A body leaving a certain point “O” moves with a constant acceleration. At the end of the 5^th second its velocity is 1.5 m/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“?

Given, velocity after 5s = 1.5 m/s

Velocity after 6s = 0 m/s

Let the acceleration be ‘a’ and initial velocity be ‘u’.

For t = 5s,

1.5 = u + 5a …(1) (using v = u + at)

For t = 6s

0 = u + 6a …(2) (using v = u + at)

Subtracting equation (2) from (1), we get

1.5 = -a

Or, a = -1.5 m/s²

Putting value of a in equation (1), we get

u = 9 m/s

So, Distance traversed before stopping = Distance covered in 6s

Body will again move with same acceleration in opposite direction,

So here initial velocity = 0, acceleration a = -1.5 m/s² and

displacement s = -27m (direction is opposite)

putting all these values in we get

-27 =

Velocity at point “O” = velocity after 6s

= 0 + (-1.5)x6 (using v = u + at)

= -9 m/s