<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt

{\bf Question}

An aeroplane has velocity ${\bf v}_A$ relative to still air.  If
it flies in a wind with velocity ${\bf v}_W$ what is the velocity
of the aeroplane relative to the ground?

\medskip

An aeroplane moves in a northwesterly direction at 500kmh$^{-1}$
due to the fact that there is an easterly wind (i.e. from the
east) of 50kmh$^{-1}$.  Determine how fast and in what direction
the aeroplane would have traveled if there was no wind.

\vspace{.25in}

{\bf Answer}

$\ds {\bf v} = {\bf v}_A + {\bf v}_W$ as ${\bf v}_A$ is relative
to still air; when there is a wind the plane has ${\bf v}_A$
relative to the wind.

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\put(1,4){\line(1,0){2}}

\put(3,4){\vector(-1,0){1}}

\put(2,4.35){\makebox(0,0){50}}

\put(2,2){\makebox(0,0){500}}

\put(4.5,2){\makebox(0,0){{\bf v}$_A$}}

\put(4.2,1.2){\makebox(0,0){$\alpha$}}

\put(4.3,0.3){\makebox(0,0){$\frac{\pi}{4}$}}

\put(1.75,3.6){\makebox(0,0)[l]{$\frac{\pi}{4}$}}

\end{picture}
\end{center}
\bigskip

cosine rule: \begin{eqnarray*} \ds {\bf v}_A^2 &amp; = &amp; 500^2 + 50^2
- 2 \times 50 \times 500 \cos \frac{\pi}{4} \\ \Rightarrow {\bf
v}_A &amp; = &amp; 466 {\rm kmh}^{-1}\end{eqnarray*}

sine rule:
\begin{eqnarray*} \ds \frac{50}{\sin \alpha} &amp; = &amp; \frac{466}{\sin
\frac{\pi}{4}} \\ \Rightarrow \sin \alpha &amp; = &amp; \frac{25 \sqrt
2}{466} \\ \Rightarrow \alpha &amp; = &amp; 0.076  \textrm{ rads}
\end{eqnarray*}




\end{document}
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