A particle is pushed along a horizontal surface in such a way that it starts with a velocity 12m/s. Its velocity decreases at a rate of 0.5m/s2. (a) Find the time it will take to come to (b) Find the distance covered by it before coming to rest.
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by v = u + at
0 = 12 -0.5×t (-ve sign as it is retardation)
t=24sec.
b). s = ut + 1/2 at^2
s= 12×24 – 6
s = 282m.
0 = 12 -0.5×t (-ve sign as it is retardation)
t=24sec.
b). s = ut + 1/2 at^2
s= 12×24 – 6
s = 282m.
Sunandit:
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