Question 185061
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Your problem statement left out an important detail:


"[J]unior's boat will go 15 miles per hour in still water.  If he can go 12 miles downstream in the same amount of time as it takes to go <b>???</b> miles upstream, then what is the speed of the current[?]"


However, all is not lost.


Remember that distance equals rate times time, that is:


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


Which, for this problem, will be convenient to express as:


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


Let the speed of the current be represented by *[tex \Large r_c] and your missing upstream distance be represented by <i>m</i>.


When the boat is going downstream, the boat is actually moving the speed through still water <b>plus</b> the speed of the current.  When going upstream, it is the speed in still water <b>minus</b> the speed of the current.


For the downstream trip, which we know to be 12 miles, we can write:


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


For the upstream trip, which we are saying is <i>m</i> miles, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ t = \frac{m}{15 - r_c}]


Now remembering that the problem said the amount of time for the downstream and upstream trips was the same, we can equate the two right sides of the above equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{12}{15 + r_c} = \frac{m}{15 - r_c}]


Giving us a proportion.  Step 1:  Cross-multiply:


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


Step 2: Simplify, collect terms, and solve for *[tex \Large r_c]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 180 - 12r_c = 15m + mr_c]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ - 12r_c - mr_c = -180 + 15m]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 12r_c + mr_c = 180 - 15m]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_c(12 + m) = 180 - 15m]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r_c = \frac{180 - 15m}{12 + m}]


Step 3: Once you determine the correct value for <i>m</i> you can substitute and do the arithmetic to determine *[tex \Large r_c].


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