Question

a hobo wishes to catch a freight train that stands in front of a train station. he waits 50 ft behind the train, whose last car is 50 ft from the station. as the train starts to leave the station with an acceleration of 2 ft/s^2, the hobo runs toward the train at a constant speed of 15 ft/s. will he catch the train before the last car reach the station or will he have to run past the station to catch the train?

Answer 1

Hobo try to decrease distance with speed $v=15ft\cdot s^{-1}$according $\Delta S_h=v\cdot t$. The train increase the distance as $\Delta S_{th}=\frac{a\cdot t^2}{2}$ , $a=2ft\cdot s^{-2}$ . The distance between hobo and last car of freight train is $S(t)=S_0+\Delta S_{tr}(t)-\Delta S_{h}(t)=S_0+\frac{a\cdot t^2}{2}-v\cdot t$ . $S_0=50 ft$ . If hobo will be lucky and catch the train $S(t)=0$ at a time moment t. Thus we should to solve equation:

$S_0+\frac{a\cdot t^2}{2}-v\cdot t=0$ This is quadratic equation, and we have

(1) $t_{1,2}=\frac{v\pm\sqrt{v^2-2aS_0}}{a}$ . We can see that hobo catch the train only if $\sqrt{v^2-2aS_0}>0;$

$v^2-2aS_0=15^2-2\cdot 2 \cdot 50=225-200=25ft^2s^{-2}>0$ Hobo may be lucky. Find (1)

$t_{1,2}=\frac{15\pm 5}{2}= \begin{cases} 10s &\text{if }+ in \ numerator\\ 5s &\text{if } - in \ numerator \end{cases}$

What do these two times mean? The train is speeding up and if hobo does not jump on the train at time $t_1=5s$ he will run ahead of the last car. But at time $t_2=10s$ last car will still catch up with him and this will be the last opportunity to sit in it.

Determine the distance between last car and the station at the time moment $t_1=5s$.

$\Delta S_{th}=\frac{a\cdot t_1^2}{2}=\frac{2\cdot 5^2}{2}=25ft$ Thus the last car not reach the station as initial distance was $50ft$.

Answer: Hobo catch the train before the last car reach the station.