Question

A car moving along a straight highway with of 126kmhinverse is brought to a stop with in a distance of 200m what is the retardation of the car

Answer 1

U = 126×5/18
u = 35m/s
use Newton's third law of motion.
v^2 – u^2 = 2as
0 – 35×35 = 2 × a × 200
a = –35×35/400
a = –1225/400
a = –49/16 m/s^2

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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