Fugitive tries to hop on a freight train traveling at a constant speed of 4.8 m/s . just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 m/s2 to his maximum speed of 6.2 m/s , which he then maintai

The figitive reaches the train

Explanation:

Let us separate in problem in two parts a first where the fugitive has acceleration and see if it reaches the train and another part where the fugitive has constant speed and the train will arrive.

Fugitive with acceleration, let's calculate the time and distance it travels, using the kinematic equations with constant acceleration.

V = Vo +  a t

The fugitive part of the rest Vo = 0, we look for the time to reach the maximum speed and the distance

t= V/a = 6.2/1.4= 4.43 s

Xf = Vo t + ½ a t²  

Xf=  ½ 1.4 4.432 = 13.7 m

Let's see how much the train travels at this time

XT = VT t = 4.8 4.43 = 21.3 m

XT > Xf    so the train does not reach this route.

For the second part two bodies with constant speed, starting from different points

Train       XoT= 21.3 m

Fugitivo  Xof = 13.7 m

At the meeting point the position of the two bodies is the same

Train      Xf = XoT + VT t

Fugitive Xf= Xof + Vf t  

         XoT + VT t = Xof + Vf t      

Grouping terminus

         t = (XoT – Xof) / (Vf – VT)

          t = (21.3 -13.7)/(6.2-4.8) = 5.43 s

          XT = XoT  + VT t

          XT = 21.3 +4.8 5.43

          XT = 47.4 m

At this point the fugitive reaches the train

Hello I have reviewed the problem, the solution is correct

The problem is divided into two parts in the first we see if the fugitive reaches the train, as it does not reach it. We made a second part at constant speed starting the fugitive from different points where it was at the end of the acceleration stage. We find the time for when they reach the meeting point and with this time we look for the total distance

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