Chapter 8, Problem 109A

To determine

The heading and airspeed needed to reach the destination on time.

Answer to Problem 109A

The heading needed to reach the destination on time is $\theta =14.24°$ .

The airspeed needed to reach the destination on time is $v_{\text{p}}=122\text{km}/\text{h}$ .

Explanation of Solution

Given:

A wind is blowing from the west at $v_{\text{w}}=30\text{km}/\text{h}$ .

Distance is 325 km due north.

Time taken will be 2.75 h.

Formula used:

The expression for the velocity of the plane is,

  $v=\frac{d}{t}$

Here, $d$ is the distance and $t$ is the time period.

Calculation:

Consider the heading as the angle $\theta$ travelled eastward from the northward path.

The velocity of the plane with respect to the wind is,

  $\begin{array}{l}{v}_{\text{pw}}=\frac{d}{t}\\ v_{\text{pw}}=\frac{325\text{ km}}{2.75\text{ h}}\\ v_{\text{pw}}=118.2\text{km}/\text{h}\end{array}$

The heading is,

  $\begin{array}{l}\tan \theta =\frac{v_w}{v_{pw}}\\ \theta ={\tan }^{-1}\left(\frac{30\text{km}/\text{h}}{118.2\text{km}/\text{h}}\right)\\ \theta =14.24°\end{array}$

The magnitude of the velocity of the plane as follows:

  $\begin{array}{l}{v}_{\text{p}}=\sqrt{v_{\text{pw}}^2+v_{\text{w}}^2}\\ v_{\text{p}}=\sqrt{{\left(118.2\text{km}/\text{h}\right)}^2+{\left(\text{30 }\text{km}/\text{h}\right)}^2}\\ v_{\text{p}}=122\text{km}/\text{h}\end{array}$

Conclusion:

Thus, the heading needed to reach the destination on time is $\theta =14.24°$ . Also, the airspeed needed to reach the destination on time is $v_{\text{p}}=122\text{km}/\text{h}$ .