SOLUTION: John paddled for 4hrs with a 4 km/hr current to reach a campsite. The return trip against the same current took 10hrs. Find the speed of John's canoe in still water.
I tried d=(
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I tried d=(
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Question 414612: John paddled for 4hrs with a 4 km/hr current to reach a campsite. The return trip against the same current took 10hrs. Find the speed of John's canoe in still water.
I tried d=(r-4)4 and d=(r+4)10
10r+40 = 4r-16
6r+40 = -16
6r = -56
r=-9.333
Doesn't seem right? a - in still water? Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! John paddled for 4hrs with a 4 km/hr current to reach a campsite. The return trip against the same current took 10hrs. Find the speed of John's canoe in still water.
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With Current DATA:
time = 4 hrs ; rate = c+4 km/h ; distance = t*r = 4(c+4) = 4c+16 km
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Against Current DATA:
time = 10 hrs ; rate = c-4 km/h ; distance = t*r = 10(c-4) = 10c-40 km
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Equation:
distance with = distance against
4c+16 = 10c-40
6c = 56
c = 9 1/3 km/h (speed of the canoe in still water)
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Cheers,
Stan H.
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