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Physics Vector Question - Finding direction

  1. Jan 27, 2010 #1
    Two airplanes taxi as they approach the terminal. Plane 1 taxies with a speed of 11.3 m/s due north. Plane 2 taxies with a speed of 7.6 m/s in a direction 22.2° north of west. What is the direction of plane 1 relative to plane 2? Give the angle counterclockwise relative to the north. What is the direction of plane 2 relative to plane 1?

    *I solved for magnitude first then velocity*

    What is the magnitude of the velocity of plane 1 relative to plane 2?

    To solve for the magnitude I found the vector components of plane 2:

    Y component: 7.6 * sin(22.2°) = 2.87 m/s
    X component: 7.6 * cos(22.2°)= 7.04 m/s
    To find the final magnitude of the resultant vector I just used the pythagorean theorem after subtracting the plane 1 components FROM the plane 2 components.

    2.87 m/s - 11.3 m/s = -8.43 m/s
    7.04 m/s - 0 m/s = 7.04 m/s

    Now to find the magnitude I used the pythagorean theorem.
    Using the pythagorean theorom I got 10.98m/s as the relative velocity. This velocity is the same for plane 2 relative to plane 1.

    To find the direction of plane 1 to plane 2 I just used the arctan formula

    arctan (7.04/-8.43) = -39.86°
    Now since its counterclockwise from north I just added to 360 to get 320° which is the correct answer.

    I can't seem to figure out how to get the direction of plane 2 relative to plane 1 however and I've tried using the law of cosines and various other methods but I'm not sure where to go.

    I've tried -320 however it is not correct. Any ideas on why?
     
  2. jcsd
  3. Jan 27, 2010 #2
    Since the plane is headed west, its x component is negative. Thinking rationally about it, a direction "counterclockwise from due north" should be smaller then 90 degrees, because the two planes are _sort of_ moving in the same general direction.
     
  4. Jan 27, 2010 #3
    If it was 40° then it would be in the wrong quadrant wouldn't it? And I've tried -40° which hasn't worked.
     
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