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Just as forces can be summed to find the resultant (net force), so too can velocities be summed to find the net velocity.

That is, a net velocity is a single velocity that represents the contributions of multiple velocities, i.e., an aircraft flying with or against the wind.

Example 1

An airplane heading due north at \(400\) km/h encounters a \(50\) km/h wind from north \(60^\circ\) east. Determine the resultant velocity (magnitude and heading) of the plane.

Solution

We first draw the vector diagram.

The resultant velocity of the plane will be the vector sum of both the wind's velocity and the plane's velocity.

This equates to solving for the resultant, \({\color{BrickRed}\vec{r}}\).

Solving for the magnitude of \({\color{BrickRed}\vec{r}}\) using the cosine law:

\[\begin{align*} \lvert {\color{BrickRed}\vec{r}}\rvert^2 &= 50^2+400^2-2(50)(400)\cos(60^\circ) \\ \lvert {\color{BrickRed}\vec{r}} \rvert &\approx 377.5 \text{ km/h}\end{align*}\]

Example 1 — Continued

An airplane heading due north at \(400\) km/h encounters a \(50\) km/h wind from north \(60^\circ\) east. Determine the resultant velocity (magnitude and heading) of the plane.

Solution

\[\begin{align*} \lvert {\color{BrickRed}\vec{r}}\rvert^2 &= 50^2+400^2-2(50)(400)\cos(60^\circ) \\ \lvert {\color{BrickRed}\vec{r}} \rvert &\approx 377.5 \text{ km/h}\end{align*}\]

Solving for the direction using the sine law:

\[\begin{align*}\dfrac{\sin(\theta)}{50} &= \dfrac{\sin(60^\circ)}{377.5} \\ \theta &\approx 7^\circ \end{align*}\]

Therefore, the resultant velocity of the plane is approximately \(377.5\) km/h on a course of north \(7^\circ\) west.

The heading is the direction the nose of the plane is pointed. The course or track is the resultant.

We call \(\theta\) the drift angle.

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Möbius

University of Waterloo