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The Classic Airplane Problem

  1. Feb 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A small plane departs from point A heading for an airport 490 km due north at point B. The airspeed of the plane is 210 km/h and there is a steady wind of 50 km/h blowing directly toward the southeast.
    (a) Determine the proper heading for the plane.
    ° west of north

    (b) How long will the flight take?

    2. Relevant equations

    Law of Sines/Cosines?

    vf = vi + a*t

    3. The attempt at a solution

    So I drew a picture.

    You need a velocity vector. This vector must be the sum of the velocity vectors of the wind and the plane, which should also be drawn on the sketch. So I drew a simple sketch showing the plane's northward trek, an angle [tex]\Theta[/tex], another vector showing the wind and the resulting vector. So:

    The first vector points due North (up), from point A.
    The wind vector points toward point A, at an angle [tex]\Theta[/tex].
    The resultant vector points from point B to the non-pointed end of the wind vector.

    ...so where in the world does the trig come in? I just don't know where to start plugging in numbers, since this isn't a right triangle it isn't a simple sin/cos problem with the Pythag. Theorem, right?
  2. jcsd
  3. Feb 8, 2010 #2


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    Homework Helper

    In this problem, AB is the resultant velocity (Vr) of wind velocity (Vw)and proper velocity of the plane (Vp)
    To make the problem easy, consider wind direction as x-axis.
    Angle between wind direction and the resultant is ( 90 + 45) degrees.
    Let θ be the angle between Vp and Vw. Then if you draw the vectors, you can see that,
    Vr*sin(135) = Vp*sinθ .....(1)
    Vr*cos(135) = Vp*cosθ - Vw .......(2)
    Now solve for Vr and θ.
  4. Feb 9, 2010 #3
    Ok I solved for Vr and for theta, but I can't use either of them since there's still the other variable in the equation. IE theta is still in the Vr equation (and I don't know it) and I have Vr in the theta equation. And plugging either equal to each other makes it unsolvable. I end up with a cos (-) plus sin (-).

    Vr = Vr

    [tex]\frac{210 * sin \Theta}{sin (135)}[/tex] = [tex]\frac{210 (cos\Theta) - 50}{cos 135}[/tex]

  5. Feb 9, 2010 #4


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    Homework Helper

    Divide eq. 1 by 2 . You get
    (Vp*sinθ) = Vp*cosθ - Vw
    Square both the sides. You get
    Vp^2*sin^2(θ) =( Vp*cosθ - Vw)^2
    Vp^2*[1- cos^2(θ)] = ( Vp*cosθ - Vw)^2
    Now simplify and solve the quadratic to find cosθ. Then find Vr.
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