
∴retardation = 3.0625 m/s²

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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