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Topic: Canoing the River (Read 582 times) |
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smull
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A woman starts paddling upstream in a canoe, and after one mile, encounters a log floating with the current. She continues to paddle upstream for onehour, then turns around and paddles downstream, until she returns to the dock where she started. If the woman and the log reach the dock at exactly the same time, how fast was the current flowing? Assume all speeds are constant.
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ThudnBlunder
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Re: Canoing the River
« Reply #1 on: Jul 1st, 2004, 10:03am » |
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1/2 mph
Got any Easy ones?
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smull
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THUDandBLUNDER,
could you be so kind as to elaborate. How did you arrive at the answer.
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ThudnBlunder
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Re: Canoing the River
« Reply #3 on: Jul 5th, 2004, 3:22pm » |
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Let all times be in hours, all distances in miles, and all speeds in mph.
Let dock be at D
Let canoe meet log at L
Let canoe turn around at T
Let speed of the canoe in still water be c
Let speed of the stream be s
Then
DL = 1
Speed of canoe when going upstream = c-s
Time for canoe to travel from L to T = 1
Hence LT = c-s
Speed of canoe when going downstream = c+s
Hence time for canoe to travel from T to D = 1+c-s / c+s
And total time from canoe meeting log to docking = 1 + (1+c-s / c+s)
This equals time for log to travel from L to D = 1/s
Putting 1 + (1+c-s / c+s) = 1/s we find that the c's cancel to give s = 1/2
(I suspect that this is your child's homework. Am I right?) 
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| « Last Edit: Jul 5th, 2004, 3:38pm by ThudnBlunder » |
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Grimbal
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Re: Canoing the River
« Reply #4 on: Jul 5th, 2004, 3:36pm » |
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The trick is that if the woman has been paddling one hour away from the log, she must have paddled the same time to get back to the log. If it happens at the very time when the woman (and the log) reach the dock, that means the log took 2 hours to move 1 mile.
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rmsgrey
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Re: Canoing the River
« Reply #5 on: Jul 6th, 2004, 5:40am » |
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A minor quibble: If all speeds are constant, then the woman travels at the same speed in both directions?
And to amplify Grimbal's trick: treat the log as fixed and the current as 0, but let the dock drift upstream at the speed the current would have been going downstream...
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