In what direction should the pilot aim the plane in Problem 44 so that it flies due south?

Relative velocity may be defined as the time rate of change of the relative position of a particle with respect to another particle. In the standard international system, it is measured in meter second inverse.

The velocity of the plane relative to the air is$v_{pa}=688\mathrm{km}/\mathrm{h}$.

The velocity of the air relative to the ground is$v_{ag}=90.0\mathrm{km}/\mathrm{h}.$

The horizontal component of the velocity of the plane relative to the air can be calculated as

$\begin{array}{c}{\left(v_{pa}\right)}_x=-v_{pa}\sin \theta \\ {\left(v_{pa}\right)}_x=-\left(688\mathrm{km}/\mathrm{h}\right)\sin \end{array}$

The horizontal component of the velocity of the air relative to the ground can be calculated as

$\begin{array}{c}{\left(v_{ag}\right)}_x=v_{ag}\cos 45°\\ {\left(v_{ag}\right)}_x=\left(90.0\mathrm{km}/\mathrm{h}\right)\cos 45°\\ {\left(v_{ag}\right)}_x=63.63\mathrm{km}/\mathrm{h}\end{array}$

The horizontal component of the velocity of the plane relative to the ground can be calculated as

${\left(v_{pg}\right)}_x=0$.

Equating the horizontal components of the velocities,

$\begin{array}{c}{\left(v_{pg}\right)}_x={\left(v_{pa}\right)}_x+{\left(v_{ag}\right)}_x\\ 0=-\left(688\mathrm{km}/\mathrm{h}\right)\sin \theta +63.63\mathrm{km}/\mathrm{h}\\ \theta =5.28°\end{array}$

Thus, the pilot should aim the plane at $5.28°$west of south.