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Vector, Direction question

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A primate research station attaches a tiny radio transmitter to a chimpanzee born in captivity before releasing it into the wild. One day, the station picks up a signal indicating the chimp is 5000 m from the station in a direction 21° S of W. Over the next day, the chimp wanders 3000 m in a direction 38° N of E. At this point,

    (a) how far from the station and
    (b) in what direction (as a positive angle measured counterclockwise from east) is the chimp? (Hint: Take east as the +x direction, and north as the +y direction.)

    2. Relevant equations
    Ax = A cos θ
    Ay = A sin θ

    For Magnitude
    [tex]\stackrel{\rightarrow}{r}[/tex] = [tex]\sqrt{A^{2}_{x} + A^{2}_{y}}[/tex]

    For Direction
    θ tan-1 [tex]\frac{A_{y}}{A_{x}}[/tex]

    3. The attempt at a solution

    (a) 2303 m
    (b) What does it mean?

    I use the direction equation but thats not correct. I had to do with counterclockwise direction that I do not know how to do.
     
    Last edited: Feb 8, 2010
  2. jcsd
  3. Feb 8, 2010 #2
    The station is your origin in the axis they are refering to, they are asking you for the angle you need to rotate when facing the +x (East) direction in order to see the chimp.

    One way you can get there is by solving the obtuse triangle you get from the sides you now have (5000, 3000 and 2303).
     
  4. Feb 8, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi Calhoun295! Welcome to PF! :wink:
    It means use angles from 0 to 360º, starting with East …

    so East is 0º, North is 90º, South is 270º, and so on. :smile:
     
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