Question 193839
<font face="Garamond" size="+2">


Considering the term 'kid' actually refers to the offspring of a goat, this is a rather interesting problem indeed.  However, regardless of the actual organisms that are operating the bicycles, the answer is the same.


Begin with:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = rt]


To describe the first one, you would say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d = 15t]


Now the distance traveled by the second bicyclist is 25 minus the distance traveled by the first, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 25 - d = 12t]


Since we are ultimately concerned with the time that they meet, the time they traveled has to be the same for each of them.


So solve each of the above equations for *[tex \LARGE t]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{d}{15}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{25 - d}{12}] 


Since *[tex \LARGE t = t], we can set these two fractions equal to each other:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d}{15} = \frac{25 - d}{12}]


Then cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12d = 15(25 - d) \ \ \Rightarrow\ \ 12d = 325 - 15d \ \ \Rightarrow\ \ 27d = 325 \ \ \Rightarrow\ \ d = \frac{125}{9}]


Which is the distance traveled by the first bicyclist.


Since the first bicyclist's speed was 15 km/hr, the elapsed time must be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{125}{9}\times\frac{1}{15}=\frac{25}{27}] hours.


Converting that to minutes and seconds to be added to the 9:00 AM start time is left as an exercise for the student.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>