Andy was rowing a boat upstream. As he passed under a low bridge, his hat was knocked off, but Andy only noticed his hat missing five minutes later. He turned his boat around immediately and rowed with the same effort downstream, fishing his hat out of the water at a point exactly one km downstream from the offending bridge. How fast was the current flowing?
Once Andy stops rowing, his relative velocity and direction are the same as the hat. Rowing away from the hat for five minutes means it took five minutes for him to return to it, for a total of ten minutes. In that ten-minute length of time, the hat traveled one kilometer.
Current flow: 1 km/10 min = 100 meters per minute
= 1.67 meters per second.
Just wanted to prove GA's answer in an algebraic manner
r = rate of current (in m/s) = rate of hat
R = Andy's normal rate (in m/s)
Distance from the bridge hat has traveled in 5 min since being knocked off = 300 (r) = 300r m
Distance from bridge Andy travels in 5min = 300(R - r) m
So....when Andy turns around ...his new rate is R + r
And the distance that Andy travels to retrieve the hat after he turns around = [300(R - r) + 1000] m
Meanwhile.....while Andy travels this distance, the hat travels [1000 - 300r ] m
So....equating times, we have that
Distance Andy travels after turning around / His Rate =
Distance hat has traveled / hat's rate
So we have that
[ 300(R - r) + 1000 ] / [R + r] = [ 1000 - 300r ] / r
[ 300R - 300r + 1000] r = [ 1000 - 300r ] [ R + r ]
300Rr - 300r^2 + 1000r = 1000R - 300Rr + 1000r - 300r^2
600Rr - 1000R = 0
3Rr - 5R = 0
R ( 3r - 5) = 0 ⇒ 3r = 5 ⇒ r = 5/3 = 1.67 m/s
Just as GA found !!
We can prove that this will always be so no matter the current rate or Andy's rate [ as long as his rate is greater than the current rate ]
Let Andy's normal rate be A and the current rate, C
Call T the time elapsed between when his hat is knocked off and he notices that it's gone
In this time the hat has traveled C * T
And Andy has traveled (A - C)T
When he turns around his rate becomes A + C
And the distance that he travels to retrieve the hat is (A + C) T
And this equals the distance he was from the bridge when he turned around plus the distance the hat has traveled from the bridge = C *2T ....so we have......
(A + C ) T = (A - C) T + C*2T
AT + CT = AT - CT + 2CT
AT + CT = AT + CT
Note to GA....I verified this result with Lancelot Link......he gave it a "thumbs up" ... [ at least I think that was the digit he was using....]
Yes, Lancelot uses his opposable thumbs for multiple purposes, including thumbs up or down. Nauseated, however, may give paws for dewclaw consideration, but his central tendency is toward his central digit with adjustments for any ambiguity.