SOLUTION: During a kayaking trip, a kayaker travels 12 miles upstream (against the current) and 12 miles downstream (with the current). The speed of the current remained constant during the

Question 387018: During a kayaking trip, a kayaker travels 12 miles upstream (against the current) and 12 miles downstream (with the current). The speed of the current remained constant during the trip. It took 3 hours going upstream and 2 hours going downstream. Find the average speed of the kayak in still water and the speed of the current.
I have tried solving the upstream and the downsteam, separately, using D=RT that I found in a tutor's solution to a similar question. I keep getting the current speed is 0. Can you help me?
Answer by CharlesG2(834) (Show Source): You can put this solution on YOUR website!
"During a kayaking trip, a kayaker travels 12 miles upstream (against the current) and 12 miles downstream (with the current). The speed of the current remained constant during the trip. It took 3 hours going upstream and 2 hours going downstream. Find the average speed of the kayak in still water and the speed of the current.
I have tried solving the upstream and the downsteam, separately, using D=RT that I found in a tutor's solution to a similar question. I keep getting the current speed is 0. Can you help me?"

Distance = Rate * Time
upstream current will slow speed, downstream current will increase speed
let KU = kayak rate upstream, KD = kayak rate downstream
upstream: 12 miles = KU * 3 hours
12/3 = 4 mph = KU
downstream: 12 miles = KD * 2 hours
12/2 = 6 mph = KD
rate upstream is 4 mph, rate downstream is 6 mph
so rate in still water must be 5 mph since subtracting or adding same number from the rate in still water
then current speed is 1 mph
upstream: 5 mph - 1 mph = 4 mph
downstream 5 mph + 1 mph = 6 mph

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