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 = -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
O = 9 – 2 t
We get, t = 4.5s
Now let us find the distance covered in 1/2 second i.e. from 4.5 to 5second
Let u = 0 at t = 4.5 sec.
Then distance covered in 1/2 s
`s = 1/2 at^2`
` s = 1/2 xx 2 xx [1/2]^2`
` = 1/4m`
Total distance covered in fifth second of its motion is given by
`S_0 = 2s = 2 (1//4) = 1//2 m`.