Question 202390
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Use distance equals rate times time, *[tex \Large d = rt]


Let *[tex \Large r] represent the speed of the boat in still water, and let *[tex \Large r_c] represent the speed of the current.  On the downstream trip, the boat is going with the current, so the total speed is *[tex \Large r + r_c].  On the upstream trip, the total speed is *[tex \Large r - r_c]


So, the downstream trip can be described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 168 = (r + r_c)7]


which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r + r_c = 24] (Eq. 1)


And the upstream trip can be described by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 168 = (r - r_c)42]


which simplifies to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r - r_c = 4] (Eq. 2)


Add Equation 1 to Equation 2:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2r + 0r_c = 28 \ \ \Rightarrow\ \ r = 14]


Hence the speed of the boat in still water is 14, and:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 14 + r_c = 24 \ \ \Rightarrow\ \ r_c = 10]


So the speed of the current is 10.


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