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Need help on a vector problem by tomorrow.

  1. Oct 27, 2007 #1
    Need help on a vector problem by tomorrow. :(

    1. The problem statement, all variables and given/known data
    A woman walks 277. m in the direction 41.6° east of north, then 171 m directly east. Find
    (a) her resultant vector displacement from the starting point

    There were other parts to the question which I got right, I can't seem to get this one.

    2. Relevant equations

    None except the pythagorean theorem:
    In this case, D=[tex]\sqrt{Dx^2 + Dy^2}[/tex]


    3. The attempt at a solution

    D1X=cos(41.6°) * 277
    D1X=207.1400711

    D2X=171

    D1Y=sin(41.6°) * 277
    D1Y=183.9075609

    D2Y=0

    Dx=378.14
    Dy=183.9075604

    D=sqrt(378.14^2+183.9075609^2)
    D=420.49 =~ 420.5

    Ok, so I got 420.5 and plugged in it. Nope, the computer says it's wrong. I tried rounding it to 421 and that was wrong also. I don't think I made any math errors, what am I doing wrong?
     
  2. jcsd
  3. Oct 27, 2007 #2

    PhanthomJay

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    You used the wrong angle..problem says east of north but you used north of east.....otherwise, your work looks good once you correct for the proper angle.
     
  4. Oct 27, 2007 #3
    Now I'm confused, what is the difference between east of north and north of east? >_> Which angle would I use?

    EDIT: The only other angle I can think of is 48.4 degrees, found by: 90+41.6+x=180
     
    Last edited: Oct 27, 2007
  5. Oct 27, 2007 #4

    cristo

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    x degrees "east of north" means x degrees from north in a clockwise manner. x degrees "north of east" means x degrees from east in an anticlockwise manner. Unless x=45, the angles will not be the same.
     
  6. Oct 27, 2007 #5

    PhanthomJay

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    And don't forget to get the resultant angle.
     
  7. Oct 27, 2007 #6
    Thanks for explaining that, I changed the angle and it worked. :)
     
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