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Math Help - resultant velocity

  1. #1
    Senior Member euclid2's Avatar
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    resultant velocity

    A plane is flying E40^oN at a speed of 600km/h. The wind is blowing at 10km/h in the direction N20^oW. Determine the resultant velocity of the plane.

    i got 601.81 km/h and 0.93^o

    correct?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    I've got 603.49 km/h and bearing 049.1^o

    What I did was:

    Resultant = \sqrt{10^2 + 600^2 - 2(600)(10)cos(110)}

    Last edited by Unknown008; September 13th 2010 at 11:20 AM. Reason: Typo with the magnitude which is in fact 603.4933651
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  3. #3
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    Hello, euclid2!

    I got different answers . . .


    \text{A plane is flying }E40^oN\text{ at a speed of 600 km/h}.
    \text{The wind is blowing at 10 km/h in the direction }N20^oW
    \text{Determine the resultant velocity of the plane.}

    \text{i got:  601.81 km/h and }0.93^o

    In the diagram, "d" represents "degrees".


    Code:
                         C
                         o           Q
                        *  *  10     :
                       *     *       :
                      *        *     :
                     *           *20d:
          P         *          70d * :
          :        *    W - - - - - -o B
          :       *           40d *
          :      *             *
          :     *           *
          :    *         * 600
          :   *       *
          :  *     *
          : *   *
          :* * 40d
        A o - - - - - - - - - - - E

    The plane starts at \,A and flies to \,B.
    . . \angle BAE \,=\, 40^o \,=\,\angle W\!BA
    . . AB \,=\,600

    The wind blows from \,B to \,C,
    . . BC \,=\,10
    . . \angle CBQ \,=\,20^o \quad\Rightarrow\quad \angle CBW = 70^o


    In \Delta ABC\!:\;\;AB = 600,\;BC = 10,\;\angle ABC = 110^o

    Law of Cosines: . AC^2 \:=\:600^2 + 10^2 - 2(600)(10)\cos110^o \:=\:364,\!204.2417

    Therefore: . AC \:=\:603.4933651 \quad\Rightarrow\quad \boxed{AC \:\approx\:603.5}


    \cos(\angle C\!AB) \;=\;\dfrac{600^2 + 603.5^2 - 10^2}{2(600)(603.5)} \;=\;0.999878832

    Hence: . \angle C\!AB \;=\;0.891941392^o  \;\approx\;0.89^o


    Then: . \angle P\!AC \;=\;90^o - 0.89^o - 40^o \;=\;49.11^o


    Therefore, the heading of the plane is: . N49.11^oE

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