The man in the boat is separated from his hat for a total of twenty minutes; ten minutes rowing upstream until he decides to go back for his hat, then ten minutes rowing downstream to get back to his hat. During this twenty minutes his hat has drifted downstream for one mile. Therefore the river current is three miles per hour. I fail to see why this is considered a puzzler.
No, it says you row upstream for 1 mile (however long that takes), lose your hat, row an additional 10 mintues at unknown speed, then turn around and row at the same, unknown speed until you catch your hat at the start (it having travelled 1 mi in the time…unspecified, but necessarily longer than 20 minutes).
I’m not sure how to solve it other than to work backwards and input current and rowing speeds that make it work out. Three MPH would be a top limit for the current-it has to be somewhat less than three, because the time the hat floats is somewhat more than 20 minutes. (:10 out, :10 back, plus the time to row the additional mile.)
There are too many variables for me to give a cognizant answer. First how fast can you row upstream vs downstream. What if the hat got caught up in an eddy, or slow spot, sure there is a practical answer, but will it work in the real world? Probably not, but for the correct answer you could be the proud owner of something!
Assume the man and his hat were on a conveyor belt moving downstream (say to the left). The man places his hat on the conveyor belt and walks at a constant speed (relative to the conveyor belt) upstream (to his right) for ten minutes. He then turns around, and walks at the same constant speed (relative to the conveyor belt) downstream (to the left) back towards his hat that is riding on the belt. If he walked for ten minutes on the belt away from his hat, it will take him ten minutes walking time on the belt to return to his hat. So it has taken him twenty minutes to leave his hat and return to it. But his hat has travelled one mile in the twenty minutes, and hence the river current is three miles per hour.
Mechaniker was right, for the reason he stated.
I logged back in to correct myself, but you beat me to it, M. The rowing speed is irrelevant.
I agree with Mechaniker about the solution, but not about it not being a good puzzler. People tend to make it more complicated than it is and try to solve it with algebra. It requires some thought to realize that the rowing speed is irrelevant, just the time.
I’m baffled, mechanichiker. The speed IS 3 mph, but how did you get that it takes him 10 minutes from the time he turns around to get the hat? It wasn’t stated in the problem.
Anyway, I did it with algebra at first. It’s not that hard. Set the time equal for both of them from the time he turns around. The hat has traveled 1-1/6 S by the time it gets to the dock at a speed of S (where S is the speed of theh river) and the man has traveled 1 and 1/6 at a speed of (2S + 1). Set the time equal so distance / speed = distance (man) / speed (man) or (1-1/6S) / S = (1+1/6) / (2S + 1).
I didn’t realize you could do it a simpler way. The hat is moving away from the man at a speed of S + 1. S for the hat and 1 for the man. The man’s speed against the current is 1, so his rowing speed is S + 1 and when he goes against the current it nets to 1.
They move apart at a net of S+1 for 10 minutes. Then he turns around and GAINS ground on the hat at a speed of (2S+1) - S. IE the man’s new speed is his rowing speed PLUS the speed of the current. So again, S + 1. Or 10 minutes.
So either mechanic hiker, you saw something immediatly or you read something in to the problem which wasn’t there but was right anyway.
I guess the part about them meeting at the dock wasn’t necessary.
“I guess the part about them meeting at the dock wasn’t necessary.”
The statement was necessary; it shows that the hat traveled one mile in the twenty minutes it was floating.
But how did you know it was floating for 20 minutes? The problem said the rower turned around after 10 minutes and met the hat at the dock. It never said how long it took to get from the turnaround point to the dock though. What did you see that says otherwise?
You don’t actually need that line about the dock. The hat was moving away from the rower at the same speed the rower was gaining on the hat (after he turned around), namely S+1 where S is the speed of the river. So the time before turnaround and after the turnaround to the meet up was the same, 10 minutes.
But it didn’t explicity state that in the problem, that the time after turnaround was also 10 minutes. If it did, then it wouldn’t be much of a problem.
“So the time before turnaround and after the turnaround to the meet up was the same, 10 minutes.”
Which is what I concluded without using algebra. I tried to illustrate this phenomenon by describing a man walking on a moving conveyor belt. The speed of the belt is analogous to the speed of the river current. If the man places his hat on the conveyor belt and then walks away from it for ten minutes, obviously it will take him another ten minutes to walk back to his hat on the belt. And as the hat on the belt (river) has traveled one mile back to its starting point during those twenty minutes, the conveyor belt (or river) must be moving at three miles per hour.
“If it did, then it wouldn’t be much of a problem.”
Which is what I said originally.
Here is another thought-provoking problem regarding relative motion: Consider a stationary wire with an electrical current flowing through it, and a charged particle moving parallel to the wire. From the standpoint of a stationery observer, a circular magnetic field surrounds the wire, and the charged particle is forced away or toward the wire according to the equation of classical physics: F = qv x B, where F dotes the force on the particle, q the magnitude of the charge, v the velocity of the particle, an B the strength of the magnetic field generated by the current flowing through the wire.
Now consider the same experimental setup from the point of view of an observer moving at the same speed as the charged particle. From his point of view, the velocity v of the charged particle is zero, but he still observes the particle to be attracted or repulsed (depending on the direction of current flow) by the wire. How can this be?
It seems to me that any truly observant physicist from the time of publication of Maxwell’s equations in 1865 to the publication of Einstein’s theory of special relativity in 1905 should have been aware that there was something seriously amiss with classical physics. It took Einstein 40 years to make the observation and to be bold enough to publish it. Therein lies his genius.
The key concept here is thate, with respect to the surface of the water, the hat is NOT moving. It is just sitting there on the surface of the water. The entire surface is moving but that is true for both the hat and the man. The only thing that matters is that the man is away from the hat for 20 minutes. It does not matter how far he went away or if he changed speed! ALL that matters is that he dropped the hat one mile from the dock, was away from the hat for 20 minutes, and got back to the hat at the dock. That means the river took 20 minutes to move the hat one mile. Mechaniker’s conveyer belt analogy illustrates the same concept. The man could have gone to the side of the river, had a picnic, and gone back to the hat. 20 minutes and one mile are the only inputs that matter!