Question

A light plane is headed due south with a speed of 185 km/h relative to still air. After 1.00 h, the pilot notices that they have covered only 135 km, and their direction is not south but$\mathbf{15}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$east of south. What is the wind velocity?

Short answer

The velocity of the wind is $64.82\frac{\mathrm{km}}{\mathrm{h}}$at $32.62°$east of north.

Step-by-step solution

Step 1. Given data

The resultant velocity of the light plane is the vector sum of the aircraft's velocity relative to the wind and speed of the wind.

Given data:

The speed of the plane relative to still air is $v_{\mathrm{pa}}=185\frac{\mathrm{km}}{\mathrm{h}}$.

The time for the flight is $t=1.00\mathrm{h}$.

The traveled distance is $d=135\mathrm{km}$.

Assumptions:

Let $v_{\mathrm{wg}}$be the wind velocity relative to the ground ${\theta }^o$east of north, and$v_{\mathrm{pg}}$ be the plane's velocity relative to the ground.

Step 2. Calculation of the wind velocity and direction

Now, if the plane moves 135 km in 1.00 h, the plane's velocity relative to the ground is $v_{\mathrm{pg}}=135\frac{\mathrm{km}}{\mathrm{h}}$.

From the diagram,

$\begin{array}{c}{\overrightarrow{v}}_{\mathrm{pg}}={\overrightarrow{v}}_{\mathrm{pa}}+{\overrightarrow{v}}_{\mathrm{wg}}\\ {\overrightarrow{v}}_{\mathrm{wg}}={\overrightarrow{v}}_{\mathrm{pg}}-{\overrightarrow{v}}_{\mathrm{pa}}\end{array}$

Then,

$\begin{array}{c}{v}_{\mathrm{wg}}=\sqrt{{\left(v_{\mathrm{pg}}\right)}^2+{\left(v_{\mathrm{pa}}\right)}^2-2v_{\mathrm{pg}}{v}_{\mathrm{pa}}\cos 15°}\\ =\sqrt{{\left(135\frac{\mathrm{km}}{\mathrm{h}}\right)}^2+{\left(185\frac{\mathrm{km}}{\mathrm{h}}\right)}^2-\left(2\times \left(135\frac{\mathrm{km}}{\mathrm{h}}\right)\times \left(185\frac{\mathrm{km}}{\mathrm{h}}\right)\cos 15°\right)}\\ =64.82\frac{\mathrm{km}}{\mathrm{h}}\end{array}$

Now, using the sine formula,

$\begin{array}{c}\frac{v_{\mathrm{wg}}}{\sin 15°}=\frac{v_{\mathrm{pg}}}{\sin \theta }\\ \sin \theta =\frac{v_{\mathrm{pg}}}{v_{\mathrm{wg}}}\times \sin 15°\\ \theta ={\sin }^{-1}\left(\frac{135\frac{\mathrm{km}}{\mathrm{h}}}{64.82\frac{\mathrm{km}}{\mathrm{h}}}\times \sin 15°\right)\\ \theta =32.62°\end{array}$

Hence, the velocity of the wind is $64.82\frac{\mathrm{km}}{\mathrm{h}}$at $32.62°$east or north.