SOLUTION: Every 15 minutes, buses leave from a certain point in both directions. A traveler who is on foot and leaves at the same time as one of the buses will meet a bus driving towards h

In this problem, it is assumed that all buses move with the same rate in each direction.


Let "d" be the distance the bus covers in 15 minutes.

    I use the term "the bus", meaning any (or every) of the buses.


Let's consider everything on the coordinate line, and let the bus station is located at x= 0 (point O), 
at the origin of this coordinate line. Let A be the point at positive distance "d" from the point O.


At the time moment t= 0, the traveler started from O to A; at the same time moment the bus X, 
    which moves from A towards O, started from A to O.

The bus spends 15 minutes to get from A to O; and the bus meets the traveler at t = 12.5 minutes.

So, the bus covered   =  of the distance d, while the traveler covered only  of the distance d.



Hence, a) the speed of the bus is 5 times the speed of the traveler, and

       b) at the moment they meet each other (t= 12.5 minutes), the traveler is at the distance  from O.



Now, consider the traveler who continues moving from O to A (and, possible, farther), and the bus Y, which started from O at the time t= 15 minutes.

When the bus Y started from O at t= 15 minutes, the traveler position is   multiplied by the time ratio , 

    i.e.  =  from A.


    +-----------------------------------------------------------------------+
    |  Let u be the speed of the traveler; then 5u is the speed of any bus. |
    +-----------------------------------------------------------------------+


For these two objects, we have standard catching up problem.


So, we write equation typical for catching up problem (here "t" is the time moment of catching up, 
                                                                counted from t= 15 minutes).

    5u*t = u*t + .

In this consideration,    is the head start for the traveler.

Left side is the distance, covered by bus Y; right side is the distance, covered by the traveler to the catching up moment.


Simplify the equation

    5ut - ut = 

    4ut      = 

      t      = 


Transform the right side this way   = .

Now notice that 5u is the bus speed;  therefore the ratio   is 15 minutes (!).


Therefore,  t =  =  minutes = 3.75 minutes.


Resume:  The traveler starts from O at t= 0;  he meets the opposite bus X at t= 12.5 minutes; 

         then the bus Y starts from O at t= 15 minutes, and 3.75 minutes later the bus Y catches up the traveler.


ANSWER.  The next bus from the bus station will catch up the traveler at t = 15 + 3.75 = 18.75 minutes.

         counting the time from the moment the traveler left the bus station.