f the worker stops pushing at the end of 5.20 s, how far does the block move in the next 4.40 s?

Question

A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force is negligible. The block starts from rest and moves a distance 13.0 m in a time 5.20 s.

Answer 1

Let the block speed at the end of 5.20s to be V, we have:
0.5*v*t = 0.5*a*t^2 = S
So: V = 2S /t = 2*13.0m/5.20s = 5.00 m/s
The block would move in the next 4.40 s: 5.00m/s * 4.40s = 22m

Answer 2

11 meters