" Clearly she will reach the log again at 2:00." - this is absolutely incorrect because the canoe speed is different in case of upstream and downstream travelling.
Here is the solution to the canoe problem:
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A---------B----------C----------------------D
A - start/end point
B - point where the log is at after the canoe wraps around the point D
C - point where the canoe meets the log
D - wrap around point of canoe
AC = 1 mile
CD = 1 hour
V - ground speed of the canoe
V1 - river stream speed
(V+V1) - canoe speed upstream
(V-V1) - canoe speed downstream
From the above drawing, it's obvious that the time it takes from log to get travel from B to A is the same as the time it takes canoe to travel from point D to A. So, all we need to get the answer is solve this equation:
AB/V1 = DA / (V+V1)
DA = CD + AC
AC = 1 miles
CD = 1hour * (V-V1)= V-V1
AB = 1 - V1
So, DA=1+V-V1.
After substituting the above values and solving the equation, we get the log speed as 0.5 miles/hour.
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