Question

A particle is pushed along a horizontal surface in such a way that it starts with velocity of
12 m/s, it retards at a rate of .5 m/s2
? Find (i) the time it will take to come to rest & (ii)
distance covered by it before coming to rest.

Answer 1

Answer:

1)v=u+at     2)s=u+1/2at ^21/2

Explanation:

12=0+.5t  t=120\5    t=24s

s=1/2at^2  S=1\2*0.5*24^2  S=144M