resultant velocity

**euclid2** · September 13th 2010, 10:07 AM

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?

**Unknown008** · September 13th 2010, 11:06 AM

I've got 603.49 km/h and bearing $049.1^o$

What I did was:

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

**Soroban** · September 13th 2010, 11:16 AM

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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