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Math Help - Vector airplane question

  1. #1
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    Vector airplane question

    An airplane is flying at 500km/hr in a wind blowing 60 km/hr toward the SE. In what direction should the plane head to end up going due east? What is the plane's speed relative to the ground?

    I assume that the plane is heading due east to start and the wind would add to the i component of speed. I got that the wind was 42.4 i and - 42.4 j so tangent of angle is 42.4 over 542.4 so angle would be arctan of this which is 4.4 degrees. Therefore I have that the plane would need to head north of east by 4.4 degrees. The answer is 4.87 degrees north of east I must be missing another part of the calculation???? The plane's speed is 540.63 km/hr but by my calculations it should be 542.4. Can someone tell me how i am going wrong this time, please.
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  2. #2
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    the plane will have to steer a heading north of east to maintain an easterly track over the ground.

    basic vector equation ...

    air vector + wind vector = ground vector

    let \theta = angle of the air vector relative to east.

    east/west components ...

    500\cos{\theta} + 60\cos(-45) = G , where G is the groundspeed east

    north/south components ...

    500\sin{\theta} + 60\sin(-45) = 0

    solve for \theta in the second equation, then use that angle in the first equation to find G.
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  3. #3
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    Hello, Frostking!

    An airplane is flying at 500km/hr in a wind blowing 60 km/hr toward the SE.
    In what direction should the plane head to end up going due east?
    What is the plane's speed relative to the ground?
    Code:
                                  B
                                  * 
                              *   | *
                   500    *       |   * 60
                      *           |h    *
                  *               |       *
              *  θ                |     45 *
        A *   *   *   *   *   *   *   *   *   * C
          : - - - - - - - - b - - - - - - - - :

    The plane flies from A to B at 500 km/hr.
    The wind is blowing from B to C at 60 m/hr.
    The result flight is AC, directly east.

    In the right triangle at the right: . \sin45^o \:=\:\frac{h}{60}\quad\Rightarrow\quad h \:=\:60\sin45^o [1]

    In the right triangle at the left: . \sin\theta \:=\:\frac{h}{500} [2]

    Substitute [1] into [2]: . \sin\theta \:=\:\frac{60\sin45^o}{500} \:=\:0.984852814

    Hence: . \theta \;=\;4.867561141^o

    The direction should be about: . 4.868^o north of east.



    We have: . \angle B \:=\:180^o - 45^o - 4.868^o \:=\:130.132^o

    Law of Sines: . \frac{b}{\sin130.132^o} \:=\:\frac{500}{\sin45^o} \quad\Rightarrow\quad b \:=\:540.6266461

    Its ground speed is about: . 540.63 km/hr.

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