A particle moving with constant acceleration of 2m/`s^2` due west has an initial velocity of 9 m/s due east. Find the distance covered in the fifth second of its motion.

Explanation

Initial velocity u = +9 m/s
Acceleration a = -2 m/`s^2`

In this problem, acceleration’s direction is opposite to the velocity’s direction.
Let “ t” be the time taken by the particle to, reach a point where it makes a turn along the straight line.
We have. v u+at
0=9 - 2t
We get, t = 4.5s
Now let us find the distance covered in 1/2 second i.e. from 4.5 to 5 second
Let u = 0 at t = 4.5 sec.
Then distance covered in 1/2s.
`s=1/2at^2`
`s=1/2xx2xx[1/2]^2`
`=1//4m`
Total distance covered in fifth second of its motion is given by
`S_0=2S=2(1//4)=1//2m`