Two Different Averages for Motion
A particle moving along a parametric path has a position at each time. Over a time interval, two natural questions are: how quickly did its position change overall, and how much distance did it cover per unit time? These questions lead to average velocity and average speed.
They are not generally the same. Average velocity depends only on the starting and ending positions, so it records net displacement and direction. Average speed depends on the entire route, including any bends or reversals.
Definitions on a Parametric Interval
Let \(\mathbf{r}(t)\) be the position vector of a particle for \(a\le t\le b\), where \(a<b\) and \(t\) represents time. The displacement is the change in position from the beginning to the end of the interval. Dividing by the elapsed time gives the average velocity vector.
The total distance traveled along a differentiable path is its arc length. If the motion is piecewise differentiable, the lengths of the pieces are added. In the differentiable case, the velocity is \(\mathbf{r}'(t)\), the speed is \(\|\mathbf{r}'(t)\|\), and the arc length is the integral of speed.
Average velocity is a vector; average speed is a nonnegative scalar. The magnitude \(\|\mathbf{v}_{\mathrm{avg}}\|\) is the magnitude of average velocity, not the average speed. In physical applications, if position is measured in meters and time in seconds, both have units of meters per second, but they can have different values.
Worked Example: A Semicircular Path
Worked Example: A Semicircular Path
Consider motion along the unit semicircle \(\mathbf{r}(t)=(\cos t,\sin t)\) for \(0\le t\le\pi\). Find the average velocity and average speed.
The elapsed time is \(\pi-0=\pi\). The initial and final positions are \(\mathbf{r}(0)=(1,0)\) and \(\mathbf{r}(\pi)=(-1,0)\), so the displacement is \[ (-1,0)-(1,0)=(-2,0). \] Therefore, $$ \mathbf{v}_{\mathrm{avg}} = \frac{(-2,0)}{\pi} = \left(-\frac{2}{\pi},0\right). $$
To find the length, differentiate the position: \(\mathbf{r}'(t)=(-\sin t,\cos t)\). Its magnitude is \[ \|\mathbf{r}'(t)\| = \sqrt{\sin^2 t+\cos^2 t} = 1. \] Thus $$ L=\int_0^\pi 1\,dt=\pi, \qquad \text{average speed}=\frac{\pi}{\pi}=1. $$
Check: the derivative calculation gives speed \(1\) throughout, and the semicircle has arc length \(\pi\), consistent with that speed over a time interval of length \(\pi\). The average velocity has magnitude \(2/\pi\), which is less than the average speed \(1\).
The particle travels a curved route of length \(\pi\), but its net displacement has length only \(2\). Average speed reflects the full semicircle; average velocity reflects the straight-line change between endpoints.
Why Average Speed Is at Least the Magnitude of Average Velocity
For any differentiable motion on \([a,b]\), total distance cannot be smaller than the straight-line distance between the endpoints. Dividing that comparison by the same positive elapsed time gives a useful relationship between the two averages.
Here is a direct justification. Write \(\mathbf{v}(t)=\mathbf{r}'(t)\), so the displacement is the integral of velocity. If the displacement \(\mathbf{D}=\mathbf{r}(b)-\mathbf{r}(a)\) is zero, its magnitude is automatically no greater than the nonnegative arc length. If \(\mathbf{D}\ne\mathbf{0}\), let \(\mathbf{u}=\mathbf{D}/\|\mathbf{D}\|\), a unit vector in the direction of displacement. Then:
The inequality follows from \(\mathbf{u}\cdot\mathbf{v}(t)\le\|\mathbf{u}\|\,\|\mathbf{v}(t)\|=\|\mathbf{v}(t)\|\). Dividing by \(b-a>0\) gives the stated comparison. The path’s bends or reversals can increase total distance without increasing the endpoint displacement.
Worked Example: Motion That Reverses Direction
Worked Example: Motion That Reverses Direction
Let a particle move on a line with position \(\mathbf{r}(t)=((t-1)^2,0)\) for \(0\le t\le2\). Take the position units to be meters and the time units to be seconds. Find its average velocity and average speed.
The endpoint positions are \[ \mathbf{r}(0)=(1,0),\qquad \mathbf{r}(2)=(1,0). \] The displacement is \((1,0)-(1,0)=(0,0)\). The elapsed time is \(2\) seconds, so $$ \mathbf{v}_{\mathrm{avg}} = \frac{(0,0)}{2} = (0,0)\ \text{m/s}. $$
The velocity is \[ \mathbf{r}'(t)=(2(t-1),0), \] so the speed is \(2|t-1|\). The particle reverses direction at \(t=1\), where its speed is zero. Compute the total distance on the two parts of the interval: $$ \begin{aligned} L &=\int_0^1 2(1-t)\,dt+\int_1^2 2(t-1)\,dt\\ &=\left[2t-t^2\right]_0^1+\left[(t-1)^2\right]_1^2\\ &=(2-1-0)+\bigl(1-0\bigr)\\ &=2\ \text{m}. \end{aligned} $$ Therefore the average speed is $$ \frac{L}{2\ \text{s}} = \frac{2\ \text{m}}{2\ \text{s}} = 1\ \text{m/s}. $$
Check: the particle moves from position \(1\) m to \(0\) m, then back to \(1\) m. Those two one-meter trips give total distance \(2\) m and zero net displacement, matching the calculations.
This example shows why the endpoint formula cannot give average speed. The particle ends exactly where it started, but it has not remained still: it traveled a positive distance.
Worked Example: Cycloid Motion
Worked Example: A Cycloid Arch
Consider the parametric motion \(\mathbf{r}(t)=(t-\sin t,1-\cos t)\) for \(0\le t\le2\pi\). Find its average velocity and average speed over this interval.
The initial position is \(\mathbf{r}(0)=(0,0)\). Since \(\sin(2\pi)=0\) and \(\cos(2\pi)=1\), the final position is \(\mathbf{r}(2\pi)=(2\pi,0)\). Thus the displacement is \((2\pi,0)\), and the elapsed time is \(2\pi\). Therefore, $$ \mathbf{v}_{\mathrm{avg}} = \frac{(2\pi,0)}{2\pi} = (1,0). $$
Differentiate the position: \[ \mathbf{r}'(t)=(1-\cos t,\sin t). \] The speed is $$ \begin{aligned} \|\mathbf{r}'(t)\| &=\sqrt{(1-\cos t)^2+\sin^2 t}\\ &=\sqrt{1-2\cos t+\cos^2t+\sin^2t}\\ &=\sqrt{2-2\cos t}\\ &=2\sin(t/2), \end{aligned} $$ because \(\sin(t/2)\ge0\) for \(0\le t\le2\pi\). Hence $$ L=\int_0^{2\pi}2\sin(t/2)\,dt =\left[-4\cos(t/2)\right]_0^{2\pi} =4+4=8. $$ The average speed is \(8/(2\pi)=4/\pi\).
Check: the average velocity calculation uses the endpoint displacement and gives \((1,0)\); its magnitude is \(1\). The average speed is \(4/\pi\), which is greater than \(1\). The arc length \(8\) is indeed greater than the displacement length \(2\pi\).
A Common Mistake: Using Displacement for Average Speed
A frequent error is to compute average speed as displacement magnitude divided by elapsed time. That expression is instead the magnitude of average velocity. For the reversing motion above, the incorrect calculation would be $$ \frac{\|\mathbf{r}(2)-\mathbf{r}(0)\|}{2-0} = \frac{0}{2} = 0. $$ This is not the average speed: it ignores the trip away from the starting point and the return trip.
The correction is to calculate total arc length, splitting the interval at a reversal when needed, and then divide by elapsed time: $$ \text{average speed} = \frac{\text{total distance traveled}}{\text{elapsed time}}. $$ For a curved path, the same principle applies: use the integral of speed over the full interval, not the straight-line distance between endpoints.
Choosing the Right Quantity
Find \(\mathbf{r}(a)\) and \(\mathbf{r}(b)\), subtract to obtain displacement, and divide by \(b-a\).
Find the speed \(\|\mathbf{r}'(t)\|\), integrate it over the interval to obtain total length, and divide by \(b-a\).
The magnitude of average velocity cannot exceed average speed. If a calculation gives the reverse, revisit the displacement or arc-length calculation.
Both quantities describe motion over an interval, but they answer different questions. Average velocity summarizes net change in position, including direction. Average speed summarizes the rate of distance traveled. A curved route or a reversal makes that distinction especially important.
Check Your Understanding
Use endpoint displacement for average velocity and total arc length for average speed.
- State the formula for average velocity of \(\mathbf{r}(t)\) on \([a,b]\).
- What quantity must be integrated to find arc length for a differentiable parametric path?
- A particle returns to its starting point after traveling a nonzero distance. What is its average velocity, and can its average speed be zero?
- For the unit semicircle example, why is the magnitude of average velocity less than the average speed?
- Explain why the magnitude of average velocity cannot exceed average speed.