Is this word problem solvable?

[madnak]

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d1 = 2 miles + (v-s)(1 hour) = 2 + v - s

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Where did the units go here? Canoeist travels upstream 2 miles and encounters log, an hour later he turns around. How does the distance before the log correlate to the time after the log? Unless we read "after *another* hour" to mean that the first 2 miles took an hour, I don't see how this is solvable.

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This isn't a process of reaching the answer, but I think it proves the answer and hopefully it will clarify the issue:

Edit - I meant to say here, we're using the first time the swimmer reaches the log as the starting point. Forget about the 2 miles the swimmer went before reaching the log, that's irrelevant. Just remember that the point at which the swimmer and log meet for the second time is 2 miles downstream from the point at which they meet for the first time, that's the relevant information being presented in the problem. What makes this problem confusing is that the starting point is presented as being 2 miles before the swimmer reaches the log.

The swimmer swims at a rate of n mph. The stream moves at a rate of x mph. During one hour of swimming upstream, the swimmer move n-x miles upstream. The log moves x miles downstream. In the next hour, the swimmer moves downstream n+x miles. The log moves downstream x miles. At the end of hour 2, the swimmer has a net downstream movement of n+x - (n-x) miles - his downstream movement in hour 2 minus his upstream movement in hour 1. That's 2x miles. The log has gone downstream x + x miles - the x miles downstream in hour 1 plus the x miles downstream in hour 2, also 2x miles. Therefore, after 2 hours the swimmer and log are at the same point, having traveled 2x miles downstream. We know that when the swimmer and log reach the same location, they are 2 miles downstream from the starting location. Thus, 2x=2 and x=1. The stream travels at 1mph.

[BruceZ]

It becomes obvious when you consider things from the point of view of the log. The log is fixed relative to the stream. If the canoe speed is Vc relative to the stream, then the log observes the canoe moving away from it at a speed Vc for 1 hour, and then coming back with a speed Vc for 1 hour. This is because an observer on land would first see the canoe moving away from the log with speed Vc - Vs relative to the land, where Vs is the velocity of the stream relative to the land, and since the log moves with speed Vs in the opposite direction, the canoe and log move apart with speed Vc - Vs + Vs = Vc . Then in the second hour, the observer on land would see the canoe moving toward the log with speed Vc + Vs relative to the land, and since the log moves with speed Vs in the same direction, the canoe and log move together with speed Vc + Vs - Vs = Vc. Since they moved apart at this same speed for 1 hour, it will take another 1 hour for them to move together. Since the log traveled 2 miles in these 2 hours, the speed of the current must be 1 mph.

[dazraf69]

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d1 = 2 miles + (v-s)(1 hour) = 2 + v - s

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Where did the units go here? Canoeist travels upstream 2 miles and encounters log, an hour later he turns around. How does the distance before the log correlate to the time after the log? Unless we read "after *another* hour" to mean that the first 2 miles took an hour, I don't see how this is solvable.

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The formula is the same as I came up with but the idea of the stream speed being relative was the missing link. Thanks for the in depth discussion/explanation. Probably would have been much easier if I knew no math at all. [img]/images/graemlins/grin.gif[/img]