Question

a car is moving wiht a velocity of 40km/h can be stopped by applying brakes atleast after 2m.If the sam car is mobing with the speed of 80km/h what is the minimum stopping distance

Answer 1

v2  = u2 + 2as

a = (40×40)/ 2 ×2 /1000= 400/1000 = 0.4 km/h^2

V2 = u2 + 2as

Acceleration remains constant

Therefore

S = 80^2/ 2a = 80^2 / 0.8 = 8 m.........answer


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Answer 2

v2 =u2 +2as
a=(40×40)/2x2/1000= 400/1000
=80^2/2a=80^2/0.8= 8m ans.