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Relative Velocity - Canoe vs. River

  1. Aug 16, 2014 #1
    1. The problem statement, all variables and given/known data

    A canoeist wants to travel straight across a river that is 0.10km wide. However, there is a strong current moving downstream with a velocity of 3.0km/h. The canoeist can maintain a velocity relative to the water of 4.0km/h.

    a) In what direction should the canoeist head to arrive at a position on the other shore directly opposite to his starting position?

    b) How long with the trip take him?

    2. Relevant equations

    a2 + b2 = c2

    sin a = (sin b/b) X a

    cos θ = adjacent/hypotenuse

    3. The attempt at a solution

    a) r2 = 42 + 32 = 25
    r = 5km
    sin θ = (sin90/5)X 3
    sin θ = 0.6
    θ = 37

    He should aim 37° upstream, from the destination reference.

    The book's answer is: He should aim upstream at an angle 41° with respect to the riverbank.

    b) cos 37 = 0.1/hyp
    hyp = 0.1/0.79
    hyp = 0.125
    0.125/4 = 0.03 = 1.9min

    The book's answer is 2.3 min.

    Any help that you can provide is greatly appreciated!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 16, 2014 #2

    Doc Al

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    Staff: Mentor

    First things first. Draw yourself a vector diagram. Yes, a right triangle is involved, but what is the hypotenuse?
     
  4. Aug 16, 2014 #3
    Thank you.

    I've done that and thought the hypotenuse was calculated from the angle as well as the length of the adjacent side:
    cos 37 = 0.1/hyp
    hyp = 0.1/0.79
    hyp = 0.125
    0.125/4 = 0.03 = 1.9min

    is this the wrong way? I figured my answer must be wrong because the angle that I calculated for question a) is different than the book's answer.
     
  5. Aug 16, 2014 #4

    Nathanael

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    Homework Helper

    You want to find the angle in which he will have no upstream/downstream velocity.

    This means the up-stream-component of his velocity* must equal the downstream velocity of the water.

    At what angle does this happen?


    *(his velocity relative to the water)
     
  6. Aug 16, 2014 #5

    Doc Al

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    You are mixing up which side is the hypotenuse. (You won't calculate the hypotenuse--it's given.)

    Draw the triangle representing this vector addition:
    V(water/shore) + V(boat/water) = V(boat/shore)

    This forms a right triangle. But the trick is correctly identifying the sides.
     
  7. Aug 17, 2014 #6
    I'm so confused.
    vws=3km/h
    vcw=4km/h

    32+42=52
    vcs=5km/h

    My brain is mush. That number doesn't seem right.

    I don't understand how this question is messing me up so much.
     
  8. Aug 17, 2014 #7

    phinds

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    2016 Award

    And we likely won't understand how to help you unless you show your work. Your work INCLUDES the vector diagram, so draw it and show it.
     
  9. Aug 17, 2014 #8

    Doc Al

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    It's not right. Show us that vector diagram.

    Your analysis assumes that vws and vcw are perpendicular. They are not.
     
  10. Aug 17, 2014 #9
    I apologize

    DSC_0321.jpg
     
  11. Aug 17, 2014 #10
    ohhhh!

    I will redo my work considering that but I don't think I know where to start.
     
    Last edited: Aug 17, 2014
  12. Aug 17, 2014 #11

    Doc Al

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    Staff: Mentor

    Finally, a diagram! :smile:

    You are told that the canoeist wants to go directly across to the other shore, so Vcs must be straight across. The angle that Vcw makes is what you need to find. (Using a corrected diagram.)
     
  13. Aug 17, 2014 #12
    talking to me about this is harder than teaching a rock how to swim!

    so vcw=4
    and vws=3
    so vcs=2.65

    finally!

    DSC_0322.jpg

    so (0.1/2.65)x60 = 2.3 min

    and sin Θ = (sin 90/4) X 2.65
    = 41°

    Hooray!!!!

    thank you very much!!!
     
  14. Aug 17, 2014 #13

    Doc Al

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    Yay! :approve:
     
  15. Aug 17, 2014 #14

    phinds

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    Nah, you did fine. The trick to these, as I think you now see, is to get the vector diagram right. The math is trivial but you need to be sure you are actually solving the right problem.
     
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