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#1
10-16-2007, 12:32 AM
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Is this word problem solvable?
[ QUOTE ]
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? [/ QUOTE ] Having trouble coming up with this solution. Any help would be appreciated. 
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#3
10-16-2007, 01:09 AM
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Re: Is this word problem solvable?
[ 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. [/ QUOTE ] Seems correct (speed is relative, etc).
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#4
10-16-2007, 01:15 AM
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Re: Is this word problem solvable?
[ 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. [/ QUOTE ] Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream?
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#5
10-16-2007, 01:22 AM
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Re: Is this word problem solvable?
[ QUOTE ]
[ 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. [/ QUOTE ] Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream? [/ QUOTE ] Speed is relative.
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#6
10-16-2007, 02:54 AM
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Re: Is this word problem solvable?
[ QUOTE ]
[ 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. [/ QUOTE ] Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream? [/ QUOTE ] 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.
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#7
10-16-2007, 03:35 AM
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Re: Is this word problem solvable?
[ QUOTE ]
[ 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. [/ QUOTE ] Wouldn't his speed be faster on the return as his paddling is combined with the speed of the stream? [/ QUOTE ] You can think of the stream standing still and the land moving all around it. PairTheBoard
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#8
10-16-2007, 07:29 AM
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Re: Is this word problem solvable?
[ QUOTE ]
[ QUOTE ] 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? [/ QUOTE ] Having trouble coming up with this solution. Any help would be appreciated. [/ QUOTE ] 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.
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#9
10-16-2007, 10:43 AM
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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
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#10
10-16-2007, 06:14 PM
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Re: Is this word problem solvable?
[ QUOTE ]
d1 = 2 miles + (v-s)(1 hour) = 2 + v - s [/ QUOTE ] 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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