From Velocity Samples to Motion Estimates
A velocity tells us both how quickly an object is moving and in which direction. If the velocity is positive, motion is in the chosen positive direction; if it is negative, motion is in the opposite direction. To estimate how far an object moves over a time period, we combine velocity information across that period.
When velocity is constant over a time interval, displacement is velocity multiplied by elapsed time. If velocity varies, we can divide time into shorter intervals, use a representative velocity on each one, and add the resulting contributions. Left, right, and midpoint sums differ only in where that representative velocity is sampled.
These sums estimate signed displacement, not necessarily total distance. If velocity is measured in metres per second and time in seconds, every term has units \((\text{metres per second})(\text{seconds})=\text{metres}\). A positive sum indicates estimated net motion in the positive direction; a negative sum indicates estimated net motion in the opposite direction.
Why the Sums Represent Motion
On a short interval, a sampled velocity can stand in for the changing velocity over that interval. Multiplying it by the interval width gives the displacement predicted by that constant-velocity approximation. Adding the predictions gives the estimated displacement across the full time period.
The choice of sample matters. A left sum uses the velocity at the start of each interval, a right sum uses the velocity at the end, and a midpoint sum uses the velocity halfway through. No one of these choices is guaranteed to be closest to the actual displacement for every motion. The sums are estimates based on the specified samples, not exact answers merely because they have been calculated correctly.
Worked Example: Comparing Three Estimates from Velocity Data
Worked Example: Comparing Three Estimates from Velocity Data
An object moves along a straight line from \(t=0\) to \(t=6\) seconds. The following table gives sampled velocities in metres per second. Estimate its displacement and distance using three intervals of width \(2\) seconds.
The intervals are \([0,2]\), \([2,4]\), and \([4,6]\). Each width is \(2\) seconds. The left samples are at \(t=0,2,4\), the right samples are at \(t=2,4,6\), and the midpoint samples are at \(t=1,3,5\).
Left estimate of displacement:
Right estimate of displacement:
Midpoint estimate of displacement:
The three signed displacement estimates are \(10\text{ m}\), \(4\text{ m}\), and \(6\text{ m}\). To estimate distance, use the magnitudes of the sampled velocities: distance counts motion in either direction positively.
Check: Each term is a velocity in \(\text{m/s}\) multiplied by \(2\text{ s}\), so each result is in metres. For example, the midpoint displacement calculation is \(4+4-2=6\), while its distance calculation is \(4+4+2=10\); changing the negative contribution to a positive one is appropriate for distance, not signed displacement.
Unequal Time Intervals
Intervals need not all have the same width. The width used in each term must match that particular interval. In particular, a velocity sample from a three-second interval contributes three times its sampled velocity, not the same amount as a sample from a one-second interval.
Worked Example: Using Unequal Intervals
Worked Example: Using Unequal Intervals
Suppose the motion is sampled on the intervals \([0,1]\), \([1,4]\), and \([4,6]\) seconds. The left-endpoint velocities are \(2,-1,-1\) metres per second, and the right-endpoint velocities are \(-1,-1,3\) metres per second. Suppose the midpoint velocities are \(1,-2,1\) metres per second. Estimate displacement and distance with each rule.
The widths are \(1\), \(3\), and \(2\) seconds, respectively. Applying the left rule gives
The right rule gives
The midpoint rule gives
For distance estimates, take the absolute value of each sampled velocity before multiplying by its interval width:
Check: The interval widths add to \(1+3+2=6\) seconds, the full time from \(0\) to \(6\). The midpoint displacement arithmetic is \(1-6+2=-3\), and its distance arithmetic is \(1+6+2=9\). The negative displacement means estimated net motion in the negative direction; it does not mean negative distance.
When Velocity Changes Sign
When velocity is positive on some parts of a motion and negative on others, the contributions to displacement have different signs. If the sampled velocities include both positive and negative values, their signed contributions can cancel. The distance estimate does not cancel these contributions, because it uses \(|v|\) at each sample.
A sign change in the actual velocity does not guarantee that a particular sum will have both positive and negative sampled contributions. A sign change may occur between sample locations, or even within an interval whose chosen sample has velocity zero. Thus, a sum describes the signs of its sampled values; it does not automatically reveal every change in velocity between them.
Worked Example: A Sign Change Missed by the Midpoint Sample
Worked Example: A Sign Change Missed by the Midpoint Sample
Let velocity on \(0\leq t\leq4\) be \(v(t)=t-2\) metres per second. Use one interval, \([0,4]\), to find the left, right, and midpoint estimates. Then compare them with the exact displacement and distance.
The interval width is \(4\) seconds. At the left endpoint, \(v(0)=0-2=-2\text{ m/s}\); at the right endpoint, \(v(4)=4-2=2\text{ m/s}\); and at the midpoint, \(v(2)=2-2=0\text{ m/s}\). Therefore,
The midpoint sample is zero even though velocity is negative for \(0\leq t<2\) and positive for \(2<t\leq4\). It therefore produces no positive or negative sampled contribution to cancel against another one.
For the exact displacement, the velocity graph is a straight line from \((0,-2)\) to \((4,2)\), crossing zero at \(t=2\). The negative triangular region has base \(2\text{ s}\) and height \(2\text{ m/s}\), so its signed area is \(-\tfrac12(2)(2)=-2\text{ m}\). The positive triangular region has the same base and height, so its signed area is \(+\tfrac12(2)(2)=2\text{ m}\). Their sum is \(-2+2=0\text{ m}\).
For distance, both triangular areas count positively, giving \(2+2=4\text{ m}\). The midpoint distance estimate based on one sample is \(|v(2)|(4)=0(4)=0\text{ m}\), which misses both portions of the motion.
Check: The exact displacement is zero because the two signed triangular areas cancel. The exact distance is \(4\text{ m}\) because the two areas are added as positive amounts. The midpoint estimates are zero because the midpoint velocity is zero, not because the object remains still throughout the interval.
A Common Mistake: Taking the Absolute Value Too Late
Suppose two consecutive intervals each have width \(1\) second, with sampled velocities \(3\text{ m/s}\) and \(-2\text{ m/s}\). The signed sum is \(3(1)+(-2)(1)=1\text{ m}\). Taking its absolute value still gives \(1\text{ m}\), but the distance estimate is \(|3|(1)+|-2|(1)=3+2=5\text{ m}\). The two calculations answer different questions.
The first estimates net change in position; the second estimates total motion counted in either direction. Also, even the sum of absolute sampled velocities is only a distance estimate: if a velocity changes sign between samples, the samples may fail to capture how much motion occurred between them.
A Reliable Setup for Motion Problems
Record each interval's endpoints and width. Do not replace unequal widths with one common width.
Use left endpoints for a left sum, right endpoints for a right sum, or interval midpoints for a midpoint sum.
Keep negative signs when estimating displacement; use absolute values for a distance estimate.
Check the arithmetic and units, then state whether the result estimates signed displacement or distance.
Why the Distinction Matters
In applications, displacement and distance describe different aspects of motion. Displacement records the net change in position, so movement in opposite directions can cancel. Distance records how much travel occurred altogether, so it never decreases when direction reverses. A valid estimate must match the quantity being asked for.
When velocity changes sign, inspect the samples and interval sizes rather than assuming the sum automatically captures the change. Finer intervals can provide more information, but a sample sum remains an estimate unless additional facts justify an exact calculation. A sign change inside an interval can be especially important: neither a single endpoint sample nor a single midpoint sample is guaranteed to represent both directions of motion.
Check Your Understanding
For each question, distinguish carefully between the sampled sum and the motion quantity it estimates.
- On intervals of widths \(2\) and \(3\) seconds, the left-sampled velocities are \(4\) and \(-1\) metres per second. What is the left estimate of displacement?
- Using the same widths and samples, what is the left estimate of distance?
- Why must each velocity sample be multiplied by its own interval width when intervals are unequal?
- Can velocity change sign within an interval even if its midpoint sample is zero? Explain briefly using \(v(t)=t-2\) on \([0,4]\).
- Why does the absolute value of a displacement estimate not generally equal a distance estimate?
