1) Brakes applied to a car produce an acceleration of 5m/s2 in the opposite direction to the motion. If car takes 1.5s to stop after applying brakes,calculate the distance travelled by it.
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The acceleration is negative means it is retardation
Thus
a(acceleration) =-5m/s2
T(time)=1.5s
So v(final velocity)= u(initial velocity) +at
0=u-7.5
u. = 7.5m/s
now s=ut+1/2at^2
s =(7.5)(1.5) -1/2(5)(1.5*1.5)
solving this will give
s = 5.625 m approx.
hope it will help you .$
Thus
a(acceleration) =-5m/s2
T(time)=1.5s
So v(final velocity)= u(initial velocity) +at
0=u-7.5
u. = 7.5m/s
now s=ut+1/2at^2
s =(7.5)(1.5) -1/2(5)(1.5*1.5)
solving this will give
s = 5.625 m approx.
hope it will help you .$
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