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anonymous
 5 years ago
while on vacation Kevin went for a swim. Swimming against the current took 8 min. to go 200 meters coming back took half as long. Find Kevin's average swimming speed and the water current.
anonymous
 5 years ago
while on vacation Kevin went for a swim. Swimming against the current took 8 min. to go 200 meters coming back took half as long. Find Kevin's average swimming speed and the water current.

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radar
 5 years ago
Best ResponseYou've already chosen the best response.0Let x = current speed in meters/sec Let y = Kevin's speed in same units then we can use the s=rt formula where s=distance, r = speed, and t = time Then kevins speed swimming against current is yx kevins speed swimming with the current is y+x

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The distance was the same both ways 200 meters. Here are two equations for each trip 200 = 8(yx) (8 minutes when swimming against current ) eq1 200 = 4(y+x) (4 minutes half the time when swimming with the current) eq2 200=8y8x eq 1 200=4y+4x eq 2 400=8y+8x eq 2 multiplied by 2 200=8y8x eq 1 600=16y adding y=37.5 meters/min substituting in eq 2 200=4(37.5) + 4x becomes 200=150+4x becomes 4x=50 or x =12.5 meter/min Change the units in my first post to meters/min as minutes were the units given in the problem.
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