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A car is moving with a velocity of 45km/hr. The driver applies brake and the car comes to rest in 3 second. What is the retardation and how far does it move before coming to rest?

A car is moving with a velocity of 45km/hr. The driver applies brake and the car comes to rest in 3 second. What is the retardation and how far does it move before coming to rest?

Solution:

Initial velocity of car (u) = 45 km/hr = 45 * 1000/3600 = 12.5 m/s

Time taken (t) = 3 s

Since the car stops finally, final velocity of car (v) = 0 m/s

Now,

Retardation (-a) = (u - v)/t

= (12.5 - 0)/3

= 4.17 m/s2

Hence, the magnitude of retardation of the car is 4.17 m/s2.

Let the car move a distance 's' before coming into rest than from equations of motion, we have,

s = ut - 1/2 (a)t2 [acceleration is negative due to retardation]

= (12.5)3 - 1/2 (4.17) (3)2

= 18.375 m

Hence, the car travelled 18.375 m before it came into rest.

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