the brakes applied to appa produce an acceleration of 6 metre per second square in the opposite direction to the motion if the car takes 2 seconds to stop after application of break calculate the distance it travel during the time
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v= final velocity of car = 0m/s
u=initial velocity = ?
s= distance travelled after applying breaks
t= time period of break=2s
acceleration = 6m/s^2
by using kinematic equation,
v=u+at
0m/s= u -6m/s^2×2s
here acceleration is retardation so taken as negative
so,
u=12m/s
We know that
s=ut+1/2(a)t^2
s=12×2 - 1/2×6×(2)^2
s=12m
u=initial velocity = ?
s= distance travelled after applying breaks
t= time period of break=2s
acceleration = 6m/s^2
by using kinematic equation,
v=u+at
0m/s= u -6m/s^2×2s
here acceleration is retardation so taken as negative
so,
u=12m/s
We know that
s=ut+1/2(a)t^2
s=12×2 - 1/2×6×(2)^2
s=12m
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