In a certain time interval the velocity reduces from 72 km per hour to 36 kilometre per hour of car if the car covers a distance of 150 metre during this time interval find the retardation of the car

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Solution

Assuming the acceleration and retardation remain constant in both case.

Case I:

Initial velocity u=36 km/h=361000/3600=10 m/s.

Final velocity v=72 km/h=721000/3600=20 m/s.

v=u+at

a=(v-u)/t=(20–10)/5=2 m/s^2

So acceleration = 2 m/s^2

Case II:

Initial velocity u =72 km/h= 20 m/s

Final velocity v= 0 m/s

a=(v-u)/t=(0–20)/20=-1 m/s^2

acceleration= -1 m/s^2

retardation/deceleration= +1 m/s^2

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