Question

A body leaving a certain point “O” moves with a constant acceleration. At the end of the 5th second its velocity is 1.5m/s. At the end of the sixth second the body stops and then begins to move backwards. Find the distance traversed by the body before it stops. Determine the velocity with which the body returns to point “O”?(27m, -9 m/ s)

“ Pardon me for such a long question”

Answer 1

Answer:

Explanation:

At time t = 5s, u= 1.5m/s, at time t = 6s, v = 0

Using v = u + at

0 = 1.5 + a x1

Therefore, a = -1.5 m/s ^2

At time t= 0s, u= ?, at time time t=6s, v = 0 and a = -1.5m/s^2

Using v= u+at

0 = u - 1.5 x 6

u = 9m/s

Using formula,

S = {(u + v) x t}/2

S= (9 x6)/2 = 27 m

As the body has to travel 27 m to reach again at O, for same distance, velocity is also same but in negative directionas at time = 0

Therefore u' = -9 m/s