Chapter 10, Problem 50RE

To determine

To calculate: To solve the navigation of an airplane.

Answer to Problem 50RE

The flying direction of the airplane is approximately ${7.141088}^{\circ }$ north of east.

Explanation of Solution

Given information: The given curve as:

An airplane, flying in the direction east of $80°$ north at $\text{325}\text{mph}$ in still air , encounters a $\text{55}\text{mph}$ tail wind acting in the direction $100°$ east of north.

The airplane maintains its compass heading but, because of the wind, acquires a new ground speed and direction. What are they?

Calculation:

The component form of a vector $v$ are as follows:

  $v=〈\mid v\mid cos\theta ,\mid v\mid sin\theta 〉$

Where $\left|v\right|$ is the magnitude of a vector $v$ and $\theta$ is the direction angle of a vector $v$ with the positive $x-\text{axis}$ .

Now, also know that the positive $x-$ axis represents the east, and the positive $y-\text{axis}$ represents the north.

Since we have that the flying direction of the airplane in the still air is $80°$ east of north, then have that the flying direction of the airplane in the still air is

  ${90}^{\circ }-{80}^{\circ }={10}^{\circ }$ north of east, which means that the flying direction of the airplane in the still air makes an angle of ${10}^{\circ }$ with the positive $x-\text{axis}$ . The speed of the airplane in the still air is equal to $\text{325}\text{mph}$

The speed is magnitude of a velocity vector, then we have that the velocity vector of the flying direction of the airplane in the still air is equal to as:

  $v_{a_0}\text{}\text{}=〈325cos({10}^{\circ }),325sin({10}^{\circ })〉\approx 〈320.0625,56.4356〉$

The positive $y-\text{axis}$ represents the east. The direction of the wind is ${100}^{\circ }$ east of north. Then, the direction of the wind is ${90}^{\circ }-{100}^{\circ }=-{10}^{\circ }$ north of east, which means that the direction of the wind makes an angle of ${10}^{\circ }$ with the negative $x-\text{axis}$ . The speed of the wind is equal to $55-$ mph. Since we know that the speed is magnitude of a velocity vector, then we have that the velocity vector of the direction of the wind is equal to

  $v_w\text{}=〈55cos(-{10}^{\circ }),55\text{sin}(-{10}^{\circ })〉\approx 〈54.1644,-9.5506〉.$

Then, if,

  

According to the previous results, we can conclude that the velocity vector of the flying direction of the airplane is equal to

  $\begin{array}{l}{v}_{a_0}+v_w\text{}=〈54.1644,-9.5506〉+〈320.0625,56.4356〉\\ v_{a_0}+v_w=〈374.2269,46.885〉\end{array}$

The magnitude of a vector $v=〈v_1\text{},v_2〉\text{}$ is equal to

  $\mid v\mid =\sqrt{v_1{}^2\text{}+v_2{}^2\text{}}.$

The speed of the airplane can be calculated as:

  $\begin{array}{l}\mid v_a\text{}\mid \approx \sqrt{{(374.2269)}^2+{(46.885)}^2}\text{}\\ \left|v_a\right|\approx \sqrt{140045.77268361+2198.203225}\\ \left|v_a\right|\approx \sqrt{142243.97590861}\\ \left|v_a\right|\approx \text{377}\text{.15245 mph}\text{.}\end{array}$

  $\begin{array}{l}\because \text{sin}\theta =\frac{v_2}{\mid v\mid }\text{}\text{}\text{and} \text{cos}\theta =\frac{v_1}{\mid v\mid }\text{}\text{}\\ \therefore \text{tan}\theta =\frac{v_1}{v_2}\text{}\text{}\end{array}$

The angle of the velocity vector is:

  $\begin{array}{l}\text{tan}\theta \approx \frac{46.885}{374.2269}\text{}\\ \theta \approx {\text{tan}}^{-1}(0.1252849\text{})\\ \theta \approx {7.141088}^{\circ }\end{array}$

because have that $\sin \theta >0\text{and}\cos \theta >0$ .

Then we have that the flying direction of the airplane is approximately ${7.141088}^{\circ }$ north of east