A particle moving with constant acceleration of 2m/`s^(2)` due west has an initial velocity of 9 m/s due east. Find the distace 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
O = 9 - 2 t
We get , t = 4.5s
Now let us find the distace convered in `(1)/(2)` second i.e. from 4.5 to 5 second Let u at t = 4.5 sec.
Then distance convered in `(1)/(2) ` s.
` s= (1)/(2) at^(2)`
s `= (1)/(2) xx 2 xx [ (1)/(2)]^(2)`
`= (1)/(4)` m
Total distance covered in fifth second of its motion is given by
`S_(0) = 2s = 2 ((1)/(4)) = (1)/(2) `m .