Case Study 1 — The Speed Camera and the Instant of Speeding

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Case Study 1 — The Speed Camera and the Instant of Speeding

Field: Physics / traffic engineering Calculus used: Average vs. instantaneous rate of change; the derivative of position as velocity, computed from the limit definition (Sections 5.1, 5.2, 5.4)

A ticket in the mail

Three weeks after driving home along a coastal motorway, you receive a notice. A "section control" camera system photographed your license plate as you entered an enforcement zone and again as you left it. The two photographs are time-stamped. The zone is exactly $1\,500$ meters long, and the stamps are $54.0$ seconds apart. The posted limit is $25$ m/s (90 km/h). The notice claims you were speeding.

You did not feel like you were speeding. You remember slowing for a curve in the middle of the zone. So how can a camera that only knows where you entered and where you left — two snapshots, three-quarters of a minute apart — possibly prove anything about your speed at any single instant?

This is not a legal question dressed up as a math problem. It is a math problem dressed up as a legal one, and the entire case rests on the distinction this chapter is built around: the difference between an average rate of change and an instantaneous rate of change.

What the two photographs actually measure

Let $s(t)$ be your car's position along the motorway, measured in meters from the entry gantry, with $t$ in seconds from the moment of the first photograph. The two data points the camera owns are

$$s(0) = 0 \text{ m}, \qquad s(54.0) = 1500 \text{ m}.$$

From these — and only these — the system computes one number, the average velocity over the zone, which is exactly the average rate of change of position (Section 5.1):

$$\bar v = \frac{s(54.0) - s(0)}{54.0 - 0} = \frac{1500 - 0}{54.0} \approx 27.78 \text{ m/s}.$$

That is $100$ km/h, comfortably over the $25$ m/s limit. Geometrically, $\bar v$ is the slope of the secant line connecting the two photographed events on the position-versus-time graph. It is a single number summarizing the whole trip across the zone, and — crucially — it is blind to everything that happened in between. You could have driven at a perfectly constant $27.78$ m/s, or you could have crawled at $10$ m/s for the first half and sprinted at $45$ m/s for the second. Both stories produce the same two photographs and the same secant slope. The average "can't see the middle," exactly as Section 5.1 warns.

So the camera knows your average speed was $27.78$ m/s. The legal question is sharper: was your instantaneous speed — the reading a speedometer would have shown at some single moment — ever above $25$ m/s?

Instantaneous velocity is a derivative

Instantaneous velocity at a time $t_0$ is not an average over an interval; it is the limit of averages over shrinking intervals — that is, the derivative of position (Section 5.4):

$$v(t_0) = s'(t_0) = \lim_{h \to 0} \frac{s(t_0 + h) - s(t_0)}{h}.$$

To make the case concrete, suppose an accident-reconstruction engineer later models your motion with the position function

$$s(t) = 30t - \tfrac{1}{2}t^2 + \tfrac{1}{180}t^3, \qquad 0 \le t \le 54,$$

fit from the curve geometry and a few intermediate camera frames. (You need not trust the exact coefficients; they are chosen so $s(0) = 0$ and $s(54) = 1500$, and so the car slows in the middle and speeds up at the ends — matching your memory.) Let us find the velocity from the definition, the honest Chapter-5 way, with no power rule.

Form the difference quotient at a general time $t$:

$$\frac{s(t+h) - s(t)}{h}.$$

Expand each piece. For the linear term, $30(t+h) - 30t = 30h$. For the quadratic term,

$$-\tfrac12(t+h)^2 + \tfrac12 t^2 = -\tfrac12\big(2th + h^2\big) = -th - \tfrac12 h^2.$$

For the cubic term, using $(t+h)^3 = t^3 + 3t^2h + 3th^2 + h^3$,

$$\tfrac{1}{180}\big[(t+h)^3 - t^3\big] = \tfrac{1}{180}\big(3t^2 h + 3t h^2 + h^3\big).$$

Add the three changes, divide by $h$ (legal because $h \neq 0$ throughout the limit), and the $h$ in the denominator cancels one power from every term:

$$\frac{s(t+h)-s(t)}{h} = 30 - t - \tfrac12 h + \tfrac{1}{180}\big(3t^2 + 3th + h^2\big).$$

Now let $h \to 0$. Every term still carrying a factor of $h$ vanishes, leaving

$$v(t) = s'(t) = 30 - t + \tfrac{1}{60}t^2.$$

This is the instantaneous velocity at every instant of your passage — one limit computation that replaces infinitely many. Notice that we never "reached" the instant; we approached it, and the $0/0$ that would have stopped us cold (Section 5.2) dissolved the moment the $h$ cancelled.

Reading the speeding off the velocity function

Evaluate $v(t)$ at a few moments:

Entry, $t = 0$: $\quad v(0) = 30 - 0 + 0 = 30 \text{ m/s}.$
Middle, $t = 27$: $\quad v(27) = 30 - 27 + \tfrac{27^2}{60} = 3 + 12.15 = 15.15 \text{ m/s}.$
Exit, $t = 54$: $\quad v(54) = 30 - 54 + \tfrac{54^2}{60} = -24 + 48.6 = 24.6 \text{ m/s}.$

The model confirms your memory: you entered fast at $30$ m/s, slowed to about $15$ m/s for the curve, and accelerated back up near the exit. At entry your instantaneous speed was $30$ m/s — fully $5$ m/s over the limit. The camera's average of $27.78$ m/s understated your worst moment.

This already settles the case in the prosecution's favor, but it does so using an exact model. The deeper point — the one that makes section-control enforcement legally robust even without a detailed model — is that the average alone forces the conclusion.

Why the average alone is enough (a preview of the Mean Value Theorem)

Here is the structural fact. Your position $s(t)$ is continuous (you do not teleport) and differentiable (your velocity never jumps to infinity — that would require infinite force). For such functions, a theorem we will prove in Chapter 9, the Mean Value Theorem, guarantees that

$$\text{there exists some instant } c \in (0, 54) \text{ where } \quad s'(c) = \frac{s(54) - s(0)}{54 - 0} = \bar v.$$

In words: at some moment inside the zone, your instantaneous velocity equalled your average velocity exactly. Since the average was $27.78$ m/s and that exceeds the $25$ m/s limit, your instantaneous speed must have hit $27.78$ m/s at some point — whether or not anyone can say precisely when. The average being over the limit is, by itself, a mathematical proof of an instant of speeding.

This is the principle on which every section-control system rests, and you can see it is nothing but the bridge from average rate to instantaneous rate that this whole chapter constructs — promoted, in Chapter 9, from intuition to theorem.

import numpy as np
import matplotlib.pyplot as plt

t = np.linspace(0, 54, 400)
s = 30*t - 0.5*t**2 + t**3/180          # position model (m)
v = 30 - t + t**2/60                     # velocity = s'(t), from the definition

fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(12, 4))
ax1.plot(t, s, 'b-', lw=2)
ax1.plot([0, 54], [0, 1500], 'k--', lw=1, label='secant (average velocity)')
ax1.set_xlabel('t (s)'); ax1.set_ylabel('position s (m)')
ax1.set_title('Position; secant slope = average velocity'); ax1.legend(); ax1.grid(alpha=0.3)

ax2.plot(t, v, 'r-', lw=2, label="instantaneous velocity s'(t)")
ax2.axhline(25, color='gray', ls='--', label='speed limit 25 m/s')
ax2.axhline(1500/54, color='green', ls=':', label='average ≈ 27.8 m/s')
ax2.set_xlabel('t (s)'); ax2.set_ylabel('velocity (m/s)')
ax2.set_title('Velocity crosses the limit near entry and exit'); ax2.legend(); ax2.grid(alpha=0.3)
plt.tight_layout(); plt.show()
# Figure: the velocity dips below the limit mid-zone but exceeds it at the ends;
# by the Mean Value Theorem the average (green) is achieved instantaneously somewhere inside.

The engineering layer: what the sensors really compute

Real enforcement hardware does not have your tidy cubic $s(t)$. Doppler-radar guns measure instantaneous velocity directly by the frequency shift of microwaves bouncing off your car — physically, the derivative computed by electromagnetism rather than by algebra. Section-control cameras, by contrast, measure only the two endpoints and rely on the Mean Value Theorem argument above. And the airbag controller buried in your dashboard does something closer to what we did by hand: it samples the accelerometer thousands of times per second and forms finite difference quotients — $\frac{s(t+h) - s(t)}{h}$ with a tiny fixed $h$ — to estimate acceleration $a(t) = s''(t)$ in real time, firing the bag when that estimate crosses a threshold within milliseconds. The chain position $\to$ velocity $\to$ acceleration is just the derivative applied once, then twice (Section 5.5), and engineers compute it with the very difference quotient this chapter defines.

Discussion Questions

Section-control cameras are sometimes called unfair because the Mean Value Theorem proves only that some instant of speeding occurred, never when. Is "we can prove it happened but not pin the moment" a sound basis for a fine? Argue both sides.
If your average speed across the zone had been exactly $25$ m/s, would you definitely not have been speeding? Explain using the average-vs-instantaneous distinction. (No — the average could sit on the limit while instantaneous speed rises above it for part of the zone and falls below for the rest.)
The MVT argument needs $s(t)$ to be continuous and differentiable. Is a real car's position function ever genuinely non-differentiable? What physical quantity would a corner in $s(t)$ require? (A corner means a velocity jump, i.e. infinite acceleration and infinite force — physically impossible, so real $s(t)$ is differentiable.)
Recompute $v(t)$ for the model above using the equivalent $x \to a$ form of the derivative (Section 5.2) and confirm you obtain the same $30 - t + \tfrac{1}{60}t^2$.
A Doppler gun gives instantaneous speed at one moment; section control gives a guaranteed-true average. Which is the fairer enforcement tool, and which is the more informative measurement? They need not have the same answer.

Your Turn — Mini-Project

A car's position is $x(t) = t^3 - 6t^2 + 9t$ (meters, $t \in [0, 4]$ seconds).

Find $v(t) = x'(t)$ from the limit definition (do not use the power rule yet).
Solve $v(t) = 0$ to find the instants when the car is momentarily stopped.
Determine the intervals where $v(t) > 0$ (moving forward) and $v(t) < 0$ (moving backward).
Plot $x(t)$ and $v(t)$ on the same time axis and relate the sign of $v$ to the rise and fall of $x$.
Compute the average velocity over $[0, 4]$, then locate (by solving $v(c) = \bar v$) at least one instant guaranteed by the Mean Value Theorem where the instantaneous velocity equals it.

Case Study 1 — The Speed Camera and the Instant of Speeding

A ticket in the mail

What the two photographs actually measure

Instantaneous velocity is a derivative

Reading the speeding off the velocity function

Why the average alone is enough (a preview of the Mean Value Theorem)

The engineering layer: what the sensors really compute

Discussion Questions

Your Turn — Mini-Project

Further Reading