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Math Help - Vectors

  1. #1
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    Vectors

    A plane is flying 650 mph in a direction of 12 degrees south of west. Meanwhile, a wind is blowing form the north at 30 mph. Find the resulting speed and direction of the plane. Help please, thanks
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  2. #2
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    Re: Vectors

    Attachment 26076
    Quote Originally Posted by limar96 View Post
    A plane is flying 650 mph in a direction of 12 degrees south of west. Meanwhile, a wind is blowing form the north at 30 mph. Find the resulting speed and direction of the plane. Help please, thanks
    Please consult the color-coded diagram I made for you.
    Note that there are two velocity vectors in play here -- the airplane velocity vector is in RED and the wind velocity vector is in GREEN.
    Further, note that I have defined West as the positive X direction and South as the positive Y direction.

    The first step is to determine the X and Y coordinates for each velocity vector. Let's start with the RED airplane velocity vector. Note that its angle is 12 degrees SOUTH OF WEST. That is, find the West direction, and move toward the South direction by 12 degrees. My diagram illustrates this. The X-component of the airplane velocity is adjacent to the 12 degree angle, and the Y-component of the airplane velocity is opposite the 12 degree angle. Thus, V_1_x=\text{Hyp}\cdot\cos{12^{\circ}}=650\cdot\cos  {12^{\circ}} , and V_1_y=\text{Hyp}\cdot\sin{12^{\circ}}=650\cdot\sin  {12^{\circ}} . Succinctly, V_1 = (635.796, 135.143).

    The velocity components for the wind vector (green) are simple -- the X-component is 0 and the Y-component is 30 mph. That is, V_2=(0, 30).

    Summing our forces, we get V_1 + V_2 = (635.796+0, 135.143+30) = (635.796, 165.143)

    The resultant speed is simply then \sqrt{635.796^2 + 165.143^2} \approx 656.9 MPH.

    The resultant direction is \theta = \tan^{-1}{\tfrac{165.143}{635.796}} \approx 14.11^{\circ} SOUTH OF WEST. In other words, 284.11^{\circ}.


    EDIT: The diagram was formerly upside-down. Fixed now.
    Attached Thumbnails Attached Thumbnails Vectors-mhf_physics_diag.jpg  
    Last edited by abender; December 4th 2012 at 10:25 PM.
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  3. #3
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    Re: Vectors

    Hello, limar96!

    Do you have to use vectors?


    A plane is flying 650 mph in a direction of 12 degrees south of west.
    Meanwhile, a wind is blowing form the north at 30 mph.
    Find the resulting speed and direction of the plane.

    Note: 12o represents 12^o.
    Code:
                              |
                              |O
        W - + - - - - - - - - o - - - -
            :      12o  *   * |
            :     *    θ  *   |
          A o     650   *     |
            |102o     *       |
            |       * x       |
         30 |     *           |
            |   *             |
            | *               |
          B o                 |
                              |
    The plane is flying from O to A.
    . . OA = 650,\;\angle WOA = 12^o

    The wind is blowing from A to B.
    . . AB = 30,\;\angle OAB = 102^o

    Let x = AB,\;\theta = \angle AOB


    Law of Cosines:

    . . x^2 \;=\;650^2 + 30^2 - 2(650)(30)\cos102^o \;=\;431,\!508.5559

    Hence: . x \:\approx\:656.89


    Law of Cosines:

    . . \cos\theta \;=\;\frac{650^2 + 656.89^2 - 30^2}{2(650)(656.9)} \;=\;0.99900185

    Hence: . \theta \;\approx\;2.56^o


    Therefore: . \begin{Bmatrix}\;\text{Speed} = 656.89\text{ mph} \\ \text{Direction} = 14.56^o\text{ S of W} \end{Bmatrix}
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