A body leaving a certain point "O" moves with a constant acceleration. At the end of the fifth 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". (27m, -9m/s)

Explanation

Velocity in `5^(th)` sec = 1.5m/sec , The body comes to rest in `6^(th)` sec.
`therefore` Final velocity in `6^(th)` sec, v = 0
`therefore` Acceleration in `6^(th)` sec is `v=u+atrArr0=1.5+a*1rArra=-1.5m//sec^(2)`
[The velocity in `5^(th)` sec becomes the initial velocity for `6^(th)` sec]
After 6 sec, the body comes to rest.
`thereforev=0,a=-1.5m//sec^(2),u=?t=6sec.`
`v=u+atrArr0=u-1.5xx6rArru=9m//sec.`
`therefore` Distance traversed by the body in 6 sec. i.e., before it stops.
`s=ut+1/2at^(2)=9xx6+1/2xx-1.5xx6^(2)=54-27=27m`.
For backward journey,
`u=0m//sec,t=6sec,a=-1.5m//sec^(2)`
`v=u+atrArrv=0-1.5xx6rArrv=-9`
`therefore` Velocity for backward journey is -9m/sec.