The breaks applied to a car produce a negative acceleration of 6m/s 2 the car takes 2 seconds to stop after applying brakes, calculate the distance it travels during this time.
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Retardation = 6 m/s^2
t = 2 sec
v = 0
Since retardation,
(u-v)/t = 6
u/2 = 6
u = 12 m/s^2
We know,
S= ut - (1/2)at^2
= 12*2 - (1/2)*6*2*2
= 24-12
= 12 m (Answer)
t = 2 sec
v = 0
Since retardation,
(u-v)/t = 6
u/2 = 6
u = 12 m/s^2
We know,
S= ut - (1/2)at^2
= 12*2 - (1/2)*6*2*2
= 24-12
= 12 m (Answer)
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