Your average speed was
(100 m) / (13.8 s) = 7.25 m/s .
If you finished 0.001s ahead of him, then at your average speed, that corresponds to
(7.25 m/s) x (0.001 s) = 0.00725 m
That's 7.25 millimeters ... about 0.28 of an inch !
NOTE:. I think this is only valid if your speed was a constant ~7.25 m/s all the way.
I would say B but I have no clue
Answer:
0.68 s
Explanation:
We are given that
Initial velocity of box=
Final velocity of box=v=11.5 m/s
Distance=d=8.5 m
We have to find the time taken by box to slow by this amount.
We know that

Substitute the values




We know that
Acceleration=
Substitute the values



Hence, the time taken by box to slow by this amount=0.68 s