Answer:
The figitive reaches the train
Step-by-step 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