Question

A pilot with an airspeed (speed with respect to air) of 120 km/hr wishes to fly due north. A 40 km/h wind is blowing from the northeast. In what direction should she head and what will her ground speed be (speed with respect to the ground)?

Answer 1

Let $\vec v_{P/G}$ denote the vector for velocity of the plane relative to the ground, and $v_{P/G}$ the corresponding speed. The pilot wants to be heading due north, taking the effect of the wind into account, so that this vector has components

$\vec v_{P/G}=v_{P/G}\,\vec\jmath$

The wind (denoted by $A$ for air) is blowing from the northeast, which means its direction is southwest, so that the vector for its velocity relative to the ground is

$\vec v_{A/G}=\left(40\,\dfrac{\rm km}{\rm h}\right)\cos225^\circ\,\vec\imath+\left(40\,\dfrac{\rm km}{\rm h}\right)\sin225^\circ\,\vec\jmath$

The plane has an airspeed of 120 km/h, but its direction $\theta$ is unknown, so the vector for its velocity relative to the air is

$\vec v_{P/A}=\left(120\,\dfrac{\rm km}{\rm h}\right)\cos\theta\,\vec\imath+\left(120\,\dfrac{\rm km}{\rm h}\right)\sin\theta\,\vec\jmath$

We have

$\vec v_{P/G}=\vec v_{P/A}+\vec v_{A/G}$

The $\vec\imath$ components tell us that we should have

$\left(120\,\dfrac{\rm km}{\rm h}\right)\cos\theta+\left(40\,\dfrac{\rm km}{\rm h}\right)\cos\225^\circ=0\implies\theta=76.4^\circ\text{ or }\theta=284^\circ$

(both relative to east) By drawing a diagram like the one attached, you'll see that the smaller angle is the correct choice. Then by matching the $\vec\jmath$ components, we can solve for $v_{P/G}$:

$v_{P/G}=\left(120\,\dfrac{\rm km}{\rm h}\right)\sin76.4^\circ+\left(40\,\dfrac{\rm km}{\rm h}\right)\sin225^\circ\approx88\,\dfrac{\rm km}{\rm h}$