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Relative velocity of plane

  1. Jun 2, 2006 #1
    A jet airliner moving initially at 300 mi/h due east enters a region where the wind is blowing at 100 mi/h in a direction 27 degress north of east. What is the new velocity of the aircraft to the ground?

    This is how i did it: Vag=Vaw + Vwg

    Vaw(aircraft relative to wind)= 300
    Vwg(wind relative to ground) = 100

    drawing out the vectors i have Vag as my hypotonuse, Vaw as my base and Vwg as my (oppositte angle of 27 degrees)....solving for Vag I get 316
    but that's not the right answer. I also need to find at what degrees the aircraft is going (north of east) but i dk how to do that?

    any suggestions?
    i was wondering if someone could draw the vectors out for me if possible
  2. jcsd
  3. Jun 2, 2006 #2


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    Consider north up (+j) and east to the right (+i),

    So the aircraft is traveling 300 mi/h i, when it encounters a wind which has a velocity of 100 mi/h in the direction which is 27° N from east.

    Now the angle of 27° is between the axis pointing east (+i) and the wind vector.

    To obtain the new aircraft vector with respect to ground, one adds the aircraft vector and the wind speed vector (with respect to ground), which then gives a resultant, which is the new aircraft speed with respect to ground.

    So 300 i + Wi is the new Vi, and the Northward component is Vj = Wj since the northward wind component carries the plane northward.

    W = Wi i + Wj j

    Then use the Pythagorean theorem to calculate the magnitude of resultant vector, and the angle is the arctangent of the Vj/Vi.
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