Chapter 6.3, Problem 93E

To determine

Find the direction and speed of the wind.

Answer to Problem 93E

The direction of wind is $N21.37°E$

The speed of the wind is,138.74 km/hr

Explanation of Solution

Given:

An airplane travels in the direction of ${148}^{°}$ with an airspeed of 875 kilometers per hour. Due to the wind, its groundspeed and direction are 800 kilometers per hour and ${140}^{°}$ .

Let's consider the following figure

Let the velocity of the airplane is $v_1$ .

The vector $v_1$ in the component form is,

  $\begin{array}{l}{v}_1=875\langle cos148,sin148\rangle \hfill \\ =\langle –742.04,463.68\rangle \hfill \end{array}$

Let the velocity of airplane in the wind is $v$ .

The vector $v$ in the component form is,

  $\begin{array}{l}v=800\langle cos140,sin140\rangle \hfill \\ =\langle –612.84,514.23\rangle \hfill \end{array}$

Let the velocity of the wind is $v_2$ .

  $v=v_1+v_2$

Therefore, the wind velocity is,

  $v_2=v–v_1$

Substituting all the values,

  $\begin{array}{l}v2=\langle –612.84–\left(–742.04\right),514.23–463.68\rangle \hfill \\ =\langle 129.2,50.55\rangle \hfill \end{array}$

Therefore, the first of wind is

  $\begin{array}{cc}\tan \theta & =\frac{50.55}{129.2}\\ & =0.39\\ \theta & =21.37\end{array}$

The direction of wind is $N21.37°E$

The speed of the wind is equal to its magnitude.

Therefore,

  $\begin{array}{cc}\parallel v_2\parallel & =\sqrt{{129.2}^2+{50.55}^2}\\ & =138.74\end{array}$

Therefore, thespeed of the wind is,

138.74 km/hr