Question

A particle is pushed along a horizontal surface in such a way that it starts with velocity of 12m/s and its velocity decreases at te rate of 0.5 m/s

then find the time taken to come at rest

find the distance covered by it before coming to rest

Answer 1

Answer:

t = 24 sec

s = 144 m

Explanation:

u = 12 m/s

v = 0 m/s

a = -0.5 m/s²

v = u + at

0 = 12 - 0.5t

t = 12 / 0.5

t = 24 sec

v² = u² + 2as

0 = 144 - s

s = 144 m

Hope this helps! :)