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Vector Question

  1. Sep 13, 2007 #1
    1. The problem statement, all variables and given/known data

    A carrier drives from her office 22.0 km northerly direction to town. Drives in direction of 60.0 degrees south of east for 47.0 km to another town. Whats her displacement from office?

    2. Relevant equations



    3. The attempt at a solution

    I drew a diagram of her route but I do not know what to do next.
     
  2. jcsd
  3. Sep 13, 2007 #2

    learningphysics

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    What type of shape do you get?
     
  4. Sep 13, 2007 #3
    A Traingle
     
  5. Sep 13, 2007 #4

    learningphysics

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    cool. so do you know which side it is whose length you need? do you know the cosine rule?
     
  6. Sep 13, 2007 #5
    Hypotunese is what I need, right?
    Cosine rule is to determine the sides/anlges of a non right angle.
    Could I use Pythagorean theorem?
     
    Last edited: Sep 13, 2007
  7. Sep 13, 2007 #6

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    In this case you don't have a right triangle, so no hypoteneuse... so you can't use pythagorean theorem. But you can use the cosine rule to get the length of the third side of that triangle...

    Then you need the direction also... so you need the angle of that third side (measured from north, south, east or west... whichever is most convenient).
     
  8. Sep 13, 2007 #7
    Okay. Let me try to work it out.
     
  9. Sep 13, 2007 #8
    a^2=b^2+c^2-2bccosA
     
  10. Sep 13, 2007 #9
    40.731
     
  11. Sep 13, 2007 #10

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    careful... what angle do you have in the triangle... opposite the side you're calculating?
     
  12. Sep 13, 2007 #11
    I did a^2 = 22^2 + 47^2 - 2(22)(47)cos60 to get 40.731 and then I did
    40.731^2 = 47^2 + 22^2 - 2(47)(22)cosB to get me 60
     
  13. Sep 13, 2007 #12

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    Yes, but in my diagram, the 60degrees, isn't the angle in the triangle... I get 90-60=30 degrees...
     
  14. Sep 13, 2007 #13
    How do you get 90 degrees?
     
  15. Sep 13, 2007 #14

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    90 degrees is the angle between the path (22.0 km north) and the east axis... she drives 60 degrees south of east... that means her path is 60 degrees below the east axis... or in other words 30 degrees east of south...

    For example... 10 degrees south of east, is 80 degrees east of south...
     
  16. Sep 13, 2007 #15
    Oh okay. So I should replace 60 with 30 in the formula I showed you earlier.
     
  17. Sep 13, 2007 #16

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  18. Sep 13, 2007 #17
    Ok. I see. Make so much more sense now.
    Is the side 30.530 and is that the answer?
     
  19. Sep 13, 2007 #18

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    I'm getting 30.034km... you should also get the direction.
     
  20. Sep 13, 2007 #19
    Yes. You are right. 30.034 km.

    How would you find the direction?
     
  21. Sep 13, 2007 #20

    learningphysics

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    You can find the angle of the displacement from any of the 4 axes (north south east west). I've drawn in one here as x:

    http://www.imagevimage.com/gallery.php?entry=images/triangle2.gif

    So when you find x... the direction would be x degrees west of south... hope that makes sense...

    To find x... you'd use the sine law...

    I've drawn x in two places... because those two angles are equal (since the lines are parallel)... hope this makes sense...
     
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