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

Explanation

Initial velocity u = +9m/s , Acceleration a = `-2m//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 = 4.5 s
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)=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`.