Is this word problem solvable?
 

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A canoeist is traveling upstream. 2 miles later he comes across a log going the opposite way.He travels for another 1 hour then turns around. He gets to his starting point the same time the log does. What is the speed of the stream?

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Having trouble coming up with this solution. Any help would be appreciated.

Re: Is this word problem solvable?
 

if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

Re: Is this word problem solvable?
 

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if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

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Seems correct (speed is relative, etc).

Re: Is this word problem solvable?
 

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if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

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Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream?

Re: Is this word problem solvable?
 

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[ QUOTE ]
if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

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Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream?

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Speed is relative.

Re: Is this word problem solvable?
 

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if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

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Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream?

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you're in a lifeboat on the ocean. you know you can swim exactly 2 hours before you drown. you always swim at a constant speed.

you've been swimming away from your boat for 1 hour. the water you and your lifeboat are in are traveling at 800 miles/hour. to your horror, you realize that you have been swimming with the current, which means you are over 800 miles from where you were when you left your lifeboat.

realizing you have no chance of getting back, you strangle yourself with the drawstring of your bathing suit. two months later a fisherman finds your watch in the stomach of a great white.

Re: Is this word problem solvable?
 

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if he spends an hour rowing away from the log, it will take him another hour to get back to it. the log travels 2 miles in 2 hours.

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Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream?

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You can think of the stream standing still and the land moving all around it.

PairTheBoard

Re: Is this word problem solvable?
 

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A canoeist is traveling upstream. 2 miles later he comes across a log going the opposite way.He travels for another 1 hour then turns around. He gets to his starting point the same time the log does. What is the speed of the stream?

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Having trouble coming up with this solution. Any help would be appreciated.

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The speed of the river is one mile per hour. The speed of the canoeist is not important (and cannot be determined). If anyone's interested I'll explain.

Re: Is this word problem solvable?
 

Algebra for those who can't follow the shortcut:

v = speed that canoeist paddles
s = speed of stream = speed of log
t = time canoeist spent traveling downstream

v-s = speed of canoeist going upstream
v+s = speed of canoeist going downstream

d1 = total distance traveled by canoeist upstream

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

d2= total distance traveled by canoeist downstream

d2 = (v+s)t = vt + st

We also know that the log traveled 2 miles in (t+1) hours

2 miles = s(t+1)
st + s = 2

also, d2 = d1

2 + v - s = vt + st
2 + v = vt + (st + s)
2 + v = vt + 2 (from above, st + s = 2)
v = vt
t = 1 hour

st + s = 2
s + s = 2
2s = 2
s = 1 mph

Re: Is this word problem solvable?
 

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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.

Re: Is this word problem solvable?
 

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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.

Re: Is this word problem solvable?
 

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.

Re: Is this word problem solvable?
 

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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]