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Canoeing accross a river!

  • Thread starter r.meghdies
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  • #1
r.meghdies
I have a test on Friday on this...now my text book, 2 physics teachers and other people say i'm wrong but i know i still think i'm right. what is the right way to answer this i explained option 2 is correct. please give explanation of why, or examples i can use to prove it.

Thank you!

A canoeist who can paddle 5km/h in still water wishes to cross a 400 m wide river, with 2km/h current. If he steers the canoe perpendicular to the current and wants to get straight across the river how long will it take him to cross the river.

Which way is right? and why?

hypotenuse = 5 km/h
................| (ignore dots)
................| < 2 km/h
__________|
^ 400 m & 4.58 km/h
Θ = 22*

option 1:

t = 0.4 km / 4.58 km/h


option 2:
cosΘ = a / h
cos 22 = 0.4 km / h
h = 0.43 km

now i do t= .43km / 4.58 km/h

t= 5.63 minutes
 

Answers and Replies

  • #2
cepheid
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Option 1 seems right to me. There is one thing in the question that, as worded, doesn't make sense. Let's say the current is to the right. If he paddles perpendicular to the current (i.e. straight across the river), then he won't make it directly across. The current will sweep him to the right. In order to make it to the point DIRECTLY across the river, he'll have to paddle diagonally, (forwards and to the left) in such a way that the leftward component of his velocity is 2 km/h. So you set up a velocity right triangle with a hypoteneuse of 5 km/h (his total velocity), and a leftward component of 2 km/h. Solving for the forwards component using pythagoras, you indeed get 4.58 km/h, and t = 0.4 km/ 4.58 km/h = 5.23 min

I don't understand option 2 at all. There is no 22 degree angle in this triangle. arccos(2/5) = 66.42 degrees.
 
  • #3
PhanthomJay
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In order to go straight across, the canoeist must steer into the current at 23.6 degrees (typo 22*?) , such that his ultimate path of travel (and velocity) will be perpendicular to the current . So the total distance he travels is 400m at 4.58km/hr, not 430 meters at that speed. I vote for option 1.
 
  • #4
cepheid
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Right, PhantomJay. Your angle is the complement of mine. It looks like our answers are consistent. His path makes an angle of 66.4 degrees with the riverbank, or 23.6 degrees with the line of sight straight across the river.
 
  • #5
r.meghdies
hmm

look at my diagram...you can say this is addition of vectors....you have to travel the hypotenuse in order to overcome the current...if not then you will not end up directly across the river bank.
|.............................|
|.............................|
|< from here to here >|
|.............................|

now if we break it up into the vectors take vector 1...the hypotenuse...it represents where your heading and where you end up...so let say no current...you swim all of the hypotenuse...now we do the addition...the current comes in and pushes you all the way down so that you end up directly across from where you started...how far did you swim...the hypotenuse...

another way of explaining it is that lets say we wanna swim against current without adding(which would really be subtracting) the vectors. We show that they would have to swim longer by extending the vector to make up for the current...this would show a greater distance traveled and would = the same time as just subtracting the vectors.

Do you still not agree with me? if so teach me...because logically what i say makes sense!
 

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  • #6
cepheid
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Dude, I think I know what your problem is. You might be having trouble with the idea of relative motion and relativity (in the classical sense).

There are TWO ways of looking at this problem. The first way is from the point of view of an observer on the river bank. In other words, in the *frame of reference* in which the riverbank is at rest. The second method is from the point of view of an observer in the river i.e. in the *frame of reference* in which the water is at rest, and the riverbanks are moving to the left at 2km/h. In physics jargon, we might call the first point of view the riverbank rest frame and the second point of view the river (or water) rest frame

Method 1: Riverbank rest frame

From this observer's point of view, your canoe moves in a *straight line* across the river a distance of 0.4 km at a velocity of 4.58 km/h and the time it takes to reach the opposite side is:

t = 60 min/hr * (0.4 km / 4.58 km/h ) = 5.23 min

Method 2: River rest frame

In this frame of reference, the canoe moves *diagonally* a distance of:

0.4 km / sin(23.6) = .436 km

at a velocity of 5 km/h (be careful here to get the right velocity in THIS frame of reference)

Therefore t = 60 min/hr * (0.436 km / 5 km/h) = 5.23 min

Either way, you get the same answer, but you cannot MIX frames of reference, which is what you did in option 2 when you used the wrong velocity for the diagonal path from the river's point of view. Do you understand?
 
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  • #7
PhanthomJay
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....you have to travel the hypotenuse in order to overcome the current...if not then you will not end up directly across the river bank.
Just to expand on Cepheid's correct response, you do not travel the hypotenuse; you paddle in that directon, but end up traveling in a straight line of 400m, perpendicular to the current. Your canoe will be 'yawed' 23.6 degrees to the left as you move, but your path of travel will be perpendicular to the current. This is what the problem specifies, perhaps unclearly: You paddle in a direction that will 'steer' you in a perpendicular path. Make sense?
 

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