http://everything2.com/?node=New%20Writeups%20Atom%20Feed&foruser=mdo 2012-06-14T21:23:55Z http://everything2.com/user/mdo/writeups/Canoe+solutionmdohttp://everything2.com/user/mdo2012-06-14T21:23:55Z2012-06-14T21:23:55Z <p>" 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.</p> <p><strong>Here is the solution to the canoe problem:</strong> <br>==========================================</p> <p>A---------B----------C----------------------D</p> <p>A - start/end point <br>B - point where the log is at after the canoe wraps around the point D <br>C - point where the canoe meets the log <br>D - wrap around point of canoe</p> <p>AC = 1 mile <br>CD = 1 hour <br>V - ground <a href="/title/speed">speed</a> of the canoe <br>V1 - river stream <a href="/title/speed">speed</a> <br>(V+V1) - canoe speed upstream <br>(V-V1) - canoe speed downstream</p> <p>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: <br> AB/V1 = DA / (V+V1) <br> DA = CD + AC <br> AC = 1 miles <br> CD = 1hour * (V-V1)= V-V1 <br> AB =&hellip;