A body moves over a horizontal surface with an initial velocity of 2 m/s. Due to friction, its velocity decreases uniformly at the rate of 0.25 m/s2. How much time will it take to stop?
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given here is
u=2m/s
v=0m/s
a=-0.25m/s^2
t=?
here the equation used by givendata is,
v=u+at
2=0+(-0.25t)
-0.25t=2
t=2÷(-0.25)or-(1/4)
t=2÷(-1/4)
t=2×4/-1
t=8seconds
so it will take 8 seconds to stop the body......
u=2m/s
v=0m/s
a=-0.25m/s^2
t=?
here the equation used by givendata is,
v=u+at
2=0+(-0.25t)
-0.25t=2
t=2÷(-0.25)or-(1/4)
t=2÷(-1/4)
t=2×4/-1
t=8seconds
so it will take 8 seconds to stop the body......
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