A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m/s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 3.8 m/s2 to his maximum speed of 8.0 m/s.

Let d = distance that the fugitive travels to get on the train.

Let t = the time to travel the distance d.

The fugitive starts from rest accelerates at a = 3.8 m/s².

Therefore

(1/2) * (3.8 m/s²) * (t s) ² = (d m)

1.9 t² = d (1)

The train travels at constant speed 5.0 m/s.

Therefore

(5.0 m/s) * (t s) = d

5t = d (2)

If the fugitive successfully boards the train, then equate (1) and (2).

1.9t² = 5t

t = 0 or t = 2.6316 s

Ignore t = 0, so t = 2.6316 s.

The speed of the fugitive after 2.6316 s, is

v = (3.8 m/s²) * (2.6316 s) = 10 s

This speed exceeds the maximum speed of the fugitive, therefore the fugitive fails to get on the train.

Answer: The fugitive fails to get on the train.