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
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
= c
So the rate of the current is miles per minute.
To change that to miles per hour we multiply by 60:
Rate of current = 60 = = 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