Chapter 6, Problem 74RE

Solution

a)

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

The component form of the velocity of the airplane is $\left(531.796,93.770\right)$ .

Given information:

An airplane is flying on a bearing of $80°$ at $540\text{ mph}$ . A wind is blowing with the bearing of $100°$ at $\text{56 mph}$ .

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 $80°$ is equivalent to a direction angle of $10°$ .

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

  $\left|v\right|=540$

Horizontal component of $v$ is $540\cos 10°$ and the vertical component is $540\sin 10°$ .

Mathematically, it is written as,

  $\begin{array}{l}\overrightarrow{v}=\left(540\cos 10°\right)\widehat{i}+\left(540\sin 10°\right)\widehat{j}\\ \overrightarrow{v}=531.796\widehat{i}+93.770\widehat{j}\\ v\approx \left(531.796,93.770\right)\end{array}$

Therefore, the component form of the velocity of the airplane is $\left(531.796,93.770\right)$ .

b)

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

The actual speed is the magnitude of resultant vector that is $594.999\text{ mph}$ and the direction of the airplane is $80°$ with the bearing.

Given information:

An airplane is flying on a bearing of $80°$ at $540\text{ mph}$ . A wind is blowing with the bearing of $100°$ at $\text{56 mph}$ .

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 $100°$ is equivalent to a direction angle of $10°$

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

  $\left|v_w\right|=55$

Horizontal component of $v_w$ is $55\cos 10°$ and the vertical component is $55\sin 10°$ .

Mathematically, it is written as,

  $\begin{array}{l}{\overrightarrow{v}}_w=\left(55\cos 10°\right)\widehat{i}+\left(55\sin 10°\right)\widehat{j}\\ {\overrightarrow{v}}_w=54.164\widehat{i}+9.55\widehat{j}\\ v_w\approx \left(54.164,9.55\right)\end{array}$

It is calculated that the component form of the velocity of the airplane is $\left(531.796,93.770\right)$

Components of resultant vector for airplane can be calculated as,

  $\begin{array}{l}{\overrightarrow{v}}_r=\overrightarrow{v}+{\overrightarrow{v}}_w\\ {\overrightarrow{v}}_r=531.796\widehat{i}+93.770\widehat{j}+54.164\widehat{i}+9.55\widehat{j}\\ {\overrightarrow{v}}_r=585.96\widehat{i}+103.32\widehat{j}\end{array}$

Compute the magnitude of resultant vector.

  $\begin{array}{c}\left|v_r\right|=\sqrt{{\left(585.96\right)}^2+{\left(103.32\right)}^2}\\ =\sqrt{343349.121+10675.022}\\ =\sqrt{354024.143}\\ =594.999\end{array}$

Calculate the direction of resultant vector.

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{b}{a}\right)\\ ={\tan }^{-1}\left(\frac{103.32}{585.96}\right)\\ =10°\end{array}$

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

  $\begin{array}{c}d=90°-10°\\ =80°\end{array}$

Therefore, the actual speed is the magnitude of resultant vector that is $594.999\text{ mph}$ and the direction of the airplane is $80°$ with the bearing.