Average Velocity Is Not Instantaneous Velocity

Why Δx/Δt from a position–time table never gives you the speed at the times in the table

The situation

A cart rolls down a ramp with uniform (constant) acceleration. A motion detector records its position every 0.20 seconds. You are asked:

“Use the data to find the velocity of the cart at t = 0.40 s, t = 0.80 s, and t = 1.20 s.”

The obvious move is to subtract neighbouring positions and divide by the time between them. That answer is always wrong — and it is wrong by exactly the same amount every single time. This page shows why, and what the number you calculated actually is.

1. Set up the motion

Every number, graph and table entry on this page is recomputed from these three values. The default cart starts at 0.40 m/s and speeds up at 2.00 m/s², sampled every 0.20 s.

Motion and sampling settings
0.40 m/s
2.00 m/s²
0.20 s
0.40 s
Which interval did you use for Δx/Δt?

Live results

True instantaneous velocity at t* (from v = v₀ + at)
1.20 m/s
Your average velocity Δx/Δt over the chosen interval
1.40 m/s
Error if you label that answer “the velocity at t*”
+0.20 m/s (16.7% too fast)
The time your average velocity actually describes (interval midpoint)
t = 0.50 s
True instantaneous velocity at that midpoint time
1.40 m/s — an exact match

2. The data, and where the velocities land

Notice the structure of the table: position values sit on the sampling times, but each Δx/Δt value sits between two rows. A run of n positions gives you only n − 1 average velocities, and not one of them belongs to a time in the left-hand column.

Position–time data and the average velocity of each interval. Shaded rows are the intervals; the average velocity of an interval belongs to the midpoint time shown, not to either endpoint.
Time t (s) Position x (m) Δx (m) Δx/Δt (m/s) Midpoint time (s) True v at midpoint (m/s)

The trap

A student asked for the velocity at t = 0.40 s looks at the table, finds the row labelled 0.40, grabs the Δx/Δt next to it, and writes it down. But that number was built from two different times. It cannot possibly be a property of one instant.

3. On the position–time graph: secant vs. tangent

Average velocity is the slope of the straight line connecting two points on the curve (the secant). Instantaneous velocity is the slope of the line that just grazes the curve at one point (the tangent). On a parabola these slopes are never equal at the endpoints.

Graph description. Two panels. Left panel — the whole run: a position versus time graph covering the whole two seconds of recording. Because the acceleration is constant, the position curve is a parabola that bends upward, getting steeper as time goes on. A shaded vertical band marks the interval you selected. Right panel — zoomed in on that interval: the same curve magnified, with three straight lines drawn across it. A red secant line joins the two data points at the ends of the interval; its slope is the average velocity you calculated. An orange tangent line touches the curve at the time you were asked about, t-star; it is clearly less steep than the secant, and the two lines fan further apart towards the right. A dashed green tangent line touches the curve at the midpoint of the interval, and it runs exactly parallel to the red secant. A dashed red right triangle under the secant shows the rise, delta x, over the run, delta t.

Read the picture this way

The red secant and the dashed green tangent are parallel. That parallel pair is the whole lesson: your average velocity is a real instantaneous velocity — just at the wrong time. Slide t* along and the two lines stay parallel; the secant never matches the orange tangent at the endpoint unless you flatten the acceleration to zero.

4. On the velocity–time graph: the offset you can measure

With constant acceleration the true velocity graph is a straight line. Your average velocity is the mean height of that line across the interval — and the mean height of a straight line occurs at its middle, never at its ends.

Graph description. A velocity versus time graph. The true velocity is a straight line rising steadily from left to right, because the acceleration is constant. The chosen interval is shaded, and a horizontal line is drawn at the height of your calculated average velocity. That horizontal line crosses the rising true-velocity line at exactly one place: the midpoint of the interval. A red marker shows where your answer would sit if you wrongly attached it to time t-star — it floats off the true line by a visible vertical gap. A green marker sits exactly on the true line at the midpoint time. The shaded region above and below the average-velocity line forms two equal triangles, showing that the overshoot at the end of the interval exactly cancels the undershoot at the start.

The algebra behind that offset

Start from the position equation for uniform acceleration and compute the average velocity from time t to time t plus delta t:

Equation: position at time t equals initial position, plus initial velocity times t, plus one half a t squared.

Equation: delta x over delta t equals the quantity position at t plus delta t, minus position at t, all divided by delta t. That simplifies to initial velocity, plus a times t, plus one half a times delta t.

The first two terms are exactly the instantaneous velocity at time t. So:

Equation: delta x over delta t equals the instantaneous velocity at time t, plus one half a delta t. And that sum is exactly equal to the instantaneous velocity at time t plus delta t over two — the midpoint of the interval.

Two consequences worth memorising

  • The error is ½aΔt, and it does not depend on t. Taking data later in the run does not help. A cart accelerating at 2.00 m/s² sampled every 0.20 s is off by 0.20 m/s at every single time in the table — at t = 0.2 s and at t = 12 s alike.
  • The error is one-sided. It is not random scatter that averages out over many trials. Forward intervals are always too fast; backward intervals are always too slow. It is a systematic error, so more data collected the same way will not fix it.

5. Why you cannot fix this by being careful

The only knob that shrinks the error is Δt. Drag the sampling-interval slider in section 1 and watch the marker move along this line.

Graph description. A graph of the size of the error against the sampling interval delta t. The relationship is a straight line through the origin with slope one half a: the error is directly proportional to how far apart your position readings are. A marker shows the current sampling interval. The line passes through the origin, which means the error only vanishes in the limit as delta t goes to zero — that limit is the definition of instantaneous velocity, and no finite table of data ever reaches it.

This is the honest reason a position–time table cannot hand you instantaneous velocities: instantaneous velocity is a limit, and your data stops at a finite Δt. Two things follow in the lab:

  • Shrinking Δt helps the physics but hurts the measurement. Halving Δt halves the ½aΔt error, but Δx also gets small, so your position uncertainty becomes a bigger fraction of it. Random noise grows as systematic bias shrinks. There is a practical floor.
  • Averaging many intervals does not help at all. Every forward interval carries the identical +½aΔt bias, so the mean of them carries it too.

6. What to do instead

You are not stuck. You just have to stop pretending an interval is an instant. Three defensible methods, in order of how much they ask of you:

Method A — Relabel the times (the midpoint rule)

Keep every Δx/Δt you already calculated, but write it next to the midpoint time instead of an endpoint. For uniform acceleration this is not an approximation — it is exact. Your seven velocities are now genuinely instantaneous velocities, just at t = 0.10, 0.30, 0.50 s and so on. Plot those against their midpoint times and the slope of the resulting straight line is the acceleration.

Method B — Use a centred interval

To get the velocity at a time that is in your table, use the point before and the point after:

Equation: the velocity at t-star is approximately the position at t-star plus delta t, minus the position at t-star minus delta t, all divided by two delta t.

The midpoint of that interval is t*, so the ½aΔt errors from the two halves cancel exactly. Select “Centred” in section 1 and watch the error readout drop to zero. The cost: you lose the first and last points of your data set, because they have no neighbour on one side.

Method C — Fit the whole curve

Fit x vs. t with a parabola and differentiate the fit, or linearise by plotting x against t² and read the slope. This uses every data point at once, is robust to random scatter, and gives you v₀ and a directly rather than one velocity at a time. This is the approach the linearization page walks through.

The sentence to take away

Δx/Δt is not an approximation to the velocity at either end of the interval. Under uniform acceleration it is the exact velocity at the middle of the interval. Treating it as an endpoint velocity introduces a systematic error of ½aΔt that no amount of extra data will remove.

7. Check yourself

Work these out before you move the sliders to confirm.

  1. A cart accelerates at 3.0 m/s². Positions are sampled every 0.10 s. By how much does a forward-difference Δx/Δt overstate the velocity at the start of each interval? (Answer: ½(3.0)(0.10) = 0.15 m/s, at every time in the table.)
  2. Your data gives Δx/Δt = 2.40 m/s for the interval from t = 0.60 s to t = 0.90 s. At what time is the cart exactly moving at 2.40 m/s? (Answer: t = 0.75 s.)
  3. Set the acceleration slider to 0. Why does every method suddenly agree? (Answer: with a = 0 the error ½aΔt is zero — for constant velocity the average over any interval equals the instantaneous value everywhere.)
  4. A classmate says “I’ll just take 50 intervals and average them to kill the error.” What is wrong with that plan? (Answer: the bias is identical and one-signed in every interval, so averaging reproduces it exactly instead of cancelling it.)