Chapter 6, Problem 75RE

Solution

a)

To find: The component form of the velocity of the airplane.

The component form of the velocity of the airplane is $\left(-463.644,124.233\right)$ .

Given information:

An airplane is flying on a bearing of $285°$ at $480\text{ mph}$ . A wind is blowing with the bearing of $265°$ at $30\text{ mph}$ .

Formula used:

Let a vector $v$ has direction angle $\theta$ , the component of $v$ can be calculated as ,

  $v=\left(\left|v\right|\cos \theta ,\left|v\right|\sin \theta \right)$

Bearing is the angle that the line of travels makes with due north and measured clockwise.

Calculation:

Let $v$ be the velocity of airplane. A bearing of $285°$ is equivalent to a direction angle of $15°$ .

Magnitude of vector $v$ is the speed of airplane, can be written as,

  $\left|v\right|=480$

Horizontal component of $v$ is $480\cos 15°$ and the vertical component is $480\sin 15°$ .

Mathematically, it is written as,

  $\begin{array}{l}\overrightarrow{v}=\left(480\cos 15°\right)\left(-\widehat{i}\right)+\left(480\sin 15°\right)\widehat{j}\\ \overrightarrow{v}=463.644\left(-\widehat{i}\right)+124.233\widehat{j}\\ v\approx \left(-463.644,124.233\right)\end{array}$

Therefore, the component form of the velocity of the airplane is $\left(-463.644,124.233\right)$ .

b)

To find: The actual speed and direction of the airplane.

The actual speed is $508.293\text{ mph}$ and the direction of the airplane is $283.843°$ with the bearing.

Given information:

An airplane is flying on a bearing of $285°$ at $480\text{ mph}$ . A wind is blowing with the bearing of $265°$ at $30\text{ mph}$ .

Formula used:

If $v\left(a,b\right)$ where $a$ and $b$ are the horizontal and vertical component respectively, then,

  $\left|v\right|=\sqrt{a^2+b^2}$

  $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$

Calculation:

Let $v_w$ be the velocity of wind. A bearing of $265°$ is equivalent to a direction angle of $5°$ .

Magnitude of vector $v_w$ is the speed of wind, can be written as,

  $\left|v_w\right|=30$

Horizontal component of $v_w$ is $30\cos 5°$ and the vertical component is $30\sin 5°$ .

Mathematically, it is written as,

  $\begin{array}{l}{\overrightarrow{v}}_w=\left(30\cos 5°\right)\left(-\widehat{i}\right)+\left(30\sin 5°\right)\left(-\widehat{j}\right)\\ {\overrightarrow{v}}_w=29.885\left(-\widehat{i}\right)+2.614\left(-\widehat{j}\right)\\ v_w\approx \left(-29.885,-2.614\right)\end{array}$

It is calculated that the component form of the velocity of the airplane is $\left(-463.644,124.233\right)$ .

Components of resultant vector for airplane can be calculated as,

  $\begin{array}{l}{\overrightarrow{v}}_r=\overrightarrow{v}+{\overrightarrow{v}}_w\\ {\overrightarrow{v}}_r=463.644\left(-\widehat{i}\right)+124.233\widehat{j}+29.885\left(-\widehat{i}\right)+2.614\left(-\widehat{j}\right)\\ {\overrightarrow{v}}_r=493.529\left(-\widehat{i}\right)+121.619\widehat{j}\end{array}$

Compute the magnitude of resultant vector.

  $\begin{array}{c}\left|v_r\right|=\sqrt{{\left(493.529\right)}^2+{\left(121.619\right)}^2}\\ =\sqrt{243570.873+14791.181}\\ =\sqrt{258362.054}\\ =508.293\end{array}$

Calculate the direction of resultant vector.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ ={\tan }^{-1}\left(\frac{121.619}{493.529}\right)\\ =13.843°\end{array}$

Let $d$ be the direction of airplane. Compute $d$ .

  $\begin{array}{c}d=270°+13.843°\\ =283.843°\end{array}$

Therefore, the actual speed is the magnitude of resultant vector that is $508.293\text{ mph}$ and the direction of the airplane is $283.843°$ with the bearing.