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
