Net Change Versus Total Change

One Motion, Two Different Totals

When an object moves, two natural questions are how far its position changed overall and how much ground it covered along the way. These questions have different answers whenever the object changes direction.

Choose a positive direction along a straight line. Motion in that direction has positive velocity; motion in the opposite direction has negative velocity. A negative velocity contributes a negative change in position, but the distance traveled during that motion is still positive.

For example, if an object moves \(8\) metres forward and then \(3\) metres backward, its position changes by \(8-3=5\) metres in the positive direction. Its total distance traveled is \(8+3=11\) metres. The first quantity preserves direction; the second counts all motion without assigning it a sign.

Interval Changes from Velocity

Suppose a motion is described on consecutive time intervals. Let \(\Delta t_i\) be the duration of interval \(i\), and let \(v_i\) be the velocity on that interval. Durations are positive. If velocity is constant on an interval, its change in position is its velocity multiplied by the interval’s duration.

The sign of \(\Delta x_i\) records direction. A positive value means position increased; a negative value means position decreased. The magnitude \(|\Delta x_i|\) records the length of that interval’s change in position.

The same displacement formula is exact if \(v_i\) is the average velocity over the entire interval. Average velocity is defined as the interval’s displacement divided by its duration, so multiplying that average by the duration recovers the displacement. However, average velocity alone does not in general determine the distance traveled within the interval. We will distinguish these cases carefully.

In formulas, if the interval displacements are \(\Delta x_1,\Delta x_2,\ldots,\Delta x_n\), then

and, when each interval’s displacement magnitude equals the distance traveled on that interval,

These formulas serve different purposes. Adding signed changes allows forward and backward motion to cancel in the net displacement. Adding magnitudes does not allow cancellation, so it measures the accumulated path length under the stated condition.

Worked Example: Constant Velocities in Both Directions

Why Distance Cannot Be Less Than Net Displacement

For any list of interval displacements, the magnitude of their signed sum cannot exceed the sum of their magnitudes. This is the triangle inequality for real numbers. It explains why, when interval magnitudes correctly represent the distances traveled, total distance is at least the magnitude of net displacement.

Here is a justification. For any two real numbers \(a\) and \(b\), both \(|a|+|b|\) and \(|a+b|\) are nonnegative. Also,

The last inequality holds because \(|ab|\geq ab\). Since the two quantities being compared are nonnegative, the inequality between their squares implies \(|a+b|\leq |a|+|b|\). Applying this two-number result repeatedly gives the inequality for any finite list.

If \(a\) and \(b\) have opposite signs and neither is zero, then \(|ab|>ab\), so the two-number inequality is strict. More generally, if a list contains both positive and negative terms, its positive and negative contributions partly cancel in the signed sum. Therefore the magnitude of the net change is strictly smaller than the sum of all magnitudes. If all nonzero changes have the same sign, there is no cancellation and equality holds.

For motion divided into intervals with no reversal inside any interval, this theorem says that total distance equals the magnitude of displacement only when the motion has no net cancellation between intervals—that is, when all nonzero interval displacements have the same sign.

Worked Example: Net Change Can Be Small After Long Travel

Average Velocity Does Not Always Give Distance

It is important not to confuse average velocity with a guarantee about motion throughout an interval. Average velocity gives the exact displacement over the whole interval. If direction changes within that interval, the distance can be greater than the magnitude of that displacement.

A Common Mistake: Taking the Magnitude of an Average

A tempting but incorrect calculation in the preceding example is to multiply the magnitude of the average velocity by the full duration and call the result distance:

This result is not the distance traveled. It is the magnitude of the net displacement, which is zero. The forward and backward motions have canceled in the average velocity. To determine distance, we need information that separates the motion into portions with no reversal, as the two \(2\ \text{s}\) intervals did in the example.

Thus, if a listed velocity is an average over a whole interval, multiplying it by the duration gives exact displacement. Summing the magnitudes of those average-velocity-based displacements gives exact distance only when direction does not reverse within each interval. If a reversal may occur, the average value alone is insufficient to determine distance.

Keeping the Two Calculations Straight

When working from interval velocities, first determine what the velocity information represents. If the velocity is constant on each interval, calculate the interval displacement and use its sign to record direction. If it is an average over the whole interval, the same product gives the exact displacement, but do not infer the distance without knowing whether direction reversed within that interval.

These quantities answer different questions, and both can be useful. Net displacement tells how much the final position differs from the starting position and in which direction. Total distance tells how much motion occurred along the path, regardless of direction. Their units are the same, but their meanings are not interchangeable.