An aeroplane can fly at \(\quantity{300}{km/h}\) in still air. If a steady \(\quantity{60}{km/h}\) wind is blowing from the direction \(030^\circ\), determine

  1. the direction in which the aeroplane must head in order that it actually flies due North,

We can draw a diagram of the velocity vectors for the wind and the aeroplane.

Velocity vectors of the aeroplane and the background wind

We can now use the sine rule to find the angle \(\alpha\). We have

\[\begin{align*} \frac{\sin \alpha}{60} &= \frac{\sin 150^\circ}{300}\\ \alpha &= \sin^{-1} \frac{\sin 150^\circ}{5} = 5.739...^\circ. \end{align*}\]

The aeroplane needs to take a bearing of \(006^\circ\) (to the nearest degree) to counteract the wind and fly due north.

  1. the aeroplane’s resultant velocity.
Velocity vectors of the aeroplane and the background wind

Using the sine rule

The angle between the velocity vectors of wind and aeroplane is \(30^\circ - \alpha = 24.261...^\circ\). Hence we find

\[\begin{align*} \frac{\sin 150^\circ}{300} &= \frac{\sin 24.261^\circ}{x}\\ x&=300 \frac{\sin 24.261^\circ}{\sin 150^\circ}. \end{align*}\]

This gives a final velocity of the aeroplane of \(v = 246.54...\approx \quantity{247}{km/h}\).


Calculating individual components

The aeroplane’s component of velocity northward is \[v_1=300 \cos \alpha.\]

The wind subtracts from this a velocity component of \[v_2 = 60 \cos 30^\circ.\]

Hence we obtain \(v= v_1-v_2 =300 \cos 5.739 - 60 \cos 30 = 246.53... \approx \quantity{247}{km/h}\), as before.

Would a similar method also work for the first part of the problem?