Average velocity and average speed are related, but they are not the same quantity. The key difference is that
average velocity is based on displacement, while average speed is based on
total distance traveled. Because displacement is a vector and distance is a scalar, average velocity is a vector
quantity and average speed is a scalar quantity.
Average velocity
If an object has displacement vector
\(\Delta \vec r\)
over an elapsed time
\(t\),
then the average velocity vector is
Main formula.
\[
\begin{aligned}
\vec v_{\text{avg}} &= \frac{\Delta \vec r}{t}
\end{aligned}
\]
Its magnitude is
\[
\begin{aligned}
|\vec v_{\text{avg}}| &= \frac{|\Delta \vec r|}{t}
\end{aligned}
\]
This quantity depends only on where the motion started and where it ended. The route taken in between does not affect the average
velocity.
Average speed
If the total distance traveled is
\(s\),
then the average speed is
\[
\begin{aligned}
\text{average speed} &= \frac{s}{t}
\end{aligned}
\]
This depends on the full path length, not on the net displacement alone.
Why the two are different
Suppose a person walks around a curved route and ends up only a short distance away from the starting point. The total distance
traveled can be large, but the displacement can be small. In that case, the average speed is much larger than the magnitude of the
average velocity.
In fact, the inequality
\[
\begin{aligned}
s \ge |\Delta \vec r|
\end{aligned}
\]
implies that
\[
\begin{aligned}
\text{average speed} \ge |\vec v_{\text{avg}}|
\end{aligned}
\]
Equality occurs only when the motion is straight-line motion without detours.
2D and 3D support
The formulas work in both two and three dimensions. In 2D, if
\(\Delta \vec r = (\Delta x,\Delta y)\),
then
\[
\begin{aligned}
|\Delta \vec r| &= \sqrt{(\Delta x)^2 + (\Delta y)^2}
\end{aligned}
\]
In 3D, if
\(\Delta \vec r = (\Delta x,\Delta y,\Delta z)\),
then
\[
\begin{aligned}
|\Delta \vec r| &= \sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}
\end{aligned}
\]
Once the displacement magnitude is known, the average velocity magnitude follows directly by dividing by time.
Worked sample
Use the sample values
\(|\Delta \vec r| = 13\,\mathrm{m}\),
\(t = 2\,\mathrm{s}\),
and
\(s = 20\,\mathrm{m}\).
Step 1. Compute the magnitude of average velocity.
\[
\begin{aligned}
|\vec v_{\text{avg}}| &= \frac{|\Delta \vec r|}{t} \\
&= \frac{13}{2} \\
&= 6.5\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
Step 2. Compute the average speed.
\[
\begin{aligned}
\text{average speed} &= \frac{s}{t} \\
&= \frac{20}{2} \\
&= 10\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
Step 3. Compare them.
\[
\begin{aligned}
10\,\mathrm{m\,s^{-1}} > 6.5\,\mathrm{m\,s^{-1}}
\end{aligned}
\]
The reason is that the total distance traveled is larger than the displacement magnitude.
Vector meaning of average velocity
Average velocity does not just have a size; it also has a direction. Its direction is the same as the direction of the displacement
vector. That is why the graph in the calculator shows the displacement arrow and the actual traveled path separately. The path
determines average speed, while the displacement arrow determines average velocity.
Important restriction
The elapsed time must be greater than zero. Otherwise, dividing by time would be undefined. Also, the total distance traveled cannot
be smaller than the displacement magnitude, because no real path can be shorter than the straight-line separation between the start
and end points.
Summary
| Quantity |
Formula |
Meaning |
| Average velocity vector |
\(\vec v_{\text{avg}} = \Delta \vec r / t\) |
Net displacement per unit time |
| Magnitude of average velocity |
\(|\vec v_{\text{avg}}| = |\Delta \vec r| / t\) |
Size of the vector average velocity |
| Average speed |
\(s / t\) |
Total path length per unit time |
| Always true |
\(\text{average speed} \ge |\vec v_{\text{avg}}|\) |
Distance is never less than displacement magnitude |
| Equality case |
\(s = |\Delta \vec r|\) |
Straight-line motion without detours |
