Question

switch wind a plane would fly due east at 150 mph. the wind blowing southeast at a rate of 50 mph. the e wind is blowing at at 45 degree angle from due east. how far off due east path does the wind blow the plane

Answer 1

Answer:

Approximately $10.8^{\circ}$.

Step-by-step explanation:

In this question, consider each velocity as a two-dimensional vector. The two components of these vectors would represent:

• Motion parallel to the east-west direction, and
• Motion perpendicular to the east-west direction.

If an object is travelling at a speed of $v$ at a direction of $\theta$ from east, the velocity vector would be:

$\displaystyle v\, \begin{bmatrix}\cos(\theta) \\ \sin(\theta)\end{bmatrix}$.

Conversely, if it is given that the velocity vector of an object is:

$\begin{bmatrix}x \\ y\end{bmatrix}$,

where $x > 0$ and $y > 0$, the angle between this motion and east would be:

$\displaystyle \theta = \arctan\left(\frac{y}{x}\right)$.

In this question, the velocity of the plane relative to the wind is $150\; \text{mph}$ at $\theta = 0^{\circ}$ from east. Hence, the vector representation of this velocity would be:

$\displaystyle 150\, \begin{bmatrix}\cos(0^{\circ}) \\ \sin(0^{\circ})\end{bmatrix}= 150\, \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}150 \\ 0\end{bmatrix}$.

It is given that the velocity of wind relative to the ground is $50\; \text{mph}$ at $\theta = 45^{\circ}$ from east. Similarly, the vector representation of this velocity would be:

$\displaystyle 50\, \begin{bmatrix}\cos(45^{\circ}) \\ \sin(45^{\circ})\end{bmatrix} = 50 \, \begin{bmatrix} \left(\sqrt{2}\right)/2\\ \left(\sqrt{2}\right)/2\end{bmatrix} = \begin{bmatrix}25\, \sqrt{2} \\ 25\, \sqrt{2}\end{bmatrix}$.

The velocity of the plane relative to the ground is the sum of the following:

• Velocity of the plane relative to the wind, and
• Velocity of the wind relative to the ground.

$\begin{bmatrix}150 \\ 0\end{bmatrix} + \begin{bmatrix}25\, \sqrt{2} \\ 25\, \sqrt{2}\end{bmatrix} = \begin{bmatrix}150 + 25\, \sqrt{2} \\ 25\, \sqrt{2}\end{bmatrix}$.

Since both components of this velocity are greater than zero, the angle between this velocity and east can be found using the inverse tangent function:

$\begin{aligned} \theta = \arctan\left(\frac{25\, \sqrt{2}}{150 + 25\, \sqrt{2}}\right) \approx 10.8^{\circ}\end{aligned}$.

In other words, when measured relative to the ground, velocity of the plane would be approximately $10.8^{\circ}$ from the east.

Answer 2

Answer:

D. 10.8°

Step-by-step explanation:

To solve this problem, we can break down the wind velocity into its eastward and southward components and then find the angle by which the plane is blown off course.

Let's denote:

• $v_p$ as the velocity of the plane without wind (due east)
• $v_w$ as the velocity of the wind (southeast)
• $\theta$ as the angle by which the wind is blowing from due east

Given:

$v_p = 150$ mph (due east)

$v_w = 50$ mph (southeast)

$\theta = 45^\circ$

First, we need to find the eastward and southward components of the wind velocity:

$\textsf{Eastward component of wind} = v_w \times \cos(\theta)$

$= 50 \times \cos(45^\circ)$

$= 50 \times \dfrac{\sqrt{2}}{2}$

$= 25\sqrt{2}$

$\textsf{Southward component of wind} = v_w \times \sin(\theta)$

$= 50 \times \sin(45^\circ)$

$= 50 \times \dfrac{\sqrt{2}}{2}$

$= 25\sqrt{2}$

Now, let's find the resultant velocity of the plane relative to the ground by adding the eastward component of the wind to the velocity of the plane:

$\textsf{Resultant velocity} = v_p + \textsf{Eastward component of wind}$

$= 150 + 25\sqrt{2}$

Using trigonometry, we can find the angle off the due east path:

$\tan(\theta') = \dfrac{\textsf{Southward component of wind}}{\textsf{Resultant velocity}}$

$= \dfrac{25\sqrt{2}}{150 + 25\sqrt{2}}$

$= 0.1907435698$

Now, we can find $\theta'$ by taking the arctan or inverse of tan of the above expression:

$\theta' = \tan^{-1}(0.1907435698) \\\\ \approx 10.79908049 \\\\ \approx 10.8 \textsf{(in nearest tenth)}$

So, the correct answer is option D. 10.8°