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.

The answer is b hope this