Average Speed And Average Velocity

One Trip, Two Different Rates

A trip can have one elapsed time but two different descriptions of how far and in what direction the traveler moved. The difference becomes important when motion changes direction: the traveler may cover substantial ground and still finish near the starting point.

For example, suppose someone walks east and then turns back west. The total distance counts both parts of the walk. The net displacement compares only the final position with the starting position, retaining the direction of the overall change. Average speed uses total distance; average velocity uses net displacement. Both divide by the same elapsed time.

To keep signs meaningful, let \(x(t)\) denote position along a line, measured relative to a chosen origin and positive direction. For times \(t_1<t_2\), the elapsed time is \(t_2-t_1\), which is positive. The displacement is the final position minus the initial position.

The formulas use the same time interval, but their numerators answer different questions. Displacement describes the overall change in position, including its sign. Distance describes how much ground was covered along the trip, regardless of direction.

Calculating Both Rates From Trip Data

For a trip described by several monotone legs, each leg's distance is the absolute value of its position change. The leg's displacement, in contrast, keeps its sign. A positive change means motion in the chosen positive direction; a negative change means motion in the opposite direction.

Why Average Speed Is at Least the Magnitude of Average Velocity

The comparison follows from a basic fact about distance and displacement: a route between two positions cannot be shorter than the straight-line separation between those positions. For movement along a line, the straight-line separation is the absolute value of the net displacement.

Here is a precise version that also covers motion that changes direction within a described portion of a trip. A partition of a time interval is a finite list of times that starts at the initial time, ends at the final time, and increases from one listed time to the next. The distance traveled over an interval can be defined as the supremum of the sums of the absolute position changes over all such partitions. If the motion has finite total distance, this supremum is that distance.

This argument does not assume that the motion is monotone between the endpoints. In particular, a portion that reverses direction may have distance greater than the absolute value of its endpoint-to-endpoint position change. The definition by partitions accounts for any such extra travel.

When Are the Two Magnitudes Equal?

For a trip consisting of finitely many straight legs, suppose each leg is monotone: it travels in one direction or stays still, without reversing direction within that leg. Let \(\Delta x_i\) be the signed position change on leg \(i\). The distance on that leg is then \(|\Delta x_i|\), so the total distance and net displacement are

The triangle inequality gives \(\left|\sum_{i=1}^{n}\Delta x_i\right|\leq\sum_{i=1}^{n}|\Delta x_i|\). Equality holds precisely when all the nonzero changes have the same sign: the legs with actual movement all point in one direction. If some nonzero legs point in opposite directions, their signed changes partially cancel in the displacement, while each leg still adds its full distance. This equality description is for the stated finite, monotone-leg trip; it does not assume that arbitrary portions of a trip have distance equal to the absolute value of their endpoint change.

When all movement is in one direction, average speed equals the magnitude of average velocity. The signed average velocity itself equals average speed if the motion is in the positive direction, and equals the negative of average speed if it is in the negative direction. A trip with no movement has both rates equal to zero.

A Common Mistake: Treating Displacement as Distance

In the first example, using the endpoint difference \(5-2=3\) metres as the total distance would give \(3/6=0.5\) metres per second. That calculation is not the average speed. It uses the displacement to calculate the magnitude of average velocity. The person actually traveled \(12\) metres out and \(9\) metres back, for \(21\) metres total, so the average speed is \(21/6=3.5\) metres per second.

The correction is to ask two separate questions before dividing by time: How far apart are the start and end positions, with direction? That gives displacement. How much distance did the trip cover, including every reversal? That gives total distance. Then divide each appropriate quantity by the same positive elapsed time.

Units provide a useful check. If distance and displacement are measured in metres and time in seconds, both rates are measured in metres per second. The units match, but the quantities are still different: average speed has no direction sign, while average velocity carries the sign of the net displacement.