4.3 Constant, Linear, and Velocity-Dependent Acceleration Models
Three Falls
Imagine three identical steel balls released from rest at the same height, at the same instant. One falls through vacuum. One falls through water. One falls through honey.
In vacuum, there is no resistance. The ball accelerates steadily, picking up speed all the way down. In water, drag pushes back against the motion, and the ball reaches a modest, steady sinking speed fairly quickly. In honey, the resistance is so fierce that the ball barely accelerates at all --- it almost immediately creeps downward at a slow, constant pace.
Same object. Same gravity. Same initial conditions. Three completely different motions. The difference is not in the setup --- it is in the acceleration law. Each medium imposes a different mathematical relationship between the ball's acceleration and its current state. That relationship --- the form of the differential equation --- determines everything about how the motion unfolds.
[Video: Three side-by-side columns, each showing a ball released from rest. Left column labeled "Vacuum," middle labeled "Water," right labeled "Honey." All three balls start falling at the same moment. The vacuum ball accelerates visibly, pulling far ahead. The water ball accelerates at first, then levels off to a steady descent. The honey ball barely accelerates --- it moves at a near-constant crawl almost from the start. Velocity readouts update in real time below each column. The video runs for about 5 seconds, by which point the vacuum ball is far below, the water ball is partway down at a steady speed, and the honey ball has barely moved.]
This section is about comparing acceleration models --- placing them side by side, solving them, and understanding how the mathematical structure of the model determines the qualitative character of the motion.
Prediction: Who Is Faster After a Long Time?
Before you read on: Two objects are dropped from rest.
- Object A falls through vacuum: $a = -g$ (constant downward acceleration, with upward positive).
- Object B falls through a resistive medium: $a = -g + bv$ (gravity minus a drag term proportional to speed, where $b > 0$ and $v < 0$ during the fall).
After a long time, which object has a larger speed?
- (a) Object A, because nothing resists its motion
- (b) Object B, because the drag eventually "pushes" it faster
- (c) They approach the same speed eventually
Commit to your answer before continuing.
[Interactive: Predict-Then-Reveal. Student selects one of three choices. After submitting, the correct answer is revealed: (a). A brief explanation appears: "Object A accelerates forever --- its speed grows without bound (until it hits the ground). Object B's drag grows with speed, so its acceleration shrinks toward zero. Object B approaches a finite limiting speed. Object A always surpasses it. The concept behind Object B's limiting speed has a name --- terminal velocity --- which we will derive shortly."]
If you chose (a), good --- your physical intuition is sound. If you chose (c), you may have been thinking of a situation where both objects fall the same distance, but the question asks about speed after a long time. Object A never stops accelerating. Object B does. That distinction is entirely a consequence of the acceleration law.
The Guiding Question
What different motion behaviors arise from different forms of acceleration laws?
In Sections 4.1 and 4.2, you learned to translate physical rules into differential equations and to solve initial-value problems. Now we put that machinery to work across several models at once. The goal is not just to solve each equation, but to compare the solutions and understand why different models produce different behaviors.
Three Models, Side by Side
Here are three acceleration models, each written as a differential equation for velocity. In every case, the object starts from rest: $v(0) = 0$.
| Model | Acceleration law | Differential equation | Physical example |
|---|---|---|---|
| Constant | $a = -g$ | $\dfrac{dv}{dt} = -g$ | Free fall in vacuum |
| Velocity-dependent (pure drag) | $a = -bv$ | $\dfrac{dv}{dt} = -bv$ | Object moving through viscous fluid, no gravity |
| Gravity plus linear drag | $a = -g + bv$ | $\dfrac{dv}{dt} = -g + bv$ | Falling through a resistive medium |
(Here we take downward as negative, so $g > 0$, $b > 0$, and a falling object has $v < 0$. The drag term $+bv$ is positive when $v < 0$, opposing the downward motion --- it acts upward, as drag should.)
These three models share the same physical scenario (an object falling under gravity through some medium) but differ in how the medium resists motion. The constant model ignores resistance entirely. The pure drag model ignores gravity. The gravity-plus-drag model includes both. Each produces a qualitatively different motion.
Pause and think: Before we solve anything, look at the three differential equations. In the constant model, the right-hand side is just a number. In the drag model, it depends on $v$. In the gravity-plus-drag model, it depends on $v$ and has a constant term. How do you think these structural differences will show up in the solutions?
Exploration: Watching the Models Diverge
[Interactive: Three-Model Comparator. Three panels arranged horizontally, each showing position $x(t)$, velocity $v(t)$, and acceleration $a(t)$ graphs for one of the three models. All share the same time axis. The initial conditions are identical: $v(0) = 0$, $x(0) = 0$. Sliders control the parameters $g$ and $b$. Students can toggle individual models on and off and overlay them on a single set of axes for direct comparison.]
Prompt 1: Set $b = 0.5$ s$^{-1}$ and $g = 10$ m/s$^2$. Play all three models. Watch the velocity graphs. Which model produces a velocity that grows without bound? Which levels off? Which decays?
Prompt 2: Focus on the gravity-plus-drag model. Increase $b$. What happens to the speed the object eventually reaches? Decrease $b$ toward zero. What model does it start to resemble?
Prompt 3: Now look at the acceleration graphs. In the constant model, acceleration is flat. In the gravity-plus-drag model, how does acceleration change over time? What value does it approach?
Prompt 4: Compare the position graphs. The constant model gives a parabola. What shape does the gravity-plus-drag model produce? Is the position graph concave up, concave down, or does its concavity change?
If you explored those prompts carefully, you saw the key pattern: the form of the acceleration law determines the shape of the solution. Constant acceleration gives straight-line velocity and parabolic position. Velocity-dependent acceleration gives curves that bend toward an equilibrium. The same initial conditions, fed into different equations, produce fundamentally different motions.
Model 1: Constant Acceleration (Review)
You solved this in Section 2.2, so we will be brief. The equation is
$$\frac{dv}{dt} = -g$$
Integrating both sides from $0$ to $t$:
$$v(t) = v(0) + (-g)t = -gt$$
(since $v(0) = 0$). Integrating again:
$$x(t) = x(0) + \int_0^t v(t')\,dt' = -\tfrac{1}{2}gt^2$$
The velocity is a straight line with slope $-g$. The position is a downward-opening parabola. The speed $|v|$ grows linearly, forever. There is no mechanism to stop the acceleration --- it is always $-g$, regardless of how fast the object moves.
This is the baseline. Every comparison in this section starts here.
Model 2: Pure Velocity-Dependent Drag
Now consider an object moving through a viscous fluid with no gravitational pull --- perhaps a puck sliding on an oily surface. The only effect is drag proportional to velocity:
$$\frac{dv}{dt} = -bv$$
where $b > 0$ is a constant that characterizes the strength of the drag. This says: the faster you move, the more the medium slows you down. If $v = 0$, the acceleration is zero --- the object at rest stays at rest.
Solving by separation of variables
This is the first velocity-dependent DE we will solve in full. The technique is separation of variables: gather all the $v$-stuff on one side and all the $t$-stuff on the other.
Step 1: Separate. Rewrite the equation as
$$\frac{dv}{v} = -b\,dt$$
(This requires $v \neq 0$. We will address $v = 0$ separately.)
Step 2: Integrate both sides.
$$\int_{v_0}^{v} \frac{dv'}{v'} = \int_0^t (-b)\,dt'$$
$$\ln|v| - \ln|v_0| = -bt$$
$$\ln\left|\frac{v}{v_0}\right| = -bt$$
Step 3: Solve for $v$.
$$\left|\frac{v}{v_0}\right| = e^{-bt}$$
Since $v$ and $v_0$ have the same sign (drag cannot reverse the direction of motion), we can drop the absolute values:
$$\boxed{v(t) = v_0\, e^{-bt}}$$
Step 4: Interpret. The velocity decays exponentially toward zero. It never actually reaches zero --- it just gets closer and closer. After a time $\tau = 1/b$, the velocity has fallen to $v_0/e \approx 0.37\, v_0$. After $2\tau$, it is $v_0/e^2 \approx 0.14\, v_0$. The constant $\tau = 1/b$ is called the time constant of the exponential decay.
Step 5: Find position. Integrate velocity:
$$x(t) = x(0) + \int_0^t v_0\, e^{-bt'}\,dt' = x(0) + v_0 \left[-\frac{1}{b}e^{-bt'}\right]_0^t = x(0) + \frac{v_0}{b}\left(1 - e^{-bt}\right)$$
As $t \to \infty$, the position approaches $x(0) + v_0/b$. The object covers a finite total distance, even though it never truly stops. This is the hallmark of exponential decay: rapid early change, slower and slower later, asymptotically approaching a limit.
What changed? What stayed the same? Compare this with the constant-acceleration model. In both cases, we integrated a DE to get velocity, then integrated velocity to get position. The technique is the same. But the solutions are structurally different: straight line vs. exponential curve, unbounded distance vs. finite distance. The difference comes entirely from the right-hand side of the DE.
Model 3: Gravity Plus Linear Drag
Now the main event. An object falls under gravity through a resistive medium:
$$\frac{dv}{dt} = -g + bv$$
Both gravity (pulling downward, contributing $-g$) and drag (opposing the motion, contributing $+bv$ when $v < 0$) are present. The object starts from rest: $v(0) = 0$.
Solving by separation of variables
Step 1: Separate.
$$\frac{dv}{-g + bv} = dt$$
Step 2: Integrate both sides. The left side requires a substitution. Let $u = -g + bv$, so $du = b\,dv$, which gives $dv = du/b$.
$$\int_0^t dt' = \int_{v(0)}^{v(t)} \frac{dv'}{-g + bv'}$$
At $v(0) = 0$: $u_{\text{initial}} = -g + b(0) = -g$.
At $v(t)$: $u_{\text{final}} = -g + bv$.
$$t = \frac{1}{b}\int_{-g}^{-g + bv} \frac{du}{u} = \frac{1}{b}\left[\ln|u|\right]_{-g}^{-g+bv} = \frac{1}{b}\ln\left|\frac{-g + bv}{-g}\right|$$
Since $v \leq 0$ during the fall and $b > 0$, we have $-g + bv \leq -g < 0$, so both numerator and denominator are negative and we can drop the absolute values (their ratio is positive):
$$t = \frac{1}{b}\ln\left(\frac{-g + bv}{-g}\right)$$
Wait --- both quantities inside the logarithm are negative, so the ratio is positive. Let us be more careful:
$$t = \frac{1}{b}\ln\left(\frac{g - bv}{g}\right)$$
(multiplying numerator and denominator by $-1$, which preserves the ratio since both were negative).
Step 3: Solve for $v$.
$$bt = \ln\left(1 - \frac{bv}{g}\right)$$
$$e^{bt} = 1 - \frac{bv}{g}$$
$$\frac{bv}{g} = 1 - e^{bt}$$
$$\boxed{v(t) = \frac{g}{b}\left(1 - e^{bt}\right)}$$
But recall that the object is falling downward, so $v < 0$. Since $b > 0$ and $e^{bt} > 1$ for $t > 0$, we have $1 - e^{bt} < 0$, confirming that $v(t) < 0$. Good --- the ball does fall.
Step 4: Check the initial condition. At $t = 0$: $v(0) = \frac{g}{b}(1 - e^0) = \frac{g}{b}(1 - 1) = 0$. Correct.
Step 5: Check the units. $g$ has units of m/s$^2$ and $b$ has units of 1/s (since $bv$ must have units of acceleration). So $g/b$ has units of m/s. The exponential is dimensionless. Everything checks out.
The emergence of terminal velocity
Now look at what happens as $t \to \infty$. The exponential $e^{bt}$ grows without bound, so
$$v(t) \to \frac{g}{b}(1 - e^{bt}) \to -\infty \text{ ?}$$
That cannot be right physically. Let me re-examine the sign conventions. We need to be careful. Let us rewrite the problem with a clearer convention.
Take downward as positive (so $v > 0$ means falling). Then gravity contributes $+g$ and drag opposes the fall, contributing $-bv$ (with $b > 0$):
$$\frac{dv}{dt} = g - bv, \qquad v(0) = 0$$
Now separate:
$$\frac{dv}{g - bv} = dt$$
Integrate from $0$ to $t$:
$$\int_0^{v} \frac{dv'}{g - bv'} = \int_0^t dt'$$
Let $u = g - bv'$, $du = -b\,dv'$:
$$-\frac{1}{b}\int_{g}^{g - bv}\frac{du}{u} = t$$
$$-\frac{1}{b}\left[\ln|u|\right]_g^{g-bv} = t$$
$$-\frac{1}{b}\ln\left(\frac{g - bv}{g}\right) = t$$
(Here $g - bv > 0$ as long as $v < g/b$, which is true throughout the motion, as we will see.)
$$\ln\left(\frac{g - bv}{g}\right) = -bt$$
$$\frac{g - bv}{g} = e^{-bt}$$
$$g - bv = g\,e^{-bt}$$
$$bv = g - g\,e^{-bt} = g(1 - e^{-bt})$$
$$\boxed{v(t) = \frac{g}{b}\left(1 - e^{-bt}\right)}$$
Now examine the long-time behavior. As $t \to \infty$, $e^{-bt} \to 0$, so
$$v(t) \to \frac{g}{b}$$
The velocity does not grow forever. It approaches a finite limiting value:
$$v_{\text{term}} = \frac{g}{b}$$
This is terminal velocity --- the speed at which the drag force exactly balances gravity, so the net acceleration drops to zero. You can verify this directly from the DE: set $dv/dt = 0$ in $dv/dt = g - bv$, which gives $0 = g - bv_{\text{term}}$, so $v_{\text{term}} = g/b$.
The solution tells us more than just the terminal speed. It tells us how the object reaches it. The exponential $e^{-bt}$ controls the approach: rapid at first, then slower and slower, asymptotically converging. The time constant is again $\tau = 1/b$. After one time constant, the velocity has reached about $63\%$ of terminal velocity. After three time constants, about $95\%$. After five, about $99\%$.
Pause and connect: Compare this with the pure drag model. In both cases, the solution involves an exponential with time constant $1/b$. In pure drag, the exponential describes decay from an initial velocity toward zero. In gravity-plus-drag, the exponential describes growth from zero toward terminal velocity. The exponential is doing the same mathematical job --- it describes the transition between two states --- but the physical stories are different.
Comparison Table: How Model Structure Shapes Motion
Let us collect all three models in one place and compare systematically. All start from $v(0) = 0$ (except the pure drag model, which needs $v(0) = v_0 \neq 0$ to have any motion at all).
| Feature | Constant: $dv/dt = g$ | Pure drag: $dv/dt = -bv$ | Gravity + drag: $dv/dt = g - bv$ |
|---|---|---|---|
| Velocity solution | $v = gt$ | $v = v_0\,e^{-bt}$ | $v = \frac{g}{b}(1 - e^{-bt})$ |
| Long-time velocity | $\to \infty$ | $\to 0$ | $\to g/b$ (terminal velocity) |
| Acceleration | Constant ($g$) | Decays: $a = -bv_0\,e^{-bt}$ | Decays: $a = g\,e^{-bt}$ |
| Long-time acceleration | $g$ (forever) | $\to 0$ | $\to 0$ |
| Velocity graph shape | Straight line | Exponential decay | Exponential approach to a limit |
| Position graph shape | Parabola | Saturating curve (finite limit) | Asymptotically linear |
| Key parameter | $g$ (sets the slope) | $b$ (sets the decay rate) | $g/b$ (sets terminal speed) |
[Interactive: Model Comparison Dashboard. The three velocity solutions are plotted on one set of axes. Sliders for $g$, $b$, and $v_0$ allow real-time adjustment. Below the velocity plot, matching acceleration and position plots update simultaneously. A table of numerical values at selected times ($t = 0, 1/b, 2/b, 3/b$) updates in real time.]
What changed? What stayed the same? All three models share the same mathematical technique (integration of a first-order DE). All produce velocity as a function of time. But the shape of that function --- linear, exponential decay, exponential approach --- is entirely determined by the right-hand side of the DE. The initial conditions pick a specific curve from a family; the model structure determines the family itself.
Variation Theory: Holding One Thing Fixed
Let us apply the variation-theory lens that runs through this course. We will hold one thing fixed and change another, and ask: what changed in the solution?
Vary the model, hold initial conditions fixed
We just did this. With $v(0) = 0$ in all three cases:
- Constant $a$: velocity grows without bound.
- Pure drag: no motion at all (you need $v_0 \neq 0$).
- Gravity + drag: velocity grows toward a finite limit.
The acceleration law alone --- not the starting state --- determines whether the object speeds up forever, slows down, or settles to a steady speed.
Vary the initial conditions, hold the model fixed
Now fix the model at $dv/dt = g - bv$ and change the initial velocity.
| Initial condition | What happens |
|---|---|
| $v(0) = 0$ (dropped from rest) | Velocity increases from zero toward $v_{\text{term}} = g/b$ |
| $v(0) = g/(2b)$ (launched downward at half of terminal speed) | Velocity increases toward $v_{\text{term}}$, but starts partway there |
| $v(0) = g/b$ (launched at terminal speed) | Velocity stays at $v_{\text{term}}$ --- acceleration is zero from the start |
| $v(0) = 2g/b$ (launched downward faster than terminal speed) | Velocity decreases toward $v_{\text{term}}$ --- drag exceeds gravity |
The general solution is $v(t) = \frac{g}{b} + \left(v_0 - \frac{g}{b}\right)e^{-bt}$. Every initial condition produces a curve that converges to the same terminal velocity. The starting point changes, but the destination does not.
Pause and think: This is a powerful result. No matter how you launch the object --- too fast, too slow, or exactly right --- it always settles to $v_{\text{term}} = g/b$. The model has a built-in attractor. Where does this property come from mathematically? (Hint: look at the sign of $dv/dt$ when $v > g/b$ versus when $v < g/b$.)
[Interactive: Initial Condition Explorer. The gravity-plus-drag velocity solution is plotted. A draggable dot on the $v$-axis lets the student choose $v(0)$. The solution curve updates in real time. A dashed horizontal line marks $v_{\text{term}} = g/b$. Every curve converges to it. Guided prompt: "Try starting above terminal velocity. Try starting below. Try starting exactly at terminal velocity. What do you notice about the acceleration in each case?"]
A Fourth Model: Position-Dependent Acceleration
The three models above all involve acceleration that depends on velocity (or is constant). What happens when acceleration depends on position?
Consider a mass on a spring, displaced from equilibrium:
$$\frac{dv}{dt} = -kx$$
where $k > 0$ is a constant related to the spring stiffness. This says: the farther the object is from the equilibrium position ($x = 0$), the stronger the acceleration pulling it back.
We cannot solve this equation by the same separation-of-variables technique, because the right-hand side involves $x$, not $v$. (The equation involves two unknown functions: $v(t)$ and $x(t)$, linked by $v = dx/dt$. This is a second-order problem, and we will return to it properly in later chapters.) But we can say something important about the behavior without solving.
If the object starts at $x > 0$ with $v = 0$: - The acceleration is $-kx < 0$: it pushes the object toward the origin. - The object accelerates toward $x = 0$, picking up speed. - When it reaches $x = 0$, the acceleration is zero, but the object has velocity --- it overshoots. - Now $x < 0$, so $a = -kx > 0$: the acceleration reverses, pushing back toward the origin. - The object slows, stops, and the cycle repeats.
This is oscillation --- the object swings back and forth around the equilibrium point, forever (in the absence of friction). The acceleration law $a = -kx$ guarantees this behavior. It is a restoring force: always directed toward $x = 0$, always proportional to the displacement.
Why Different Models Produce Different Behaviors
Here is the structural insight this section has been building toward.
Constant acceleration ($a = \text{const}$): The right-hand side of the DE does not depend on the state of the system ($v$ or $x$). The system has no feedback --- it accelerates at the same rate regardless of what is happening. This produces uniform change: straight-line velocity, parabolic position.
Velocity-dependent acceleration ($a = f(v)$): The system has feedback through velocity. In the drag model, the faster you go, the more drag slows you down. This self-correcting feedback produces exponential behavior --- either decay (pure drag) or approach to equilibrium (gravity plus drag). The feedback prevents runaway growth.
Position-dependent acceleration ($a = f(x)$): The system has feedback through position. In the spring model, the farther you are from equilibrium, the harder you are pushed back. This creates a different kind of feedback loop --- one that overshoots and reverses, producing oscillation.
| Acceleration depends on... | Feedback type | Resulting behavior |
|---|---|---|
| Nothing (constant) | None | Unbounded growth (linear $v$, parabolic $x$) |
| Velocity | Self-limiting | Exponential approach to equilibrium |
| Position (restoring) | Overshooting | Oscillation |
This is the concept reveal: the form of the acceleration law determines the qualitative character of the motion. You do not need to solve the equation to know this. The structure of the model --- what the acceleration depends on and how --- predicts whether the motion will be runaway, self-limiting, or oscillatory.
Connection to earlier work: In Section 2.2, you solved the constant-acceleration model and found parabolic trajectories. In Section 2.3, you saw time-dependent acceleration, where the right-hand side of the DE was a given function of $t$ and the solution was a direct integral. Now you have seen velocity- and position-dependent models, where the unknown function appears on the right-hand side of its own equation. The behavior is richer precisely because the system "talks to itself" --- its current state influences its own rate of change.
Practice
Layer 1: Concrete
Problem 1. Solve the initial-value problem
$$\frac{dv}{dt} = -bv, \qquad v(0) = v_0$$
by separation of variables. Express your answer in terms of $v_0$, $b$, and $t$.
Check your answer
Separate variables: $$\frac{dv}{v} = -b\,dt$$ Integrate from $t = 0$ (where $v = v_0$) to $t$: $$\int_{v_0}^{v}\frac{dv'}{v'} = \int_0^t (-b)\,dt'$$ $$\ln v - \ln v_0 = -bt$$ $$\ln\left(\frac{v}{v_0}\right) = -bt$$ $$\frac{v}{v_0} = e^{-bt}$$ $$v(t) = v_0\,e^{-bt}$$ The velocity decays exponentially from its initial value $v_0$ toward zero. The time constant is $\tau = 1/b$: after one time constant, $v$ has dropped to about $37\%$ of its initial value.Problem 2. Solve the initial-value problem
$$\frac{dv}{dt} = g - bv, \qquad v(0) = 0$$
by separation of variables. Find the terminal velocity and the time constant.
Check your answer
Separate variables: $$\frac{dv}{g - bv} = dt$$ Integrate from $t = 0$ (where $v = 0$) to $t$: $$\int_0^{v}\frac{dv'}{g - bv'} = \int_0^t dt'$$ For the left integral, substitute $u = g - bv'$, $du = -b\,dv'$: $$-\frac{1}{b}\int_g^{g-bv}\frac{du}{u} = t$$ $$-\frac{1}{b}\left[\ln u\right]_g^{g-bv} = t$$ $$-\frac{1}{b}\ln\left(\frac{g-bv}{g}\right) = t$$ $$\ln\left(\frac{g-bv}{g}\right) = -bt$$ $$\frac{g - bv}{g} = e^{-bt}$$ $$g - bv = g\,e^{-bt}$$ $$bv = g(1 - e^{-bt})$$ $$v(t) = \frac{g}{b}(1 - e^{-bt})$$ **Terminal velocity:** As $t \to \infty$, $e^{-bt} \to 0$, so $v \to g/b$. Therefore $v_{\text{term}} = g/b$. **Time constant:** $\tau = 1/b$. After one time constant, $v(\tau) = \frac{g}{b}(1 - e^{-1}) = \frac{g}{b}(1 - 0.368) \approx 0.63\, v_{\text{term}}$. You can verify by substituting back into the DE: $dv/dt = g\,e^{-bt}$ and $g - bv = g - g(1 - e^{-bt}) = g\,e^{-bt}$. Both sides match.Layer 2: Pattern
Problem 3. Below are four velocity-time graphs. Match each graph to the acceleration model that produced it.
Models: - (i) $dv/dt = g$ (constant acceleration, downward positive) - (ii) $dv/dt = -bv$, with $v_0 > 0$ (pure drag) - (iii) $dv/dt = g - bv$, with $v_0 = 0$ (gravity plus drag) - (iv) $dv/dt = -kx$ (restoring, position-dependent)
Graphs: - Graph A: Velocity increases linearly from zero, without bound. - Graph B: Velocity starts at $v_0 > 0$ and decays exponentially toward zero. - Graph C: Velocity oscillates sinusoidally around zero. - Graph D: Velocity starts at zero, increases rapidly at first, then levels off at a constant value.
Check your answer
- **Graph A** matches **(i)**, constant acceleration. Constant $a$ produces a straight-line $v(t)$. - **Graph B** matches **(ii)**, pure drag. The velocity decays exponentially: $v = v_0\,e^{-bt}$. - **Graph C** matches **(iv)**, position-dependent restoring acceleration. The $a = -kx$ model produces oscillation, so the velocity varies sinusoidally. - **Graph D** matches **(iii)**, gravity plus drag. Velocity rises from zero and asymptotically approaches terminal velocity: $v = (g/b)(1 - e^{-bt})$. The key diagnostic: Graph A is unbounded (no feedback). Graph B decays (velocity-dependent feedback, no driving force). Graph D saturates (velocity-dependent feedback with a constant driving force). Graph C oscillates (position-dependent restoring feedback).Layer 3: Structure
Problem 4. Why does $a = -kx$ produce oscillation while $a = -bv$ produces exponential decay? Both are "restoring" in some sense --- both produce accelerations that oppose the current state. What structural difference is responsible for the different behaviors?
Check your answer
The critical difference is *what the acceleration depends on*. In $a = -bv$: the acceleration opposes the *velocity*. When the object is moving, drag slows it down. As it slows, the drag decreases. The object asymptotically approaches rest. There is no mechanism to overshoot --- once the velocity reaches zero, the acceleration is also zero, and nothing pushes the object further. In $a = -kx$: the acceleration opposes the *displacement*. When the object is displaced from equilibrium, the acceleration pushes it back. But as the object approaches $x = 0$, it has been accelerating the whole time and has built up velocity. At $x = 0$, the acceleration is zero, but the velocity is at its maximum. The object overshoots past equilibrium. Now $x$ has the opposite sign, so the acceleration reverses, slowing the object and pushing it back. The cycle repeats. The structural difference: in the drag model, the quantity being "corrected" ($v$) is the same as the quantity whose derivative is being set ($dv/dt$ corrects $v$). This is direct, first-order feedback --- it produces exponential behavior. In the spring model, the quantity being "corrected" ($x$) is *not* the velocity --- it is the *integral* of velocity. The acceleration corrects $x$, but $x$ changes through $v$, and $v$ changes through $a$. There is a delay in the feedback loop: $a$ pushes against $x$, but by the time $x$ returns to zero, $v$ has built up. This two-step feedback loop (acceleration $\to$ velocity $\to$ position $\to$ acceleration) is what produces oscillation. In short: first-order feedback (correcting the derivative) gives exponential decay. Second-order feedback (correcting through an intermediate quantity) gives oscillation.Layer 4: Debug
Problem 5. A student is modeling a skydiver's fall. They use the constant-acceleration model $a = -g$ (taking upward as positive) and compute velocity and position as functions of time. At what point does their model's prediction become unreliable, and why? What observable quantity would signal that the model is breaking down?
Check your answer
The constant-acceleration model predicts that velocity grows linearly without bound: $v(t) = -gt$. In reality, as the skydiver speeds up, air resistance becomes significant. The drag force grows with speed, reducing the net downward acceleration. Eventually, the skydiver approaches terminal velocity --- at which point the net acceleration is nearly zero, and the constant-acceleration model is wildly wrong. The model becomes unreliable when the skydiver's speed is a significant fraction of the terminal velocity. At that point, drag is no longer negligible, and the assumption $a = \text{const}$ breaks down. **Observable signal:** The model predicts that speed increases linearly with time. If you measured the skydiver's actual speed, you would see it increase rapidly at first (matching the model) but then curve and flatten as drag becomes important. The moment the measured speed begins to deviate from the straight line $v = gt$, the constant-acceleration model is failing. **Quantitative estimate:** For a skydiver in a spread-eagle position, terminal velocity is roughly 55 m/s. The constant-acceleration model gives $v = gt = 9.8t$. At $t = 5$ s, the model predicts $v \approx 49$ m/s --- already close to terminal velocity. In reality, the skydiver would be noticeably slower than $49$ m/s because drag has been building throughout the fall. After about 10--12 seconds, the real skydiver is near terminal velocity at $\sim 55$ m/s, while the model predicts $\sim 100$ m/s --- nearly double the actual speed. The model is unreliable well before this point; a reasonable estimate is that it becomes significantly inaccurate after about 3--5 seconds of fall. The fix, of course, is the gravity-plus-drag model from this section. The constant-acceleration model is a special case --- valid only when the speed is much smaller than $v_{\text{term}}$.Multiple Representations: The Gravity-Plus-Drag Model
Let us look at the gravity-plus-drag solution from four perspectives at once, to build a full picture.
Algebraic: $v(t) = \dfrac{g}{b}(1 - e^{-bt})$
Graphical: An increasing curve that starts at zero with slope $g$, bends over, and asymptotically approaches the horizontal line $v = g/b$.
Numerical: For $g = 10$ m/s$^2$ and $b = 2$ s$^{-1}$ (so $v_{\text{term}} = 5$ m/s and $\tau = 0.5$ s):
| $t$ (s) | $e^{-bt}$ | $v$ (m/s) | $a$ (m/s$^2$) | $v / v_{\text{term}}$ |
|---|---|---|---|---|
| 0 | 1.000 | 0.00 | 10.0 | 0% |
| 0.5 | 0.368 | 3.16 | 3.68 | 63% |
| 1.0 | 0.135 | 4.32 | 1.35 | 86% |
| 1.5 | 0.050 | 4.75 | 0.50 | 95% |
| 2.0 | 0.018 | 4.91 | 0.18 | 98% |
| 3.0 | 0.002 | 4.99 | 0.02 | 100% |
Physical: The object accelerates rapidly at first (when the speed is low, drag is small, and gravity dominates). As speed increases, drag increases, and the net acceleration shrinks. Eventually drag equals gravity, acceleration drops to zero, and the object falls at a constant speed.
These four descriptions say the same thing in different languages. Fluency means being able to start from any one and produce the others.
Reflection
Think back over this section.
How does the mathematical form of the acceleration law let you predict the motion's character before solving?
Consider: you saw three models with the same physical setup (an object subject to downward force) but different resistance laws. Without solving a single equation, the structure of the DE told you whether the motion would be unbounded, self-limiting, or oscillatory. Where does that predictive power come from --- the constants, the variables, or the way they are combined?
You might also revisit the prediction from the beginning of this section. If it surprised you, can you now explain why Object A's speed exceeds Object B's after a long time? What structural feature of each model guarantees this?
Looking Ahead
You have now seen how the form of the acceleration law shapes the character of the motion. Constant acceleration gives parabolas. Velocity-dependent acceleration gives exponentials. Position-dependent acceleration gives oscillations. These are not just solutions to practice problems --- they are the fundamental behavior families of classical mechanics.
But real physical systems are rarely this clean. A pendulum's acceleration depends on $\sin\theta$, not $\theta$. Air drag may depend on $v^2$, not $v$. The exact equations are often too complicated to solve in closed form.
In the next section, you will learn what physicists do when the exact equation is intractable: approximate it. Linearization --- replacing a complicated function with its tangent-line approximation near a point --- transforms hard problems into solvable ones. The models you studied in this section (linear drag, linear restoring force) are themselves often the result of linearization. Understanding when and why that approximation works is the subject of Section 4.4.