River crossing and relative velocities

  • Thread starter Egoyan
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  • #1
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Hi everyone,

I've been having difficulties with this problem for a while. Here is my best attempt at solving it. If there's anything wrong, I honestly can't figure it out :). I would appreciate if anyone could go over it quickly and tell me if/what I did wrong.

Homework Statement


A boat crosses a wide river with a speed of 12km/h relative to water. The river has a uniform speed of 6 km/h due east relative to earth.
(a) Determine the speed of the boat relative to a stationary observer.
(b) In what direction should the boat be heading to reach an opposite point directly across the river?


Homework Equations


I've set up the velocities as such:
Vbe = Velocity of the boat relative to the earth (and observer),
Vbw = Velocity of the boat relative to the water, and
Vwe = Velocity of the water relative to the earth.

Therefore, Vbe = Vbw + Vwe


The Attempt at a Solution


After drawing a picture of the situation, I've determined that

Vbw = (-12sinΘ i + 12cosΘ j) km/h
Vwe = (6i + 0j) km/h
Vbe = (0i + Vbe j) km/h ← This is something I'm not certain of. Am I right to assure that since we want to go directly across the river, relative to the earth, this vector should have a 0 i-component?

Using Vbe = Vbw + Vwe,

1) -12sinΘ + 6 = 0
→ Θ = 30° (counterclockwise from positive y-axis)

2) magnitude of Vbe = 12cosΘ = 12cos30 = 10.39

To answer a) using the above, Vbe = (0i + 10.39j) km/h, or 10.39 km/h due north relative to the observer.
b) Direction should be 120° from positive x-axis.



Have I made any mistakes somewhere? For some reason, I had quite a hard time visualizing this problem.

Thank you for your time!

Egoyan
 

Answers and Replies

  • #2
haruspex
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Your answer and reasoning look right for b). Part a) is unclear - you cannot answer it without assuming a relative angle.
 
  • #3
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Hi,

Thanks for taking the time to look at the problem.

That's what I thought too - and that's part of why I had such a hard time visualizing it, I believe. But this was an exam question (I am 100% certain of the phrasing, I've got a copy of it with me), and I was wondering what was the deal here...

Some of my classmates were illustrating the problem as a triangle, setting it up such that the hypotenuse would be the resultant velocity relative to the observer; that is v=√(12^2+6^2) = 13.4 km/h. Anyone knows if this could be right?

Thanks again,

Egoyan
 
  • #4
haruspex
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Hi,

Thanks for taking the time to look at the problem.

That's what I thought too - and that's part of why I had such a hard time visualizing it, I believe. But this was an exam question (I am 100% certain of the phrasing, I've got a copy of it with me), and I was wondering what was the deal here...

Some of my classmates were illustrating the problem as a triangle, setting it up such that the hypotenuse would be the resultant velocity relative to the observer; that is v=√(12^2+6^2) = 13.4 km/h. Anyone knows if this could be right?

Thanks again,

Egoyan
Yes, I suspect part a should have said "if the boat heads straight across relative to the water". This would make 13.4 correct.
 
  • #5
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Ah, yes, I see. Thanks a lot!
 

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