Afugitive 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

M= 700 kgvalue of g( gravity) is constant = 9.8 m/s² or 10 m/s² w = mg = 700*9.8 = 6860n or 6.680*10³ n