Chapter 9.4, Problem 84AYU

To determine

To find: thedirectionheading should be maintained to head directly east and to find theactual speed of the aircraft.

Answer to Problem 84AYU

  $N{83.5}^{\circ }E;\alpha =267.7\text{mph}$ .

Explanation of Solution

Given:

Speed of windis $40mile/hr$ from the northwest.

Average speed of an airplane is $250mile/hr$ .

Calculation:

According to the question,

  $\begin{array}{l}{v}_g=v_w+v_a\\ v_a=250\left(\cos \alpha i+\sin \alpha j\right)\\ v_w=40\left(\frac{i-j}{\Vert i-j\Vert }\right)\\ v_w=40\left(\frac{i-j}{\sqrt{1+1}}\right)\\ v_w=\frac{40}{\sqrt{2}}\left(i-j\right)\\ v_w=20\sqrt{2}\left(i-j\right)\end{array}$

Therefore,

  $\begin{array}{l}{v}_g=v_w+v_a\\ v_a=250\left(\cos \alpha i+\sin \alpha j\right)+20\sqrt{2}i+20\sqrt{2}j\\ =ai\end{array}$

  $\begin{array}{l}250\sin \alpha -20\sqrt{2}=0\\ 250\sin \alpha =20\sqrt{2}\\ \sin \alpha =\frac{20\sqrt{2}}{250}=0.11314\\ \alpha ={6.5}^{\circ }\end{array}$

  $\begin{array}{l}\cos \theta =\frac{v_w\cdot j}{\Vert v_w\Vert \cdot \Vert j\Vert }\\ =\frac{-3\left(0\right)+\sqrt{391}}{20}=\frac{\sqrt{391}}{20}=0.9887\\ \theta ={8.6}^{\circ }\end{array}$

Planes heading:

  $\begin{array}{l}N{83.5}^{\circ }E\\ 250\cos \left({6.5}^{\circ }\right)i+20\sqrt{2}i=ai\end{array}$

And the planes speed is $\alpha =267.7\text{mph}$

Hence, the planes headlineand the speed of plane is $N{83.5}^{\circ }E;\alpha =267.7\text{mph}$ .