Question

distance covered by the body during the internal from 10 seconds to 20 seconds

Answer 1

I thought that question is not proper or incomplete

Answer 2

Answer:

The 5th second is from t=4s→t=5s.

Explanation:

At t=4s the velocity, after retardation:
v4=10ms−(2ms2⋅4s)=2m/s

At t=5s:
v5=10ms−(2ms2⋅5s)=0m/s

So the average speed in that 5th second:
¯v=2m/s−0m/s2=1m/s

And the distance will be ¯v⋅t=1ms⋅1s=1m