Question

A particle is pushed along a horizontal surface in such a way that starts with a velocity of 12m/s. Its velocity decreases at 0.5m/s^2. A) find the time it will take to come to rest. B) find the distance covered by it before rest.​

Answer 1

A) Using v=U+at formula put v=0 and a=-0.5 there fore the becomes 24sec .

B)Also use s=ut+1/2at*2 formula we get 144m as answer

Answer 2

A) We know, v = u + at

So here; 0 = 12m/s - 0.5 ms^-2 (t)

or, - 0.5 ms^-2 (t) = - 12 m/s

or, t = (-12/0.5) s = 24 s [time taken]

B) Also, 2as = v^2 - u^2

or, - s = 0 - 144m

or, s = 144 m [distance covered]

HOPE THIS COULD HELP!!!