Question

A car moving along a straight highway with speed of 126 km/h is brought to stop within a distance of 200 m . What is the retardationof the car ?

Answer 1

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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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Answer 2

Answer:

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:

v2 - u2 = 2as

(0)2 - (35)2 = 2 × a × 200

a = - 35 × 35 / 2 × 200 = - 3.06 ms-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 s