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
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Answered by
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
Thank u★★★
#ckc
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
Thank u★★★
#ckc
Answered by
3
v2 =u2 +2as
a=(40×40)/2x2/1000= 400/1000
=80^2/2a=80^2/0.8= 8m ans.
a=(40×40)/2x2/1000= 400/1000
=80^2/2a=80^2/0.8= 8m ans.
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