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.

Question

anonymous · Answer

help

radar · Answer

Let 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 y-x
kevins speed swimming with the current is y+x

radar · Answer

The distance was the same both ways 200 meters.
Here are two equations for each trip
200 = 8(y-x)    (8 minutes when swimming against current ) eq1
200 = 4(y+x)    (4 minutes half the time when swimming with the current) eq2

200=8y-8x  eq 1
200=4y+4x  eq 2
400=8y+8x  eq 2 multiplied by 2
200=8y-8x   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.