Converting relative velocities

If ${\overrightarrow{v}}_{\text{XY}}$ represents the velocity of X with respect to Y, then

${\overrightarrow{v}}_{\text{AB}}+{\overrightarrow{v}}_{\text{BC}}={\overrightarrow{v}}_{\text{AC}}\text{.}$

In other words, we can eliminate the B and go straight from A to C, provided we know the velocity of B relative to C. Sometimes you will have to rearrange this equation to solve for a different velocity—that’s about as complicated as these problems get.

Example

An airplane has a heading of 245 km/h [N 17º E], and the wind is coming from [S 35º W] at 89.0 km/h. What is the course?

First, we need to familiarize ourselves with airplane terminology:

heading (${\overrightarrow{v}}_{\text{PA}}$)

velocity of the plane with respect to the air

wind (${\overrightarrow{v}}_{\text{AE}}$)

velocity of the air with respect to the Earth

course/groundspeed (${\overrightarrow{v}}_{\text{PE}}$)

velocity of the plane with respect to the Earth

The velocity of the wind is a vector pointing in the direction in which it is going. Since the wind is coming from [S 35º W], it is going to [N 35º E].

Since ${\overrightarrow{v}}_{\text{PE}}={\overrightarrow{v}}_{\text{PA}}+{\overrightarrow{v}}_{\text{AE}}\text{,}$ the course is

$245\text{km/h [N 17º E]}+89.0\text{km/h [N 35º E]}=331\text{km/h [N 22º E]}\text{.}$