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Vector Airplane Question

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data
    An airplane flies at 600km/h on a bearing of W38N while a wind of 75km/h from W25S is present.
    Find the planes resultant velocity and direction




    3. The attempt at a solution

    So heres what I did:

    The Airplane: x = 600cos142 = -472.8
    y = 600sin142 = 369.39

    The Wind: x = 75cos25 = 67.973
    y = 75sin25 = 31.69

    Sum of x: -472.8 + 67.973 = -404.827
    Sum of y: 360.30 + 31.69 = 401.09

    Resultant would be using pythagorean theorem:

    this gives me 569.877

    and the angle would be theta = tan1- (401/-494.877) = -44.734 degrees


    I think what I did is right (please correct if im wrong) but I was wondering about my angle. The angle is not below the horizontal right? I just dont understand why its negative. If i were to change it to bearing form would i do......... 90-44.734 = 45.26 degrees therefore on a bearing of N45.26W? Please clarify the solution I have found
     
  2. jcsd
  3. Mar 1, 2013 #2

    tms

    User Avatar

    I didn't check your numbers, but the method is correct. The angle is negative because the arctangent returns a value between [itex]-\pi/2[/itex] and [itex]\pi/2[/itex]. You have to look at the signs of x and y to find the actual angle.
     
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