Question

a car is moving along a straight highway with speed of 126 km/h is brought to a stop within a distancen of 200 m.What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?​

Answer 1

Initial velocity of the car (u) = 126km/hr = 35m/s

Distance travelled before coming to rest (s)=200m

Final velocity of the car (v) =0 (as it comes to rest)

Using the 3rd equation of motion,

v^2 - u^2 = 2as

a = (v^2 - u^2)/2s

=(0 - 35^2)/2×200

= –1225/400

= –3.06m/s^2

Let time taken to come to rest is t sec

Using the 1st equation of motion,

v = u + at

0 = 35 – 3.06×t

t =35/3.06

=11.43 sec

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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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