The problem is as follows:
A train is cruising with a constant speed of $10\frac{m}{s}$. In the exact moment when it enters a $60\,m$ tunnel a man standing in the midpoint of the train starts running to the tunnel at rate of $4\,frac{m}{s}$ with respect to the train. If the man and the train began to leave from the tunnel the same. What will be the length of the length of the train?.
The existing alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{200 m}\\ 2.&\textrm{280 m}\\ 3.&\textrm{400 m}\\ 4.&\textrm{48 m}\\ 5.&\textrm{108 m}\\ \end{array}$
I'm very confused at this problem, and not sure exactly how to solve it. Can someone help me here?. What I've attempted so far is to assume that the elapsed time for the man running to a point will be the same for that point in the train to leave.
Hence:
Let: $\textrm{x: distance covered by the man}$
and $\textrm{d: size of half of the portion of the train}$
$L=60\,m$ , length of the tunnel
Then:
This comes the tricky part which is how to work with reference frames and speed?
I'm assuming;
$v_{man/train}=v_{man}-v_train=4$
$v_{man}=4+10=14$
Then:
$\frac{x}{14}=\frac{d-x+60}{10}$
But this equation has two unknowns and it cannot be solved. So far I have ran out from ideas. Can someone help me?. Since this situation is confusing, it would help a lot to include a sketch or diagram to understand the situation, so I appreciate someone can include it so I could understand what's happening. Any help please?.
Assuming you have to find the length of the train (the header says otherwise).
say, length of the train is $2l$.
Time taken by the person to leave the other end of the tunnel $ = \dfrac {60 + l}{14}$ (as person starts from midpoint of the train)
Time taken by the front of the train to leave the tunnel $ = \dfrac {60}{10}$
Equate the two to find $l$. The answer is $2l$.