The Harvard bridge and the Longfellow bridge are 1 mile apart.
physics/hat.in.river New
New entry, edited from material collected for the archive and not yet in a released version.
Problem
The Harvard bridge and the Longfellow bridge are 1 mile apart. The MIT crew team starts rowing upstream at the Longfellow. As the crew passes under the Harvard, the coxswain's hat falls into the river. 10 minutes later, the coxswain notices and turns the boat around instantaneously and has the crew go back to get it, rowing at the same constant rate. By the time the team reaches the hat, they are back at the Longfellow. How fast is the river flowing? -- Julien Beasley <jbb@ATHENA.MIT.EDU>, adapted from a book called "Puzzlegrams"
Show solution
Let u = speed of rowing (wrt a stationary frame of reference)
v = speed of water (wrt a stationary frame of reference)
Equating the time for which the hat travels in the river and the time
which the boat takes to catch up with it
1/v = 10 + (1 + 10*(u-v)) / (u+v)
Solving,
v = 1/20 miles/minute.
-- Vinay Gupta <gupta@cs.unc.edu>
It's the hard way.
The easy way defines speed w.r.t. the moving river:
speed of rowing is constant w.r.t. river
speed of hat is zero w.r.t. river
So, rowing back to get the hat takes another 10 minutes.
And indeed:
speed of water w.r.t. stationary frame (say, the Harvard)
= 1 mile/20minutes
-- Izak van Langevelde <iavlang@cs.vu.nl>
----------
In article <2svpj2INNvon@sat.ipp-garching.mpg.de>,
brownd@afmp01.cern.ch writes:
|> Here's a fairly simple solution.
|>
|> Presumably the velocities of the water and the boat add. We know
|> the hat drifts with the water velocity (V_w) first for 10 minutes,
|> then till the boat and it meet at the 2nd bridge (time T). For the
|> first 10 minutes the boat goes against the current a distance
|> (V_b - V_w)*10, then with the current for a distance (V_b + V_w)*T.
|> The total distance traveled by both hat and boat is 1 mile. This
|> gives;
|>
|> 1) (V_b + V_w)*T - (V_b - V_w)*10 = 1
|>
|> 2) V_w*(10+T) = 1
|>
|> Rearranging equation 1) gives
|>
|> 1') V_b*(T-10) + V_w*(T+10) = 1
|>
|> substituting 2) into 1') gives
|>
|> 3) V_b*(T-10) = 0
|>
|> Presumably the boat isn't standing still, so this gives
|>
|> T = 10 (minutes)
|>
|> Plugging back into 2) gives
|>
|> V_w= 1/20 miles/minute
|>
|> Note that the boat velocity is unconstrained (except to being
|> non-zero).
Well done! That was the hard way, though. You can do it the easy way in
your head in 2 seconds.
The key to this one is realizing that the boat takes the same amount
of time rowing away from the hat as rowing to the hat. Imagine the hat
on a conveyor belt. If you walk away from it for 10 min (you being also
on the belt), it will take you 10 min to get back. Your total trip
takes 20 min (going away from the hat and back). The hat travels 1
mile. 1mile/20min = 3 mph
-- Julien Beasley <jbb@athena.mit.edu>