Chapter 8.6, Problem 40E

To determine

To find: the ground speed and direction of the plane.

Answer to Problem 40E

  $\begin{array}{c}\text{Speed}=162.2\text{ km/h}\\ \text{Direction}=47°53\text{'}\text{ south of east}\end{array}$

Explanation of Solution

Given information:

Speed of plane = 150 km/h

Direction of plane = $30°\text{ south of east}$

Speed of wind = 50 km/h

Direction of wind = $25°\text{ west of south}$

Calculation:

Let the vector of plane and wind be $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ respectively. Express the angles in respect to the $+x$ axis.

Evaluate $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ as shown below.

  $\begin{array}{c}\overrightarrow{v_1}=\left(150\cos 330°\right)\overrightarrow{i}+\left(150\sin 330°\right)\overrightarrow{j}\\ \overrightarrow{v_2}=\left(50\cos 245°\right)\overrightarrow{i}+\left(50\sin 245°\right)\overrightarrow{j}\end{array}$

The ground speed of plane will be evaluated as follows:

  $\begin{array}{c}\left|\overrightarrow{v_1}+\overrightarrow{v_2}\right|=\sqrt{{\left(150\cos 330°+50\cos 245°\right)}^2+{\left(150\sin 330°+50\sin 245°\right)}^2}\\ =\sqrt{{\left(108.77\right)}^2+{\left(-120.32\right)}^2}\\ =\sqrt{26307.82}\\ =162.2\text{km/h}\end{array}$

Thus the ground speed of the plane is $162.2\text{km/h}$ .

The ground direction of plane will be evaluated as follows:

  $\begin{array}{c}\tan \theta =\frac{150\sin 330°+50\sin 245°}{150\cos 330°+50\cos 245°}\\ =\frac{-120.32}{108.77}\\ =-1.1062\\ \theta ={\tan }^{-1}\left(-1.1062\right)\\ =-47.88°\\ =-47°53\text{'}4\text{'}\text{'}\\ =47°53\text{'}4\text{'}\text{'}\text{ south of east}\end{array}$

Thus the ground direction of the plane is $47°53\text{'}4\text{'}\text{'}\text{ south of east}$ .