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Airplane Direction and Wind

  1. Aug 6, 2015 #1
    1. The problem statement, all variables and given/known data

    A pilot with an airspeed of 500 km / hr wants to fly from City A to City B (which is exactly 700km North of City A); however, there is a wind from the West at 50 km / hr. In which direction should the plane fly in order to arrive directly at it's destination.

    2. Relevant equations

    [tex] \cos{\theta} = \frac{\vec{A} \bullet \vec{B}} {AB} [/tex]

    3. The attempt at a solution

    [tex] \vec{V_{wind}} = < 50, 0 > km / hr [/tex]
    [tex] \vec{V_{plane}}= < 0 , 500 > km / hr [/tex]
    [tex] \vec{V_{res}} = \vec{V_{wind}} + \vec{V_{plane}} = < 50, 500 > km / hr[/tex]

    [tex] \cos({\vec{V_{plane}},\vec{V_{res}}}) = \frac{< 0 , 500 > \bullet < 50 , 500 >} {500\sqrt{50^2+500^2}} = \frac {50(0) + 500(500)} {251246.8905} [/tex]

    [tex] = 0.9950 [/tex] [tex] \theta = \arccos({0.9950}) = 5.71^\circ [/tex]

    because the wind is blowing the plane 5.71 degrees E of N, the plane should be set 5.71 degrees W of N, so that it counteract the effect of wind and will be blown directly North.

    The part that is bugging is me is whether or not that kind of logic is actually sound. Does anyone see anything wrong with my train of thought?
     
  2. jcsd
  3. Aug 6, 2015 #2

    BvU

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    Isn't it so that you want the resultant ##\vec v_{\rm plane} + \vec v_{\rm wind}## to be ##<0, {\rm something} >## ?
     
  4. Aug 6, 2015 #3

    Oh, so would it be something like?

    [tex] \vec v_{\rm plane} + \vec v_{\rm wind} = < 50\cos{\theta}, 500\sin{\theta} >[/tex]

    [tex] \theta = 90^\circ [/tex]

    so you'd want to rotate that rotate the resultant vector 90 degrees to cancel out the effect of the wind, meaning you'll set the plane at

    90 + 5.7 = 95.7 degrees
     
  5. Aug 6, 2015 #4

    billy_joule

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    I agree with that.
    tan-1 (50/500) = 5.7degrees is a quicker route.
     
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