91Ó°ÊÓ

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

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

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,

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

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