Chapter 11.2, Problem 25E

To determine

To find: The new ground speed and the direction.

Answer to Problem 25E

The new ground velocity is $346.735\text{mph}$ and the direction is $\text{1}\text{.32°}\text{East}\text{Of}\text{North}$ .

Explanation of Solution

The angle made by the plane with the positive x axis is calculated as,

  $\begin{array}{c}\theta =90°-20°\\ =70°\end{array}$

The component form of a vector is given by,

  $\langle x,y\rangle =\langle R\cos \theta ,R\sin \theta \rangle$

Thus, the velocity vector is,

  $\langle 325\cos 70°,325\sin 70°\rangle =\langle 111.157,305.400\rangle$

Now the tail wind acts in the direction $40°$ west of north. The angle made wih the positive x axis is,

  $\begin{array}{c}\theta =40°+90°\\ =130°\end{array}$

The velocity vector is,

  $\langle 40\cos 130°,40\sin 130°\rangle =\langle -25.712,30.642\rangle$

The resultant velocity vector is calculated as,

  $\begin{array}{c}\overrightarrow{v_R}=\langle 111.157+\left(-25.712\right),305.400+30.642\rangle \\ =\langle 85.445,336.042\rangle \end{array}$

The magnitude of the velocity vector is calculated as,

  $\begin{array}{c}v=\sqrt{{85.445}^2+{336.042}^2}\\ =346.735\text{mph}\end{array}$

The direction is calculated as,

  $\begin{array}{c}{\theta }^{\prime }={\tan }^{-1}\left(\frac{336.042}{85.445}\right)\\ =1.321°\text{East}\text{of}\text{North}\end{array}$

Therefore, the new ground velocity is $346.735\text{mph}$ and the direction is $\text{1}\text{.32°}\text{East}\text{Of}\text{North}$ .