6.3 Resistive Forces and Terminal Speed

The Skydiver's Puzzle

A skydiver jumps from a plane at 4,000 meters. For the first few seconds, she accelerates --- fast. The speedometer climbs: 20 m/s, 30 m/s, 40 m/s. But then something changes. The acceleration slows. The speed creeps upward more gradually. After about 12 seconds, the speedometer reads roughly 55 m/s and stops climbing. She is still falling. Gravity has not turned off. The Earth has not stopped pulling. And yet, her acceleration has dropped to zero.

This is not a special trick. It happens to every object falling through a fluid --- a raindrop, a ping-pong ball, a coffee filter, a cat. At some point, the speed levels off. The object falls at a constant velocity as if it had forgotten it was being pulled by gravity.

Why?

The answer lives in the interplay between two forces: gravity pulling down and drag pushing up. You have already met both --- weight in Section 5.4, drag in Section 5.4, and the differential equation that governs their competition in Section 4.3. This section is where the physics catches up to the mathematics. Here, you will see why that differential equation takes the form it does, what terminal velocity means as a force-balance condition, and how the approach to terminal speed connects to the exponential solutions you have already derived.

Prediction

Before you read on: Two skydivers jump together from the same altitude at the same moment. Skydiver A has mass 60 kg. Skydiver B has mass 120 kg. Assume both have the same body position and therefore the same drag coefficient and cross-sectional area.

Two questions:

Who reaches terminal velocity first?

Who has a higher terminal velocity?

Commit to your answers before continuing.

[Interactive: Predict-Then-Reveal. The student selects answers to both questions from dropdowns. Choices for question 1: "The lighter skydiver," "The heavier skydiver," "They reach it at the same time." Choices for question 2: "The lighter skydiver," "The heavier skydiver," "They have the same terminal velocity." After committing, the student sees: "The heavier skydiver has a higher terminal velocity --- the drag force must be larger to balance the greater weight, and that requires a higher speed. As for who reaches terminal velocity first: the lighter skydiver reaches hers sooner, because the drag force catches up to her smaller weight at a lower speed. We will derive both results below."]

If your intuition told you the heavier skydiver falls faster, good --- that is correct, and the reason is not "heavier things fall faster" in the naive sense. It is that a heavier object needs more drag to reach equilibrium, and more drag requires more speed. If you were unsure about the timing, that is natural --- it requires thinking carefully about how quickly the force balance evolves for each mass. We will make this precise.

The Guiding Question

Why do some moving objects stop accelerating even while forces still act on them?

The key word is "net." Forces still act --- gravity pulls, drag pushes --- but they can balance. When the net force is zero, the acceleration is zero, even though multiple forces are present and large. This section is about understanding how that balance is reached, why it is stable, and what determines the speed at which it occurs.

Exploration: Watching the Force Balance Evolve

[Interactive: Drag and Terminal Speed Explorer. A single panel shows an object falling from rest through a fluid. On the left, a force diagram displays two arrows: weight $mg$ pointing down (constant length) and drag $f_D$ pointing up (length grows with speed). A net force arrow shows the difference. On the right, three graphs update in real time: $v(t)$, $a(t)$, and a force-balance plot showing $mg$ and $f_D$ as functions of time. Sliders control: mass $m$ (1 to 10 kg), drag parameter $b$ (0.1 to 5 s$^{-1}$), and a toggle between linear drag ($f_D = bv$) and quadratic drag ($f_D = cv^2$). The terminal speed is marked with a dashed line on the velocity graph.]

Prompt 1: Set $m = 2$ kg and $b = 0.5$ s$^{-1}$ with linear drag. Watch the falling object. At $t = 0$, how large is the drag force? How large is the net force? What is the acceleration?

Prompt 2: Now watch as the object speeds up. What happens to the drag arrow on the force diagram? What happens to the net force arrow? At what moment does the net force arrow disappear?

Prompt 3: Look at the $v(t)$ graph. Describe its shape. Is the velocity increasing linearly? What happens to the rate of increase as time passes?

Prompt 4: Look at the $a(t)$ graph. What does the acceleration do over time? Does it ever become negative? What value does it approach?

Prompt 5: Now double the mass to $m = 4$ kg, keeping $b = 0.5$ s$^{-1}$. What changes on the velocity graph? Does the terminal speed increase or decrease? By how much?

Prompt 6: Reset to $m = 2$ kg, and instead increase $b$ from 0.5 to 1.0. What changes? Does the terminal speed increase or decrease?

If you worked through those prompts, you saw the central mechanism: drag starts at zero (when the object is at rest), grows as the object speeds up, and eventually equals gravity. At that moment, the net force vanishes, the acceleration drops to zero, and the speed stops changing. The speed at which this balance occurs depends on both the mass and the drag parameter. More mass means more weight to balance, requiring higher speed. More drag means the balance is reached at a lower speed.

The Physics of Terminal Speed

Let us now make this precise. Consider an object of mass $m$ falling through a fluid, starting from rest. Two forces act on it:

Weight $mg$, directed downward (constant).
Drag $f_D$, directed upward (grows with speed).

Taking downward as positive, Newton's second law gives:

$$ma = mg - f_D$$

The minus sign on $f_D$ is because drag opposes the downward motion. The drag force grows with speed, so as $v$ increases, $f_D$ increases, and the net force $mg - f_D$ shrinks. The acceleration is:

$$a = g - \frac{f_D}{m}$$

At $t = 0$, the object is at rest, so $f_D = 0$ and $a = g$ --- the object accelerates at the full gravitational rate. As speed builds, drag grows, and $a$ decreases. Eventually, drag equals weight:

$$f_D = mg$$

At this point $a = 0$, and the velocity stops changing. The object has reached terminal speed.

This is an equilibrium --- not of position, but of velocity. The object is still moving. It is still subject to two large forces. But those forces balance perfectly, and the motion continues at constant speed.

Terminal Speed for Linear Drag

For small, slow objects moving through viscous fluids, drag is proportional to speed:

$$f_D = bv$$

where $b$ is a positive constant (with units of kg/s) that depends on the object's size, shape, and the fluid's viscosity. Newton's second law becomes:

$$m\frac{dv}{dt} = mg - bv$$

or equivalently:

$$\frac{dv}{dt} = g - \frac{b}{m}v$$

To find terminal speed, set $dv/dt = 0$:

$$0 = g - \frac{b}{m}v_{\text{term}}$$

$$v_{\text{term}} = \frac{mg}{b}$$

Notice what this says. Terminal speed is proportional to mass (heavier objects fall faster at terminal velocity) and inversely proportional to the drag parameter (more drag means lower terminal speed). This answers the prediction: the 120 kg skydiver has a terminal speed twice that of the 60 kg skydiver, assuming the same $b$.

Terminal Speed for Quadratic Drag

For larger, faster objects moving through air --- skydivers, baseballs, raindrops --- the drag force is better modeled as proportional to $v^2$:

$$f_D = cv^2$$

where $c = \frac{1}{2}C_D \rho A$ combines the drag coefficient $C_D$, the fluid density $\rho$, and the cross-sectional area $A$. Setting drag equal to weight:

$$cv_{\text{term}}^2 = mg$$

$$v_{\text{term}} = \sqrt{\frac{mg}{c}}$$

Now terminal speed grows as the square root of mass. Doubling the mass does not double the terminal speed --- it increases it by a factor of $\sqrt{2} \approx 1.41$. The 120 kg skydiver's terminal speed is about 41% higher than the 60 kg skydiver's, not double.

This distinction matters. Linear drag applies to fog droplets and dust particles. Quadratic drag applies to skydivers and falling sports equipment. The physics is the same --- drag balances gravity --- but the quantitative relationship between mass and terminal speed changes with the drag law.

The Full Solution: Approach to Terminal Speed

You already solved the differential equation for linear drag in Section 4.3. Let us revisit that result, now with physical understanding behind it.

The equation is:

$$\frac{dv}{dt} = g - \frac{b}{m}v, \qquad v(0) = 0$$

Define $\beta = b/m$ for compactness. The solution (derived by separation of variables in Section 4.3) is:

$$v(t) = \frac{g}{\beta}\left(1 - e^{-\beta t}\right) = v_{\text{term}}\left(1 - e^{-t/\tau}\right)$$

where $v_{\text{term}} = g/\beta = mg/b$ and $\tau = 1/\beta = m/b$ is the time constant.

The acceleration is:

$$a(t) = \frac{dv}{dt} = g\,e^{-\beta t} = g\,e^{-t/\tau}$$

Let us read these equations physically:

At $t = 0$: $v = 0$ and $a = g$. The object starts from rest and accelerates at the full gravitational rate. Drag is zero because speed is zero.
At $t = \tau$: $v = v_{\text{term}}(1 - e^{-1}) \approx 0.63\,v_{\text{term}}$. The object has reached 63% of terminal speed. The acceleration has dropped to $g\,e^{-1} \approx 0.37g$ --- only 37% of the initial value. Drag has grown to match 63% of gravity.
At $t = 3\tau$: $v \approx 0.95\,v_{\text{term}}$. The object is within 5% of terminal speed. The acceleration is $0.05g$ --- nearly zero.
As $t \to \infty$: $v \to v_{\text{term}}$ and $a \to 0$. The force balance is complete.

The approach is exponential, not linear. The object does not gain speed at a steady rate and then suddenly stop accelerating. Instead, the acceleration itself decays exponentially. The closer the velocity gets to $v_{\text{term}}$, the slower the remaining approach. Strictly speaking, the object never quite reaches terminal speed --- it gets asymptotically closer, forever. In practice, after about $3\tau$ to $5\tau$, the speed is indistinguishable from $v_{\text{term}}$.

Connection to Section 4.3

In Section 4.3, you solved $dv/dt = g - \beta v$ as a mathematical exercise in separable differential equations. You found the exponential solution, computed the time constant, and analyzed the asymptotic behavior. All of that was correct --- but it was algebra.

Now you see where that equation comes from. The $g$ term is gravity. The $-\beta v$ term is drag. The equation says: the net acceleration is gravity minus a velocity-dependent resistance. The terminal velocity $v_{\text{term}} = g/\beta$ is not just a mathematical limit of the solution --- it is the speed at which drag equals weight. The time constant $\tau = 1/\beta$ is not just a parameter in the exponential --- it measures how quickly the drag-gravity contest approaches its conclusion.

The same equation, the same solution, the same exponential approach. But the meaning is different. In Chapter 4, the equation was a model to be solved. Here, it is a physical story: gravity pulls, drag resists, and the competition has an inevitable outcome.

Multiple Representations

The same falling object, viewed three ways.

1. Force diagram evolving in time

[Video: A single object falls downward. Beside it, a free-body diagram updates in real time. At $t = 0$: a long downward arrow (weight $mg$) and no upward arrow. As the object speeds up, an upward arrow (drag) appears and grows. The net force arrow (shown in a distinct color) starts equal to $mg$ and shrinks. At terminal speed, the two arrows are equal in length, the net force arrow vanishes, and a label reads "$a = 0$." The video loops, pausing at three key moments: early fall ($a \approx g$), mid-fall ($a \approx g/2$), and terminal speed ($a = 0$).]

At the start, gravity dominates. The net force is large, and the object accelerates rapidly. As speed increases, the drag arrow grows. The net force shrinks. At terminal speed, the two arrows match. This is what "forces are balanced" looks like --- not the absence of forces, but two large forces in perfect opposition.

2. Velocity-time graph

The $v(t)$ curve starts at zero with a steep slope (slope = $g$), bends over as drag grows, and flattens asymptotically toward $v_{\text{term}}$. The curve has the characteristic shape of $1 - e^{-t/\tau}$ --- steep at first, progressively flatter, never quite reaching the horizontal dashed line.

The slope of the $v(t)$ graph at any moment is the acceleration. Early on, the slope is steep (large acceleration). Later, the slope is nearly flat (small acceleration). At terminal speed, the slope is zero.

3. Acceleration-time graph

The $a(t)$ curve starts at $g$ and decays exponentially toward zero: $a(t) = g\,e^{-t/\tau}$. This is a pure exponential decay --- the same mathematical shape as the pure-drag velocity decay from Section 4.3, but here it describes how the acceleration fades rather than the velocity.

These three representations tell the same story in different visual languages. The force diagram shows why the acceleration changes (the drag arrow grows). The velocity graph shows the consequence (speed approaches a limit). The acceleration graph shows the mechanism (the decay of the net force toward zero). Fluency means being able to start from any one and explain the other two.

Terminal Speed as an Attractor

There is a subtle and important feature of terminal speed: it is not just a speed the object approaches from below. It is an attractor --- a speed the object approaches from any initial condition.

Consider the general solution for gravity plus linear drag, starting from an arbitrary initial velocity $v_0$:

$$v(t) = v_{\text{term}} + (v_0 - v_{\text{term}})\,e^{-t/\tau}$$

If $v_0 = 0$ (dropped from rest): the velocity increases toward $v_{\text{term}}$ from below. This is the standard case.
If $v_0 < v_{\text{term}}$ (thrown downward, but slower than terminal speed): the velocity still increases toward $v_{\text{term}}$, starting partway there.
If $v_0 = v_{\text{term}}$ (launched at exactly terminal speed): the velocity stays at $v_{\text{term}}$ forever. The exponential term vanishes.
If $v_0 > v_{\text{term}}$ (launched downward faster than terminal speed): the velocity decreases toward $v_{\text{term}}$. Drag exceeds gravity, so the net force is upward, decelerating the object until the speed drops to terminal.

In every case, the exponential term $e^{-t/\tau}$ decays to zero, and $v(t) \to v_{\text{term}}$. The terminal speed is the unique velocity where the forces balance, and the dynamics always push the system toward it. Overshoot or undershoot --- it does not matter. The system corrects.

This is what makes terminal speed an equilibrium. It is not just a value the speed happens to reach. It is a value the speed is drawn to by the structure of the forces. In the language you will encounter later in this course: terminal speed is a stable equilibrium of the velocity.

[Interactive: Attractor Demonstration. The $v(t)$ curve is plotted for gravity-plus-linear-drag. A draggable dot on the vertical axis lets the student choose $v_0$. The horizontal dashed line shows $v_{\text{term}}$. As $v_0$ is dragged above or below $v_{\text{term}}$, the curve adjusts --- always converging to the dashed line. Guided prompt: "Try launching faster than terminal speed. What happens to the acceleration? Try launching at exactly terminal speed. What does the curve look like?"]

Answering the Prediction

Now we can answer both questions from the prediction precisely.

Who has a higher terminal velocity? For quadratic drag (the realistic model for skydivers), $v_{\text{term}} = \sqrt{mg/c}$. With the same $c$, doubling the mass gives:

$$\frac{v_{\text{term,B}}}{v_{\text{term,A}}} = \sqrt{\frac{m_B}{m_A}} = \sqrt{2} \approx 1.41$$

The heavier skydiver has a terminal speed about 41% higher.

Who reaches terminal velocity first? This is trickier. The time constant for linear drag is $\tau = m/b$, which is proportional to mass. The heavier skydiver has a longer time constant --- it takes longer for the drag-gravity contest to play out. The lighter skydiver reaches her (lower) terminal speed sooner. The heavier skydiver reaches his (higher) terminal speed later.

So the lighter skydiver stabilizes first at a lower speed, and the heavier skydiver takes longer to stabilize at a higher speed. If they jump together, the heavier skydiver initially falls at the same rate (both start at $a = g$), then gradually pulls ahead as the lighter skydiver's acceleration drops off first.

Practice

Layer 1: Concrete

Problem 1. A small ball bearing of mass 0.01 kg falls through oil. The linear drag coefficient is $b = 0.05$ kg/s.

(a) Calculate the terminal speed.

(b) Calculate the time constant $\tau$.

(c) How fast is the ball bearing moving after one time constant? After three time constants?

Check your answer

**(a)** Terminal speed: $$v_{\text{term}} = \frac{mg}{b} = \frac{0.01 \times 9.8}{0.05} = 1.96 \text{ m/s}$$ **(b)** Time constant: $$\tau = \frac{m}{b} = \frac{0.01}{0.05} = 0.2 \text{ s}$$ **(c)** After one time constant ($t = \tau$): $$v(\tau) = v_{\text{term}}(1 - e^{-1}) = 1.96 \times 0.632 = 1.24 \text{ m/s}$$ This is 63% of terminal speed. After three time constants ($t = 3\tau$): $$v(3\tau) = v_{\text{term}}(1 - e^{-3}) = 1.96 \times 0.950 = 1.86 \text{ m/s}$$ This is 95% of terminal speed. After just 0.6 seconds, the ball bearing is nearly at its final falling speed.

Problem 2. A skydiver of mass 80 kg falls in a spread-eagle position. The quadratic drag parameters give $c = \frac{1}{2}C_D \rho A = 0.25$ kg/m.

(a) Calculate the terminal speed.

(b) Convert your answer to km/h.

(c) If the skydiver tucks into a head-down dive, reducing the effective area by a factor of 3 (so $c$ becomes $0.25/3 \approx 0.083$ kg/m), what is the new terminal speed?

Check your answer

**(a)** Terminal speed with quadratic drag: $$v_{\text{term}} = \sqrt{\frac{mg}{c}} = \sqrt{\frac{80 \times 9.8}{0.25}} = \sqrt{3136} = 56 \text{ m/s}$$ **(b)** Convert: $56 \text{ m/s} \times 3.6 = 202 \text{ km/h}$, or about 125 mph. This matches the well-known value for a spread-eagle skydiver. **(c)** With the reduced area: $$v_{\text{term}} = \sqrt{\frac{80 \times 9.8}{0.083}} = \sqrt{9446} \approx 97 \text{ m/s}$$ That is about 350 km/h (217 mph). Same person, same mass --- but a smaller cross-section dramatically increases the terminal speed. This is why competitive skydivers in head-down dives can exceed 200 mph.

Layer 2: Pattern

Problem 3. Three objects fall through the same fluid with linear drag coefficient $b$. They have masses $m$, $2m$, and $3m$. All start from rest.

(a) Rank their terminal speeds from lowest to highest.

(b) Rank their time constants from shortest to longest.

(c) After a very long time, all three are at terminal speed. Rank the net forces acting on them.

Check your answer

**(a)** Terminal speed is $v_{\text{term}} = mg/b$. Since $b$ is the same for all three: $$v_{\text{term},1} = \frac{mg}{b}, \quad v_{\text{term},2} = \frac{2mg}{b}, \quad v_{\text{term},3} = \frac{3mg}{b}$$ Ranking: $v_{\text{term},1} < v_{\text{term},2} < v_{\text{term},3}$. Heavier objects have higher terminal speeds. **(b)** The time constant is $\tau = m/b$. So the time constants are $m/b$, $2m/b$, and $3m/b$. Ranking: $\tau_1 < \tau_2 < \tau_3$. Heavier objects take longer to reach terminal speed. (They have both farther to go and a slower timescale.) **(c)** After a very long time, all three are at terminal speed, which means $a = 0$ for all three. By Newton's second law, $F_{\text{net}} = ma = 0$ for each. **The net force is zero for all three --- they are tied.** This is the defining condition of terminal speed. The individual forces (gravity and drag) are different for each object, but the net force is zero in every case.

Problem 4. Object A falls through air (quadratic drag, $c_A$). Object B falls through honey (linear drag, $b_B$). Both have the same mass $m$ and both reach terminal speed.

(a) Write the terminal speed for each object.

(b) If you double the mass of each, by what factor does each terminal speed change?

Check your answer

**(a)** For Object A (quadratic drag): $v_{\text{term,A}} = \sqrt{mg/c_A}$. For Object B (linear drag): $v_{\text{term,B}} = mg/b_B$. **(b)** Doubling the mass: Object A: $v_{\text{term,A}}' = \sqrt{2mg/c_A} = \sqrt{2}\,v_{\text{term,A}}$. The terminal speed increases by a factor of $\sqrt{2} \approx 1.41$. Object B: $v_{\text{term,B}}' = 2mg/b_B = 2\,v_{\text{term,B}}$. The terminal speed doubles. The drag law determines the mass-dependence of terminal speed. With linear drag, terminal speed is proportional to mass. With quadratic drag, it is proportional to the square root of mass. This is a structural consequence of the force model, not a coincidence.

Layer 3: Structure

Problem 5. Why is the approach to terminal speed exponential rather than linear?

In other words: why does the velocity curve bend toward $v_{\text{term}}$ gradually, instead of increasing at a steady rate and then abruptly leveling off?

Check your answer

The approach is exponential because **the rate of change of velocity depends on the velocity itself**. When the velocity is far below terminal speed, the drag force is small relative to gravity, and the net force (and thus the acceleration) is large. The object gains speed quickly. But as the velocity increases, drag increases, the net force decreases, and the acceleration drops. The closer the velocity gets to terminal speed, the smaller the remaining net force, and the slower the remaining velocity change. Mathematically, the equation $dv/dt = g - \beta v$ says that the rate of velocity change is proportional to the *gap* between the current speed and the terminal speed. Let $\Delta v = v_{\text{term}} - v$. Then: $$\frac{d(\Delta v)}{dt} = -\frac{dv}{dt} = -(g - \beta v) = -\beta(v_{\text{term}} - v) = -\beta\,\Delta v$$ This says: the rate at which the gap closes is proportional to the size of the gap. That is exactly the condition for exponential decay. A large gap closes quickly; a small gap closes slowly. The gap never reaches zero --- it just gets exponentially smaller. A linear approach would require the acceleration to be constant right up to the moment of terminal speed and then abruptly jump to zero. But that would require the drag force to remain constant while the speed changes --- which contradicts the drag model. The velocity-dependence of drag is what makes the approach gradual, and the proportionality between the gap and the rate of change is what makes it specifically exponential.

Layer 4: Debug

Problem 6. A student says: "An object at terminal velocity has no forces acting on it. That's why it doesn't accelerate."

What is wrong with this statement? Rewrite it correctly.

Check your answer

The student has confused "no net force" with "no forces." At terminal velocity, there are at least two large forces acting on the object --- gravity pulling downward and drag pushing upward. These forces are not zero. They are equal in magnitude and opposite in direction, so their *sum* is zero. Newton's second law says $F_{\text{net}} = ma$. Zero acceleration requires zero *net* force, not zero forces. The correct statement is: "An object at terminal velocity has zero *net* force acting on it. Gravity and drag are both present but they balance each other exactly, so the acceleration is zero." This is the same distinction that arises with any equilibrium. A book resting on a table has two forces (weight and normal force) that balance. It would be wrong to say no forces act on the book. The same logic applies to terminal velocity --- it is a dynamic equilibrium, maintained by the balance of two large and opposing forces.

Problem 7. A student models a falling tennis ball using linear drag: $dv/dt = g - bv$. They compute $v_{\text{term}} = g/b = 50$ m/s. But when they drop the ball from a tall building, the measured terminal speed is about 31 m/s. They double-check their value of $b$ and confirm it matches published data for tennis balls. What went wrong?

Check your answer

The student used the wrong drag model. A tennis ball falling through air at speeds around 30--50 m/s experiences *quadratic* drag ($f_D = cv^2$), not linear drag ($f_D = bv$). Linear drag is appropriate for very small, slow objects in viscous fluids --- dust particles, small spheres in oil. For a tennis ball in air, the Reynolds number is far too high for linear drag to apply. With quadratic drag, the terminal speed is $v_{\text{term}} = \sqrt{mg/c}$, which gives a different numerical value from $mg/b$. The student's drag data was correct, but they plugged it into the wrong formula. The physics demands matching the drag model to the physical regime --- and for a tennis ball in air, that regime is quadratic. This is a good reminder that the choice of drag model (linear vs. quadratic) is not arbitrary. It is dictated by the physics of the situation, particularly the size, speed, and fluid involved. Using the wrong model gives the wrong terminal speed even when the input data is correct.

Reflection

Think about terminal speed and the concept of equilibrium.

What is the relationship between terminal speed and equilibrium?

In Section 5.3, you learned that equilibrium means zero net force. Terminal speed is exactly that --- the velocity at which the net force on a falling object is zero. But it is an unusual kind of equilibrium: the object is not at rest. It is moving at constant velocity. The position changes; the velocity does not.

Now consider: is terminal speed the same kind of equilibrium as a book sitting on a table? Both have zero net force. Both have zero acceleration. What is different?

Finally, think about the attractor property. A book on a table, if nudged slightly, returns to rest. A falling object, if its speed is perturbed above or below $v_{\text{term}}$, returns to $v_{\text{term}}$. Both are stable equilibria. The book's equilibrium is in position. Terminal speed is an equilibrium in velocity. The mathematics --- exponential return toward a fixed point --- is the same in both cases.

Looking Ahead

You have now seen how resistive forces create a natural speed limit for falling objects, and how the competition between gravity and drag produces the exponential approach to terminal speed. This is the physical story behind the differential equation you solved in Section 4.3.

In the next section, we turn to situations where the net force is not constant and not simply velocity-dependent --- cases where force depends on position, configuration, or other changing quantities. The acceleration itself becomes a state-dependent quantity, and the simple models of this section give way to richer dynamics.