Question

An object is moving towards east with velocity 9m/s acceleration is towards west 2m/s^2 find the distance covered in 5th sec?

Answer 1

We need to calculate the distance covered in the 5th second.
Note that we have to calculate distance and not displacement.

I will consider positive sign towards East and negative sign towards west.

Note that the velocity will become 0 at t=4.5 seconds

distance covered in 4 seconds,
s=ut+0.5a$t^{2}$
s=36-16
s=20m

distance covered in 4.5 seconds,
s=40.5-20.25=20.25m

distance covered in 5 seconds,
S=ut+0.5a$t^{2}$
S=45-25=20m

Now, note how the object covered 0.25m from 4 seconds to 4.5 seconds and covered .25m once again from 4.5 seconds to 5 seconds.

Hence total distance covered here is 0.5 metres