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Homework Help: Vector Problem of two twins

  1. Jan 16, 2010 #1
    1. The problem statement, all variables and given/known data
    Two twins set out to row separately across a swiftly moving river. They have identical canoes, and can row at the same speed in still water. Twin A aims straight across the river but, due to the current, is carried downstream before reaching the opposite bank. Twin B aims her canoe upstream at an angle of 56.0 degrees relative to the riverbank, so as to arrive on the opposite side at a point that is directly across from her starting point.

    Calculate the ratio of the time it takes twin A to cross the river to the time it takes twin B to cross.

    Determine the direction of twin A's motion, expressed as an angle relative to the downstream direction.


    2. Relevant equations



    3. The attempt at a solution
    In order to find a ratio of time, I set up update of position formulas.

    Twin A:
    [tex]\Delta[/tex]X=Vavg*t

    In the case of the x direction I'll call the initial velocity V which should equal Twin B's initial velocity

    [tex]\Delta[/tex]X=V*t

    time for twin A:
    t=[tex]\Delta[/tex]X/V

    Twin B:
    [tex]\Delta[/tex]X=Vcos34*t
    t=[tex]\Delta[/tex]X/Vcos34

    Ratio of A/B for t is cos34, I got this part right.

    I don't know how to go about the second part. The drag of the water represents a Force vector, and Twin A start off with a velocity vector in the x direction.

    What I do know:
    Initial y position=0
    Initial y velocity=0
     
  2. jcsd
  3. Jan 16, 2010 #2

    Doc Al

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    Staff: Mentor

    Part 2 is a vector addition problem, where the two vectors are perpendicular:
    The velocity of A with respect to the ground = velocity of A with respect to the water + velocity of the water with respect to the ground ​

    Hint: Make use of the fact that B travels directly across the river.
     
  4. Jan 16, 2010 #3
    Alright, so can I assume that the vector downstream is Vsin(56)? And the vector to the other side of the stream is V?

    Therefore, angle relative to the upstream direction should be arctan(1/sin56)?

    Then 180-arctan(1/sin56)= direction relative to downstream direction = 150

    But that's wrong. What did I do wrong?
     
  5. Jan 16, 2010 #4

    Doc Al

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    Why sine?
     
  6. Jan 16, 2010 #5
    I assumed sine because it was the y component of the Vector V for twin B and I assumed that since he traveled 0 Y displacement that his Y vector (Vsin56) equaled the vector downstream.
     
  7. Jan 16, 2010 #6

    Doc Al

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    The stream travels along the x-direction, so why use the y-component?
     
  8. Jan 16, 2010 #7
    Sorry, I assumed the river to be in the Y direction. Upstream being higher than downstream.

    Doesn't that coincide with the way I solved the first part?

    Sorry for the misunderstanding, I'm sure I'm the messed up one though.
     
  9. Jan 16, 2010 #8
    Wait, I see my confusion now. The sin of 56 is the x component. The sin of 34 is the y component. Besides that, did I do the rest of the problem correctly?
     
  10. Jan 16, 2010 #9

    Doc Al

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    My bad for not realizing that you have the river moving along the y-direction. In any case, the angle is given with respect to the riverbank (the y-axis) and you want the component parallel to the riverbank.
     
  11. Jan 16, 2010 #10

    Doc Al

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    There you go. Other than that, I think it was OK.
     
  12. Jan 16, 2010 #11
    Ok so arctan(1/sin34)=60.8 degrees.

    Is this not 60.8 degrees relative to the riverbank from the upstream direction?

    So the downstream angle should be 119.2 degrees? but this is wrong...
     
  13. Jan 16, 2010 #12

    Doc Al

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    No, it's relative to the downstream direction. (Rivers flow downstream. :wink:)

    You're just mixing up downstream and upstream. You had the correct answer above.
     
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