the brakes applied to a car produce a acceleration of 6ms in the opposite direction to the motion if the car takes 2s to Stop after the application of brakes calculate the distance it travel during this time
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i think that it is 12 m
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stopping distance, s=u²/2a
to find u we will use. first =n of motion ; v=u-at
as v=o
then u=at =6×2=12 m/s
then stopping distance, s=12²/2×2=144/4=36m
to find u we will use. first =n of motion ; v=u-at
as v=o
then u=at =6×2=12 m/s
then stopping distance, s=12²/2×2=144/4=36m
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