1. ## 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

2. ## Re: Vectors

Attachment 26076
Originally Posted by limar96
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
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.

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