Chapter 11, Problem 50RE

To determine

To calculate: The ground speed and direction of the airplane.

Answer to Problem 50RE

The required ground speed and direction of the airplane are 591.982 mph and $81.821°$ respectively.

Explanation of Solution

Given information:

The direction of the airplane = $80°$ east of north

Speed of the airplane = 540 mph

Speed of the tail = 55 mph

Direction of the tail = $100°$ east of north

Calculation:

Consider the magnitude and direction angle of a vector be $R$ and $\theta$ respectively that is,

  $x=R\cos \theta$

  $y=R\sin \theta$

Write the component form of the vector as:

  $\langle x,y\rangle =\langle R\cos \theta ,R\sin \theta \rangle$

So the velocity vector is,

  $540\langle \cos 10°,\sin 10°\rangle =\langle 540\cos 10°,540\sin 10°\rangle$

The direction of the tail wind is $100°$ east of north. The angle between the tail’s direction and the positive x -axis is $90°-100°=-10°$ .

Write the velocity vector as:

  $55\langle \cos \left(-10°\right),\sin \left(-10°\right)\rangle \approx \langle 55\cos \left(-10°\right),55\sin \left(10°\right)\rangle$

It is seen that the sum of two velocity vectors is the resultant vector and the sum of two vectors $\overrightarrow{u}=\langle {\overrightarrow{u}}_1,{\overrightarrow{u}}_2\rangle$ and $\overrightarrow{v}=\langle {\overrightarrow{v}}_1,{\overrightarrow{v}}_2\rangle$ is,

  $\begin{array}{c}\overrightarrow{u}+\overrightarrow{v}=\langle {\overrightarrow{u}}_1,{\overrightarrow{u}}_2\rangle +\langle {\overrightarrow{v}}_1,{\overrightarrow{v}}_2\rangle \\ =\langle {\overrightarrow{u}}_1+{\overrightarrow{v}}_1,{\overrightarrow{u}}_2+{\overrightarrow{v}}_2\rangle \end{array}$

Thus, the resultant vector is:

  $\begin{array}{c}\langle 540\cos 10°,540\sin 10°\rangle +\langle 55\cos \left(-10°\right),55\sin \left(-10°\right)\rangle =\left(595\cos 10°,485\sin 10°\right)\\ =\langle 585.961,84.219\rangle \end{array}$

Also, it is known that the magnitude of the velocity vector is the speed and the magnitude of a vector $\langle \overrightarrow{a},\overrightarrow{b}\rangle$ is:

  $\left|\langle \overrightarrow{a},\overrightarrow{b}\rangle \right|=\sqrt{{\overrightarrow{a}}^2+{\overrightarrow{b}}^2}$

The magnitude of the velocity vector is:

  $\begin{array}{c}\sqrt{{\left(585.961\right)}^2+{\left(84.219\right)}^2}=\sqrt{343350.2935+7092.839961}\\ =\sqrt{350443.1335}\\ \approx 591.982\text{mph}\end{array}$

Also, it is known that the direction angle $\theta$ is $\tan \theta =\frac{a}{b}$ .

So, the direction of the airplane is:

  $\begin{array}{c}{\tan }^{-1}\left(\frac{585.961}{84.219}\right)={\tan }^{-1}\left(6.9575\right)\\ \approx 81.821°\end{array}$

Hence, the required ground speed and direction of the airplane are 591.982 mph and $81.821°$ respectively.