Chapter 6, Problem 47CLT

To determine

To find: The resultant speed and true direction of the airplane.

Answer to Problem 47CLT

Resultant speed $=R\approx 543.9$

Bearing $=\theta ={32.6}^0$

Explanation of Solution

Given information:

Speed of airplane $=500$ kilometers per hour.

Bearing of airplane $={30}^0$

Wind velocity $=150$ kilometers per hour.

Direction of wind $=$ N ${60}^0$ E

The bearing is measured clockwise from North.

Consider 2 vectors of plane and wind.

Adding tip to tail and supplementing angles, we will produce the triangle below:

  

Where

  $R=$ Resultant speed.

  $\theta =$ Bearing

Value of $\theta$ is given by $\theta ={30}^0+A$

Using law of Cosines for solving $R$

We get

  $\begin{array}{l}{R}^2={500}^2+{50}^2-2\left(50\right)\left(500\right)\cos \left({150}^0\right)\\ \Rightarrow R^2=250000+2500-50000\left(\frac{-\sqrt{3}}{2}\right)\\ \Rightarrow R^2=252500+25000\left(\sqrt{3}\right)\\ \Rightarrow R^2=252500+43302.5\\ \Rightarrow R^2=295802.5\\ \Rightarrow R=\sqrt{295802.5}\\ \Rightarrow R\approx 543.9\end{array}$

Therefore,

Resultant speed $=R\approx 543.9$

Using law of sines for solving $A$

We get

  $\begin{array}{l}\frac{\sin \left(A\right)}{50}=\frac{\sin \left({150}^0\right)}{543.9}\\ \Rightarrow \frac{\sin \left(A\right)}{50}=\frac{0.5}{543.9}\\ \Rightarrow \sin \left(A\right)=50\times \frac{0.5}{543.9}\\ \Rightarrow \sin \left(A\right)=\frac{25}{543.9}\\ \Rightarrow \sin \left(A\right)=0.046\\ \Rightarrow A={\sin }^{-1}\left(0.046\right)\\ \Rightarrow A\approx {2.6}^0\end{array}$

Therefore,

Bearing $=\theta ={30}^0+{2.6}^0={32.6}^0$