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# Andy

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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?

Feb 9, 2018

#1
+1324
+3

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.

GA

Feb 10, 2018
#2
+101228
+3

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  !!

Feb 10, 2018
#3
+101228
+3

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....]

Feb 11, 2018
#4
+1324
+3

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.

GA

Feb 11, 2018
#5
+101761
+2

Thanks Chris and Ginger.

I really like questions like this one :))

Feb 11, 2018
#6
+101228
0

I liked it, too....I'll have to say that the answer isn't exactly intuitive...it would seem that it would take him more time to turn around and retrieve the hat than the time elapsed between when he lost it and when he realizes that he lost it....but....nope  !!!

Feb 11, 2018