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*t8 = 0 + 3.6 * tt=2.222s to reach 8.0 m/sAt that time the train has moved 4.5 m/s * 2.222s = 9.999 mHe travelled (another kinematics equation)Vf^2 = Vi^2 + (2*a*d)(8.0)^2 = (0)^2 + (2 * 3.6 * d)d=8.888 mThe train is 9.999m, the fugitive is 8.888m,He still needs to travel9.999-8.888= 1.111mHe 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 trainSo if it takes 2.222 seconds to approach the train at 8.888m, it should take2.222 + .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