Question

6. An airplane whose air speed is 650km/h is supposed to fly in straight path 35° north of
east. But a steady 95 km/h wind is blowing from north. In what direction should the
plane head and what is the plane resultant speed?

Answer 1

This problems is related to the notion of addition of vectors.

The resultant velocity $\vec{v}_{rel} = \vec{v}_{wind} + \vec{v}_{plane}.$ We know that the wind counteracts to the movement of

the plane, so the resultant velocity will be smaller than the initial velocity $v_{plane}$ . Also, we conclude that

the angle in question should be larger than $35^\circ$ .

The angle between $\vec{v}_{wind}$ and $\vec{v}_{res}$ should be $\alpha = 90^\circ + 35^\circ = 125^\circ.$ We may calculate the angle $\beta$ between $\vec{v}_{res}$ and $\vec{v}_{plane}$ using the law of sines

$\dfrac{v_{plane}}{\sin\alpha} = \dfrac{v_{wind}}{\sin\beta}.$

Therefore, $\sin\beta = \dfrac{v_{wind}}{v_{plane}}\sin\alpha = \dfrac{95}{650} \sin125^\circ \approx 0.12, \quad \beta = 6.9^\circ.$ The third angle $\gamma$ between $v_{wind}$ and $v_{plane}$ is $\gamma = 180^\circ - \alpha-\beta = 180^\circ - 125^\circ-6.9^\circ= 48.1^\circ.$

Therefore, we may calculate $v_{res}$ using the law of cosines

$v_{res}^2 = v_{wind}^2 + v_{plane}^2 - 2v_{wind}v_{plane}\cos\gamma.$

$v_{res}\approx 590\,km/h.$

The angle in question is $90^\circ - \gamma = 90^\circ-48.1 = 41.9^\circ$ north of east.