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Thread: resultant velocity

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

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

    i got $\displaystyle 601.81 km/h$ and $\displaystyle 0.93^o$

    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 $\displaystyle 049.1^o$

    What I did was:

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

    Last edited by Unknown008; Sep 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 . . .


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

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

    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 $\displaystyle \,A$ and flies to $\displaystyle \,B.$
    . . $\displaystyle \angle BAE \,=\, 40^o \,=\,\angle W\!BA$
    . . $\displaystyle AB \,=\,600$

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


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

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

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


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

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


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


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

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