Question 201646
<font face="Garamond" size="+2">


Use distance equals rate times time, or *[tex \Large d = rt], or, for this problem, you would want to express time as a relationship between distance and rate, namely *[tex \Large t = \frac{d}{r}].  The speed of the current subtracts from the speed of the boat when going upstream and adds to the speed of the boat when going downstream, so the upstream trip can be described as:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{9}{(15 - r_c)}]


And the downstream trip


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{12}{(15 + r_c)}]


Since we are told it is the same amount of time, we can equate the two right-hand sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{9}{(15 - r_c)}\ =\ \frac{12}{(15 + r_c)}]


Cross-multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 9(15 + r_c) = 12(15 - r_c)]


Solve for *[tex \Large r_c]


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