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Two airplanes - vectors

  1. Sep 7, 2008 #1
    1. The problem statement, all variables and given/known data

    Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 13.4 m/s due north. Plane 2 taxies with a speed of 6.2 m/s in a direction 18.8° north of west.
    (a) What are the direction and magnitude of the velocity of plane 1 relative to plane 2?
    (b) What are the direction and magnitude of the velocity of plane 2 relative to plane 1?


    2. Relevant equations

    The law of cosines: A^2 = B^2 + C^2 - 2*B*C*cos(A)



    3. The attempt at a solution

    I set the vectors tail - to - tail and used the law of cosines to solve for the resultant vector, which I call n:

    n^2 = 6.2^2 + 13.4^2 - 2(6.2)(13.4)cos(71.2)
    n = 12.823 m/s

    So I have the magnitude. Now I need the direction. For (a), the direction will be X degrees north of east, and for (b) the direction will be Y degrees south of west. I'm have a feeling that X=Y, but I can't seem to get the right answer. Here's what I'm doing:

    I solve for the angle between the 12.823 and the 18.8 sides:
    theta = cos^-1 ([12.823^2 + 18.8^2 - 13.4^2]/[2*12.823*18.8])
    theta = 45.4385 degrees

    That leaves the remaining angle in the triangle to be 63.3615 degrees.

    So to find the direction for, let's say, part (b), shouldn't I just subtract 64.3615 degrees from 90 degrees? That would yield 25.6385 degrees. But that answer is incorrect.

    What am I failing to see?

    Thanks for your response!
     
  2. jcsd
  3. Sep 7, 2008 #2
    At the very end of my post I meant to say 63.3615 and 26.6385 degrees; it was a typo.
     
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