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Mathematics 15 Online
OpenStudy (anonymous):

a tugboat goes 140 miles upstream in 10 hours the return trip downstream takes 5 hours. Find the speed of the tugboat without a current and the speed of current

OpenStudy (shane_b):

\[\large speed = \frac{distance}{time}\]Let s be the tuboat's speed and c be the water current: \[\large s_{downstream}+c=\frac{140mi}{5h}=28mi/h\]\[\large s_{upstream}-c=\frac{140mi}{10h}=14mi/h\]The difference in speeds is:\[\large 28mi/h-14mi/h=14 mi/h\]Therefore the speed of the current must be half of that (7mi/h) since it adds to your speed downstream and retards your speed upstream. If the speed of the current is 7 mi/h then the speed of the boat without a current would be:\[\large 28mi/h-7mi/h=21m/h\]

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