Case Study 1 — Clocking a Cheetah: Velocity From Position

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Case Study 1 — Clocking a Cheetah: Velocity From Position

Field: Physics (Modeling Portfolio Track C) Calculus used: The tangent problem as instantaneous velocity (§1.1, §1.7); the area problem as accumulated distance (§1.2); a first glimpse of the Fundamental Theorem (§1.4) Key skills demonstrated: Turning a real measurement table into secant slopes; watching secant slopes converge to an instantaneous rate; recovering distance as an area under a velocity curve

The Setup

In 2012, biologists at the Cincinnati Zoo filmed an eleven-year-old cheetah named Sarah running a measured 100-metre course. High-speed cameras and a radio-tagged lure let them record her position many times per second. Sarah covered the 100 metres in 5.95 seconds, hitting a top speed near 29 metres per second — about 104 km/h. It remains one of the most carefully measured sprints of any land animal.

A photo-finish clock gives you one number: the total time. But the biologists wanted more. How fast was Sarah going at the two-second mark? When did she stop accelerating? How much of the 100 metres did she cover in the first three seconds? None of these questions can be answered by a stopwatch. Every one of them is a calculus question — and remarkably, you can answer all of them with nothing more than the two ideas from this chapter: secant slopes and Riemann sums.

We will not use Sarah's raw data (it is noisy and proprietary). Instead we model her acceleration phase with a clean curve that matches the real numbers closely. During the first three seconds, before she reaches top speed, her distance from the start is very well described by

$$s(t) = 4.6\,t^2 \qquad \text{(metres, with } t \text{ in seconds).}$$

At $t = 3$ this gives $s(3) = 4.6 \times 9 = 41.4$ m, and the implied top speed at $t = 3$ is about $27.6$ m/s — both squarely in line with the filmed sprint. This is our laboratory. Everything below is real calculus on a realistic model.

The Tangent Problem: How Fast at $t = 2$?

Speed is the rate at which distance changes — the slope of the position-versus-time graph. But Sarah is accelerating, so that slope is different at every instant. Asking "how fast was she going at exactly $t = 2$ s?" is precisely the tangent problem from §1.1: find the slope of the curve $s(t)$ at a single point.

We attack it exactly as we attacked $y = x^2$ in §1.1. Pick the instant $t = 2$ and a nearby instant $t = 2 + \Delta t$. The average velocity over that little interval is a secant slope:

$$\bar v = \frac{s(2 + \Delta t) - s(2)}{\Delta t}.$$

This is just (distance covered) ÷ (time taken) — the everyday notion of average speed. Now shrink $\Delta t$ and watch:

$\Delta t$ (s)	$s(2+\Delta t)$ (m)	average velocity (m/s)
$0.1$	$20.286$	$18.86$
$0.01$	$18.58446$	$18.446$
$0.001$	$18.418405$	$18.4046$
$0.0001$	$18.4018404$	$18.40046$

The average velocities march toward a single number: $18.4$ m/s. That limiting value is Sarah's instantaneous velocity at $t = 2$ — the speedometer reading you could never get from the photo-finish clock alone.

We can confirm the limit exactly with the §1.7 algebra, no rules required:

$$\bar v(\Delta t) = \frac{4.6(2 + \Delta t)^2 - 4.6(2)^2}{\Delta t} = \frac{4.6\,(4 + 4\Delta t + \Delta t^2 - 4)}{\Delta t} = \frac{4.6\,\Delta t\,(4 + \Delta t)}{\Delta t} = 4.6\,(4 + \Delta t).$$

As $\Delta t \to 0$, this approaches $4.6 \times 4 = 18.4$ m/s — matching the table to the last digit. We have computed an instantaneous velocity using only the tangent idea from this chapter.

# Secant velocities of s(t) = 4.6 t^2 at t = 2, as the interval shrinks.
def s(t: float) -> float:
    return 4.6 * t**2

t0 = 2.0
for dt in [0.1, 0.01, 0.001, 0.0001]:
    v_avg = (s(t0 + dt) - s(t0)) / dt
    print(f"dt = {dt:<7}  average velocity = {v_avg:.5f} m/s")

# Output:
# dt = 0.1      average velocity = 18.86000 m/s
# dt = 0.01     average velocity = 18.44600 m/s
# dt = 0.001    average velocity = 18.40460 m/s
# dt = 0.0001   average velocity = 18.40046 m/s

Running the same argument at a general time $t$ gives $\bar v(\Delta t) = 4.6(2t + \Delta t) \to 9.2\,t$. So Sarah's velocity at any instant in the acceleration phase is

$$v(t) = 9.2\,t \quad \text{m/s}.$$

This is a derivative, though we will not call it that by name until Chapter 5. It answers the biologists' second question too: her acceleration is constant at $9.2$ m/s² during this phase (the slope of $v(t)$), and she reaches her top speed of about $27.6$ m/s right around $t = 3$ s — exactly when the model is built to hand off to her steady-state cruising speed.

The Area Problem: How Far in Three Seconds?

Now run the logic in reverse. Suppose the only instrument that survived the shoot was a radar gun — it recorded Sarah's velocity $v(t) = 9.2\,t$, but not her position. Can we recover how far she travelled in the first three seconds?

Yes — and this is the area problem of §1.2. If Sarah ran at a constant speed, distance would just be speed × time, the area of a rectangle. She does not, but we can chop the three seconds into thin slices, treat the speed as roughly constant on each slice, and add up the little distances. Each term is (speed) × (a little time) = (a little distance) — a Riemann sum.

Slice $[0, 3]$ into $n$ equal pieces of width $3/n$, take the velocity at the right edge of each piece, and sum:

$$\text{distance} \approx \sum_{i=1}^{n} v\!\left(\frac{3i}{n}\right)\cdot \frac{3}{n} = \sum_{i=1}^{n} 9.2 \cdot \frac{3i}{n}\cdot\frac{3}{n}.$$

# Distance as the area under the velocity curve v(t) = 9.2 t on [0, 3].
def v(t: float) -> float:
    return 9.2 * t

def distance_estimate(n: int) -> float:
    width = 3.0 / n
    return sum(v(i * width) * width for i in range(1, n + 1))

for n in [3, 6, 100, 1000]:
    print(f"n = {n:5d}   distance estimate = {distance_estimate(n):.4f} m")

# Output:
# n =     3   distance estimate = 55.2000 m
# n =     6   distance estimate = 48.3000 m
# n =   100   distance estimate = 41.8140 m
# n =  1000   distance estimate = 41.4414 m

The estimates converge to 41.4 metres. With only three rectangles the estimate is a crude 55.2 m (the rectangles overshoot a rising line badly); with a thousand it has settled on 41.4 m, just as the Riemann sums for $y = x^2$ settled on $1/3$ in §1.2.

The Punchline: A First Sighting of the Fundamental Theorem

Look at the two answers we got, by two completely different routes:

From the position curve $s(t) = 4.6t^2$, the distance covered by $t = 3$ is simply $s(3) = 4.6 \times 9 = 41.4$ m.
From the velocity curve $v(t) = 9.2t$, the area underneath it on $[0, 3]$ is also 41.4 m.

That is not a coincidence. The velocity is the slope (derivative) of the position; the distance is the area (integral) under the velocity. Getting the same number both ways is the Fundamental Theorem of Calculus (§1.4) appearing in the wild: integrating a rate of change recovers the total change. Sarah's velocity is the rate at which her position changes; accumulate that rate over three seconds and you get back the change in her position. We previewed exactly this with $y = x^2$ and its antiderivative $x^3/3$ in §1.4 — here it is again, wearing a fur coat.

Why this matters. This single example contains both halves of calculus and the bridge between them. Differentiation took position and produced velocity (a slope). Integration took velocity and produced distance (an area). The Fundamental Theorem guarantees they undo each other — which is why, once you can do one, you get the other for free. Every chapter of this book is, in some sense, a deeper version of clocking the cheetah.

Connections to Other Chapters

Velocity as the derivative of position; acceleration as the derivative of velocity: Chapter 5 (the derivative as instantaneous rate) and Chapter 6.
The power rule that makes $\frac{d}{dt}(4.6t^2) = 9.2t$ a one-line calculation: Chapter 7.
Distance as $\int v(t)\,dt$, made rigorous as a limit of Riemann sums: Chapter 13.
The Fundamental Theorem that links the two (and makes $s(3)$ the shortcut for the area): Chapter 14.
Full projectile and orbital motion, where position is a vector $\mathbf{r}(t)$: Chapters 25 and 28.

Discussion Questions

We modelled only the first three seconds with $s(t) = 4.6t^2$. After top speed, Sarah runs at a roughly constant $\approx 27.6$ m/s. Sketch what the position graph looks like for $0 \le t \le 5.95$. Where is the curve steepest, and what does that steepness mean physically?
The three-rectangle distance estimate (55.2 m) overshot the true 41.4 m. Using a picture of right-endpoint rectangles under a rising line, explain why right-endpoint sums overshoot for an increasing velocity. Would left-endpoint rectangles overshoot or undershoot?
We found $v(t) = 9.2t$ by shrinking $\Delta t$. Suppose your radar gun can only sample every $0.1$ s. How would noisy, real-world data complicate the convergence you saw in the clean table? What might you do about it?
The Fundamental Theorem let us read the distance straight off $s(3)$ instead of summing a thousand rectangles. In your own words, why is "find an antiderivative and subtract" so much less work than "add up infinitely many rectangles"?
Inertial navigation systems in aircraft integrate measured acceleration twice to track position with no GPS. Lay out the two integration steps. Which chapter's tools would you need to carry them out by formula rather than by Riemann sum?

Your Turn — Mini-Project

Using the model $s(t) = 4.6t^2$ for $0 \le t \le 3$:

Write a function secant_velocity(t0, dt) and tabulate the instantaneous velocity at $t = 1$ and $t = 2.5$ by shrinking dt. Confirm your limits match $v(t) = 9.2t$.
Write a right_riemann(a, b, n) that estimates the area under $v(t) = 9.2t$. Estimate the distance covered on $[1, 3]$ with $n = 1000$, and check it against $s(3) - s(1)$.
Plot $s(t)$ and $v(t)$ on the same time axis. Mark the point $t = 2$, draw the tangent line to $s$ there with slope $18.4$, and shade the area under $v$ from $0$ to $3$. You will have drawn, in one figure, both ancient problems and the theorem that unites them.

Case Study 1 — Clocking a Cheetah: Velocity From Position

The Setup

The Tangent Problem: How Fast at $t = 2$?

The Area Problem: How Far in Three Seconds?

The Punchline: A First Sighting of the Fundamental Theorem

Connections to Other Chapters

Discussion Questions

Your Turn — Mini-Project

Further Reading