Afugitive tries to hop on a freight train traveling at a constant speed of 4.5 m/s. just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 3.6 m/s2 to his maximum speed of 8.0 m/s.

Let’s see if he catches before he reaches 8m/sVelocity = acceleration * time, V = at, So, 8 m/s = 3.6 m/s * t, so t = 2.22 sec in that time he moved Distance = Initial Velocity * time + 1/2 acceleration * time * time  S = 0*2.22 + 1/2*3.6*2.22*2.22 = 8.9 meters in 2.22 seconds, the boxcar moved S = V*t = 4.5 * 2.22= 9.99 meters, so he still needs to catch up 9.99 m - 8.9 m = 1.09 meters. Since he’s moving at 8 m/s and the train also is moving at 4.5 m/s, his closing speed is (8 - 4.5) = 3.5 m/s 
It will take him 1.09m/ 3.5m /s = .31 seconds more to catch up. 
So, answer for How long does it take him to catch up to the empty box car is 2.22 sec + .31 sec = 2.53 seconds. answer for What is the distance traveled to reach the box car, we use the speed of the train since it was constant, Distance = Velocity * time = 4.5 m/s * 2.53 s = 11.4 meters.

Ithink they do have the same properties just in less quantities i think

Scientific classification