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Need help with finding magnitude

  1. Apr 22, 2010 #1
    1. The problem statement, all variables and given/known data

    The eye of a hurricane passes over Grand Bahama Island. It is moving in a direction 59.3◦ north of west with a speed of 38.9 km/h. Exactly 2.89 hours later, the course of the hurricane shifts due north, and its speed slows to 27.8 km/h, as shown.
    How far from Grand Bahama is the hurricane 4.42 h after it passes over the island?

    The problem is already solved and has the solutions to the problem. It is #3. But what I am not understanding is why they used 120.7 as their theta rather than 59.3

    2. Relevant equations

    Magnitude (D) = square root ( Dx^2 + Dy^2)

    3. The attempt at a solution

    I understand how they did the problem, but can you just explain to me why they used 120.7 as their theta and not 59.3. If they use 120.7 as their theta, wouldn't sin120.7 give you your x-component, not your y-component. And when using cos120.7, wouldn't that give you your y-component, not your x-component.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Apr 22, 2010 #2
    try with theta = 59.3 and compare your answer with the one in the book. Angle simply depends on coordinates you use, magnitude of displacement is independent of this choice, at least it should be ;]
     
  4. Apr 22, 2010 #3
    i did try it with theta = 59.3. i got a displacement of 139.4, which is wrong
     
  5. Apr 22, 2010 #4
    check it again. If you put theta = 59.3 in equations for delta x and delta y in paper which you gave, then the answer for total displacement is the same as with theta = 120.7
     
  6. Apr 22, 2010 #5
    Ya, you're right. I just used the wrong numbers. But thanks for the help
     
  7. Apr 22, 2010 #6
    In a compass plot, the 0 degree point is the rightmost part of the plot. See attached:

    http://people.rit.edu/pnveme/pigf/TwoDGraphics/twod_dir_compass.gif [Broken]
     
    Last edited by a moderator: May 4, 2017
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