Airplane Direction and Wind

  • #1
138
43

Homework Statement



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.

Homework Equations



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

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?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
13,920
3,449
Isn't it so that you want the resultant ##\vec v_{\rm plane} + \vec v_{\rm wind}## to be ##<0, {\rm something} >## ?
 
  • #3
138
43
Isn't it so that you want the resultant ##\vec v_{\rm plane} + \vec v_{\rm wind}## to be ##<0, {\rm something} >## ?

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
 
  • #4
billy_joule
Science Advisor
1,200
331
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.
I agree with that.
tan-1 (50/500) = 5.7degrees is a quicker route.
 

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