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Math Help - Vector Problem

  1. #1
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    Post Vector Problem

    An airplane flies horizontally from east to west at 304 mi/hr relative to the air. If it flies in a steady 24 mi/hr wind that blows horizontally toward the southwest (45 degrees south of west), find the speed and direction of the airplane relative to the ground.
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  2. #2
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    Re: Vector Problem

    Hello, Jessica!

    An airplane flies horizontally from east to west at 304 mi/hr relative to the air.
    If it flies in a steady 24 mi/hr wind that blows horizontally toward the southwest (45o south of west),
    find the speed and direction of the airplane relative to the ground.

    Code:
                             304
            - - - B *   *   *   *   *   * A
              45o * 135o          *
                *           *
           24 *       *
            *   *
         C*
    The plane is flying from A to B at 304 mph.
    But there is wind blowing from B to C at 24 mph.
    Hence, the plane's flight is from A to C.

    Law of Cosines:
    . AC^2 \;=\;24^2 + 304^2 - 2(24)(304)\cos135^o \;=\; 103,\!210.1002

    Hence: . AC \;=\;321.4188889 \;\approx\;321.4\text{ mph.}


    Law of Cosines:
    . . \cos A \;=\;\frac{304^2 + 321.4^2 - 24^2}{2(204)(321.4)} \;=\;0.998601718

    Hence: . A \;=\;3.030300248 \;\approx\;3^o

    The direction is 3o south of west.


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Um . . . it said vectors, didn't it?


    \overrightarrow{AB} \;=\;\langle 304\cos180^o,\:304\sin180^o\rangle \;=\;\big\langle -304,\:0\big\rangle

    \overrightarrow{BC} \;=\;\langle 24\cos225^o,\:24\sin225^o\rangle \;=\;\left\langle \text{-}12\sqrt{2},\:\text{-}12\sqrt{2}\right\rangle

    \overrightarrow{AC} \;=\;\overrightarrow{AB} + \overrightarrow{BC} \;=\; \big\langle \text{-}304-12\sqrt{2},\:\text{-}12\sqrt{2}\big\rangle


    Then: . |\overrightarrow{AC}| \;=\;\sqrt{\left(\text{-}304 - 12\sqrt{2}\right)^2 + \left(\text{-}12\sqrt{2}\right)^2}

    Hence: . |\overrightarrow{AC}| \;=\;\sqrt{103,\!310.1022} \;=\;321.418889 \;\approx\;321.4\text{ mph.}


    \tan\theta \;=\;\frac{\text{-}12\sqrt{2}}{\text{-}304-12\sqrt{2}} \;=\;0.052872695

    . . \theta \;=\;3.026561265 \;\approx\;3^o


    I'm gratified that my answers match!
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  3. #3
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    Re: Vector Problem

    Thanks, this really helped!
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