Chapter 6.1, Problem 44E

Solution

(a)

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

The component form of the velocity of the airplane is equal to $〈79.88,453.01〉$ .

Given information:

An airplane is flying on a compass heading (bearing) of ${170}^{\circ }$ at $460$ mph. A wind is blowing with the bearing ${200}^{\circ }$ at $80$ mph.

Concept used:

The component form of a vector is $\overrightarrow{a}=〈\left|a\right|\cos \theta ,\left|a\right|\sin \theta 〉$ .

Calculation:

An airplane is flying on a compass heading (bearing) of ${170}^{\circ }$ at $460$ mph. A wind is blowing with the bearing ${200}^{\circ }$ at $80$ mph.

  

Apply the above concept and find the component form of the velocity vector $\overrightarrow{v}$ below:

  $\begin{array}{c}\overrightarrow{v}=〈460\cos {80}^{\circ },460\sin {80}^{\circ }〉\\ =〈79.88,453.01〉\end{array}$

(b)

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

The actual ground speed is equal to $531$ mph and the direction of the plane is equal to ${84}^{\circ }$ .

Given information:

An airplane is flying on a compass heading (bearing) of ${170}^{\circ }$ at $460$ mph. A wind is blowing with the bearing ${200}^{\circ }$ at $80$ mph.

Concept used:

The direction of a vector is $\theta ={\tan }^{-1}\left(\frac{b}{a}\right)$ .

Calculation:

An airplane is flying on a compass heading (bearing) of ${170}^{\circ }$ at $460$ mph. A wind is blowing with the bearing ${200}^{\circ }$ at $80$ mph.

  

The actual ground speed will be

  $\begin{array}{c}\text{Actual}\text{speed}=〈460\cos {80}^{\circ },460\sin {80}^{\circ }〉+〈80\cos {110}^{\circ },80\sin {110}^{\circ }〉\\ =〈79.88,453.01〉+〈-27.36,75.18〉\\ =〈79.88-27.36,453.01+75.18〉\\ =〈52.52,528.19〉\\ =\sqrt{{\left(52.52\right)}^2+{\left(528.19528\right)}^2}\\ \approx 531\end{array}$

Now, the direction of the plane will be

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{528.19}{52.52}\right)\\ \approx {84}^{\circ }\end{array}$