An object is pushed along a horizontal surface in such a way that it starts with a velocity of
12 m/s. Its velocity decreases at a rate of 0.5 m/s^2. Find:
a. The time it will take to come to rest.
b. Distance covered by it before coming to rest.
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Answered by
1
Answer:
Given
v=12m/s
acceleration =0.5m/s^2
u=0m/s
time= v-u/a
time=12-0/0.5=6seconds
diatance= ut 1/2 at^2
distance =0×6×1/2 ×0.5×6^2
distance =0.5×4.5=2.25m
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Answered by
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a. from 1st equation of motion,
v= u + at (v= 0, u=12m/s, a= -0.5m/s2, t= ?)
0=12+(-0.5* t)
t= 12/0.5
t= 24s
b. from 3rd equation of motion
2as= v2-u2
2*(-0.5)s=0^2 - 12^2
s=24/1
s= 24m
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