Afugitive 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.

The fugitive fails to get on the train.

Iknow the first part it is the average star i am not 100% sure