Question

The pilot of an airplane notes that the compass indicates a heading due west. The airplane's speed relative to the air is 150 km/h. The air is moving in a wind at 30.0 km/h toward the north. Find the velocity of the airplane relative to the ground.

Short answer

The relative velocity of the airplane with respect to the ground is 153 km/h at an angle$11.3°$ north of west.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The speed of airplane relative to the air is150 km/h.
• The speed of wind is 30.0 km/h toward the north.

The relative velocity

The velocity of the airplane relative to the ground is,

${\overrightarrow{\mathbf{V}}}_{\mathbf{\text{PG}}}\mathbf{=}{\overrightarrow{\mathbf{V}}}_{\mathbf{\text{PA}}}\mathbf{+}{\overrightarrow{\mathbf{V}}}_{\mathbf{\text{AG}}}$

The relative velocity of the airplane with respect to the ground

Consider the vector diagram given below,

The velocity of the airplane relative to the ground is,

${\overrightarrow{V}}_{\text{PG}}={\overrightarrow{V}}_{\text{PA}}+{\overrightarrow{V}}_{\text{AG}}$

Write the expression for the magnitude of the relative velocity of the airplane with respect to the ground using the above figure

$$\begin{gathered} V_{\text{PG}}^2=V_{\text{PA}}^2+V_{\text{AG}}^2 \\ {V}_{\text{PG}}=\sqrt{V_{\text{PA}}^2+V_{\text{AG}}^2} \end{gathered}$$

Here, $V_{\text{PG}}$is the velocity of the airplane relative to the ground, $V_{\text{PA}}$is the velocity of the airplane relative to the air and $V_{\text{AG}}$is the velocity of the air relative to the ground.

Substitute $150 \text{km/h}$for $V_{\text{PA}}$ and $30.0 \text{km/h}$for $V_{\text{AG}}$in above expression to find $V_{\text{PG}}$.

$\begin{array}{rcl}{V}_{\text{PG}}& =& \sqrt{{(150 \text{km/h})}^2+{(30.0 \text{km/h})}^2}\\ & =& 152.97 \text{km/h}\\ & \approx & 153 \text{km/h}\end{array}$

Write the expression for the direction of the relative velocity.

$$\begin{gathered} \tan \theta =\frac{V_{\text{AG}}}{V_{\text{PA}}} \\ \theta ={\tan }^{-1}\left(\frac{V_{\text{AG}}}{V_{\text{PA}}}\right) \end{gathered}$$

Substitute the value in the above expression, and we get,

$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\frac{30.0 \text{km/h}}{150 \text{km/h}}\right)\\ & =& 11.3°\\ & & \end{array}$

Therefore, the velocity of the airplane relative to the ground is 153 km/h at an angle $\begin{array}{rcl}& & 11.3°\end{array}$north of west.