Average and instantaneous velocity

Two position readings are the least information that can say anything about a rate. Call them (t₁, x₁) and (t₂, x₂); the average velocity over the interval is v_avg = (x₂ − x₁)/(t₂ − t₁), which on an x–t graph is the slope of the straight line, the secant, joining the two points. A glider logged at x = 0.35 m when t = 0.40 s and at x = 0.95 m when t = 1.60 s has v_avg = 0.60 m ÷ 1.20 s = +0.50 m s⁻¹. The sign belongs to the answer and is set by the positive direction you declared: had the glider gone from 0.95 m to 0.35 m in the same time, v_avg would be −0.50 m s⁻¹. And the number is blind to everything between the endpoints. The glider could have coasted steadily or raced out to 1.40 m and come back; every history that starts and ends at those two points shares one secant and one v_avg.

Average speed is total distance travelled divided by elapsed time, and distance piles up whichever way the object moves. So average speed is never less than |v_avg|, and the two agree only when nothing reverses inside the interval. A runner who goes 1.2 km out along a straight path at 6.0 m s⁻¹ and returns at 4.0 m s⁻¹ ends where she started, so v_avg = 0. Her average speed is not 5.0 m s⁻¹ either: the legs take 200 s and 300 s, giving 2400 m ÷ 500 s = 4.8 m s⁻¹. An average over time weights each speed by how long it lasts, and the slow leg lasts longer. The same weighting governs average velocity. For a run built from steady legs, v_avg = Σ vᵢΔtᵢ / Σ Δtᵢ, which here is (+6.0 × 200 − 4.0 × 300)/500 = 0, never a plain mean of the leg velocities.

Take x(t) = 0.5t³ − 4.5t² + 12t, with x in m and t in s, and ask for the velocity at t = 1.0 s, where x = 8.0 m. Secants starting there have slopes 2.000 m s⁻¹ over Δt = 1.0 s, 3.125 m s⁻¹ over 0.5 s, 4.205 m s⁻¹ over 0.1 s and 4.470 m s⁻¹ over 0.01 s. The numbers settle, and algebra says where. Expand x(t + Δt) − x(t) and divide by Δt: the secant slope is exactly 1.5t² − 9t + 12 + (1.5t − 4.5)Δt + 0.5Δt². Every term after the first three carries a factor of Δt, so as Δt → 0 they vanish and the limit is v(t) = dx/dt = 1.5t² − 9t + 12, which is 4.5 m s⁻¹ at t = 1.0 s. On the graph each secant pivots about the point (1.0 s, 8.0 m), and the line they close in on is the tangent. Near t = 1.0 s the error is about −3.0 m s⁻² × Δt: halve the window and you halve the error, but no finite window removes it.

The tangent slope tells you two things at each instant. Its sign is the direction of motion, and its magnitude |v| is the instantaneous speed. For the cart above, v(t) = 1.5(t − 2)(t − 4) m s⁻¹: positive before 2 s, negative between 2 s and 4 s, positive again after. At t = 3.0 s the cart is at x = +9.0 m but v = −1.5 m s⁻¹, so it is heading towards smaller x at 1.5 m s⁻¹; the sign of v says nothing about the sign of x. Where v passes through zero, x(t) has a peak or a trough and the cart turns round, at 10.0 m when t = 2.0 s and at 8.0 m when t = 4.0 s. Unlike the averages, instantaneous speed always equals |instantaneous velocity|, because a vanishing interval leaves no room for a reversal. Over 1.0 s to 5.0 s the same cart has |v_avg| = 0.50 m s⁻¹ but an average speed of 1.50 m s⁻¹.

Slide a secant sideways without changing its slope until it just touches the curve. For a smooth x(t) that always happens at least once inside the interval: some instant t* has v(t*) = v_avg. This is the mean value theorem, and it is how average-speed cameras work. Cameras 3.0 km apart that time a car at 100 s record an average of 30 m s⁻¹, or 108 km h⁻¹, so the speedometer read exactly 108 km h⁻¹ at some moment between them, whatever the driver says about the rest of the section. The theorem needs smoothness, and it is worth watching it fail. A shuttle that runs at +2.0 m s⁻¹ for 3.0 s and then reverses instantly to −2.0 m s⁻¹ for 1.0 s has v_avg = (6.0 m − 2.0 m)/4.0 s = +1.0 m s⁻¹, yet in that model its velocity is never +1.0 m s⁻¹. The instant reversal puts a corner in x(t), where no tangent exists. A real shuttle brakes over a short time and passes through +1.0 m s⁻¹ on the way.

A motion sensor reports positions at discrete instants, each uncertain by some u(x), so the narrowest window is one sample step, and narrowing it makes matters worse. The difference of two independent readings carries √2 u(x), so a secant across a window of width w has noise √2 u(x)/w. With u(x) = 1.0 mm, a 0.020 s step gives 0.071 m s⁻¹ of noise and a 1.0 ms step gives 1.4 m s⁻¹: the limit that perfects the model wrecks the measurement. Widening the window suppresses noise but lets curvature back in, which is why the window should be centred. For the cart above, the centred secant [x(t + h) − x(t − h)]/(2h) works out to exactly v(t) + 0.5h² m s⁻¹ with h in s, with no term in h at all. At h = 0.10 s the curvature error is 0.005 m s⁻¹ and the noise is √2 × 1.0 mm ÷ 0.20 s = 0.007 m s⁻¹. Halve h and the noise doubles; double h and the curvature error quadruples. A good estimate of v lives in that balance.