SOLUTION: A man is rowing a boat upstream. As he passes under a bridge, a bottle falls out. 20 min later the rower realizes this and rows back to pick it up. He catches up with the bottle 1

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Question 805445: A man is rowing a boat upstream. As he passes under a bridge, a bottle falls out. 20 min later the rower realizes this and rows back to pick it up. He catches up with the bottle 1 mile downstream from the bridge. How fast is the current?
Found 3 solutions by Edwin McCravy, Alan3354, AnlytcPhil:
Answer by Edwin McCravy(20064) About Me  (Show Source):
You can put this solution on YOUR website!
A man is rowing a boat upstream. As he passes under a bridge, a bottle falls out. 20 min later the rower realizes this and rows back to pick it up. He catches up with the bottle 1 mile downstream from the bridge. How fast is the current? 
Make this chart:

                                          | distance | rate |     time      |
-----------------------------------------------------------------------------
bottle floating downriver                 |          |      |               |  
man going upriver from bridge             |          |      |               |
man going downriver back to bridge        |          |      |               |
man going downriver from bridge to bottle |          |      |               |


20 minutes is a fraction of an hour, and we will be having to use
fractions to get time, so to avoid so many fractions let the rates 
be in miles per minute instead of miles per hour.  Then we'll 
convert to miles per hour at the end.


Let the rate of the man in still water = r
Let the rate of the current = c

Then 
the bottle's floating rate = c (the same as the rate of the current)
the man's upriver rate = r-c
the man's downriver rate = r+c
the bottle's floating distance = 1 mile
the man's time going upriver from bridge = 20 minutes.
the man's distance from bridge to bottle = 1 mile 


We fill those in:
                                          | distance | rate |     time      |
-----------------------------------------------------------------------------
bottle floating downriver                 |    1     |   c  |               |  
man going upriver from bridge             |          |  r-c |      20       |
man going downriver back to bridge        |          |  r+c |               |
man going downriver from bridge to bottle |    1     |  r+c |               |


Use time = distance/rate to fill in bottle's time floating downstream = 1/c
Use distance = rate·time to fill in man's distance upriver from bridge = (r-c)20 = 20(r-c)
Use time = distance/rate to fill in man's time from bridge to bottle = 1/(r+c)


                                          | distance | rate |     time      |
-----------------------------------------------------------------------------
bottle floating downriver                 |    1     |   c  |      1/c      |  
man going upriver from bridge             | 20(r-c)  |  r-c |      20       |
man going downriver back to bridge        |          |  r+c |               |
man going downriver from bridge to bottle |    1     |  r+c |    1/(r+c)    |


The man's distance upriver from bridge = man's distance downriver back to the
bridge, so we also put 20(r-c) for man's distance downriver back to the
bridge. 
 
                                          | distance | rate |     time      |
-----------------------------------------------------------------------------
bottle floating downriver                 |    1     |   c  |      1/c      |  
man going upriver from bridge             | 20(r-c)  |  r-c |      20       |
man going downriver back to bridge        | 20(r-c)  |  r+c |               |
man going downriver from bridge to bottle |    1     |  r+c |    1/(r+c)    |


Use time = distance/rate to fill in man's time downriver from bridge to bottle


                                          | distance | rate |     time      |
-----------------------------------------------------------------------------
bottle floating downriver                 |    1     |   c  |      1/c      |  
man going upriver from bridge             | 20(r-c)  |  r-c |      20       |
man going downriver back to bridge        | 20(r-c)  |  r+c | 20(r-c)/(r+c) |
man going downriver from bridge to bottle |    1     |  r+c |    1/(r+c)    |


The equation comes from 

               %28matrix%283%2C1%2C%0D%0A%22bottle%27s%22%2Cfloating%2Ctime%29%29 %22%22=%22%22 %28matrix%283%2C1%2C%0D%0A%22man%27s%22%2Ctime%2Cupriver%29%29%22%22%2B%22%22%28matrix%285%2C1%2C%0D%0A%22man%27s%22%2Ctime%2Cdownriver%2Cto%2Cbridge%29%29%22%22%2B%22%22


                     1%2Fc %22%22=%22%22 20%22%22%2B%22%2220%28r-c%29%2F%28r%2Bc%29%22%22%2B%22%221%2F%28r%2Bc%29

Multiply through by LCD = c(r+c)

                    r+c = 20c(r+c) + 20c(r-c) + c
                    r+c = 20cr+20cē + 20cr-20cē + c 
                      r = 40cr
                      1 = 40c
                     1%2F40 = c

So the rate of the current is 1%2F40 miles per minute.
To change that to miles per hour we multiply by 60:

        Rate of current = 1%2F40%22%22%2A%22%2260 = 60%2F40 = 1.5 mi/hr.

It is interesting here that since r, the rate of the man in still water, 
cancels out, that means that the rate of the current is independent of
the speed of the man in still water.

Edwin

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A man is rowing a boat upstream. As he passes under a bridge, a bottle falls out. 20 min later the rower realizes this and rows back to pick it up. He catches up with the bottle 1 mile downstream from the bridge. How fast is the current?
======================
Choose the frame of reference to simplify things.
Ignore the bridge.
========================
He rows away from the bottle for 20 minutes, so it takes him 20 minutes to row back to it, 40 minutes total.
---
In 40 minutes (2/3 hours), the bottle moved 1 mile.
1/(2/3) = 1.5 mi/hr current
---------------------------
His direction is also irrelevant, he can row up or downstream, or any direction.
The situation is similar if the bottle falls out near a buoy in open water.

Answer by AnlytcPhil(1809) About Me  (Show Source):
You can put this solution on YOUR website!
Alan is pretending that the water in the river is perfectly still and the river bed is moving under it. He is pretending that the bottle does not move at all.
What he has found is not the rate of the current, for the water is still. He has found the rate of the moving river bed under the perfectly still water. That is a legitimate, and simple, way to look at the problem, although amusing.
Edwin