A fugitive tries to hop a freight train traveling at a constant speed Of 4.5 m/s. Just as a empty box car passes him, the fugitive starts from rest and accelerates at a = 3.6 m/s squared to his maximum speed of 8.0 m/s. How long does it take him to catch up to the empty box car? What is the distance traveled to reach the box car?

We use the kinematics equation:

Vf = Vi + a*t

8 = 0 + 3.6 * t

t=2.222s to reach 8.0 m/s

At that time the train has moved

4.5 m/s * 2.222s = 9.999 m

He travelled (another kinematics equation)

Vf^2 = Vi^2 + (2*a*d)

(8.0) ^2 = (0) ^2 + (2 * 3.6 * d)

d=8.888 m

The train is 9.999m, the fugitive is 8.888m,

He still needs to travel

9.999-8.888 = 1.111m

He needs to cover the rest of the distance in a smaller amount of time, however hes at his maximum velocity, so ...

8m/s (man) - 4.5m/s (train) = 3.5 m/s more

(1.111m) / (3.5m/s) =.317seconds more to reach the train

So if it takes 2.222 seconds to approach the train at 8.888m, it should take

2.222 +.317 = 2.529 seconds to reach the train completely

Last but not least is to figure out the total distance traveled in that time frame:

(Trains velocity) * (total time)

(4.5m/s) * (2.529s) = 11.3805m