4.1 From Acceleration Laws to Differential Equations

The Loop

Imagine you drop a coffee filter from the top of a stairwell. It flutters downward, falling fast at first, then settling into a gentle drift. Gravity pulls it down, but air resistance pushes back --- and the faster the filter falls, the harder the air pushes.

Here is the trouble. To predict the motion, you need to know the acceleration. But the acceleration depends on the velocity. And the velocity depends on the acceleration, because the acceleration is what changes the velocity. You are caught in a loop:

To find $a$, you need $v$.
To find $v$, you need $a$.
To find $a$, you need $v$.
...

In Chapters 2 and 3, this loop never appeared. Acceleration was handed to you as a function of time --- $a(t) = -g$, or $a(t) = 2t$, or some other expression that depended only on the clock. You integrated once to get $v(t)$, integrated again to get $x(t)$, and the problem was done. The clock told you the acceleration at each moment, and you never had to worry about what the object was doing.

But nature does not usually work that way. The drag on the coffee filter depends on its speed. The gravitational pull on a satellite depends on its distance from Earth. The restoring force of a spring depends on how far it has been stretched. In each case, the acceleration depends on the very quantity it is trying to determine.

How do you break out of the loop?

Before you read on: Suppose an object moves along a line with an acceleration law $a = -v$ --- the acceleration is always equal to the negative of the current velocity. The object starts with some positive velocity $v_0 > 0$.

What happens to the object?

(a) It decelerates to a stop in finite time and stays there.

(b) It decelerates, reverses direction, and oscillates back and forth.

(c) It slows down forever, approaching zero velocity but never quite reaching it.

Commit to your answer. We will return to it.

[Interactive: Predict-Then-Reveal. The student selects (a), (b), or (c). Their choice is locked in and displayed alongside the actual answer later in the section.]

Why Chapter 2's Method Stalls

Let's be precise about what goes wrong. In Section 2.3, you used the general kinematic relationship

$$v(t) = v_0 + \int_0^t a(t')\,dt'$$

to find velocity from acceleration. This worked beautifully when $a$ was given as a function of time. For example, if $a(t) = 6 - 2t$, you could compute:

$$v(t) = v_0 + \int_0^t (6 - 2t')\,dt' = v_0 + 6t - t^2$$

No difficulty. The integrand was a known function of $t'$, and the integral was a routine calculus exercise.

Now try the same approach with the acceleration law $a = -v$. You want to compute:

$$v(t) = v_0 + \int_0^t a(t')\,dt' = v_0 + \int_0^t \bigl[-v(t')\bigr]\,dt'$$

Look at what happened. The unknown function $v(t')$ appeared inside the integral. You cannot evaluate this integral, because you do not know $v(t')$ --- that is the thing you are trying to find. The method has eaten its own tail.

This is not a failure of technique. It is a signal that you need a different framework.

Pause and connect: In Chapter 2, the chain was: given $a(t)$ $\to$ integrate $\to$ get $v(t)$ $\to$ integrate $\to$ get $x(t)$. Each step used a known function as input. The chain worked because $a$ depended only on $t$. Now $a$ depends on $v$ or $x$ --- the unknown is inside the equation. That changes the nature of the problem entirely.

The Idea: An Equation That Refers to the Unknown

Let's write the problematic acceleration law differently. Instead of treating it as a computational instruction ("plug in and integrate"), let's write it as a relationship that must hold at every instant.

The acceleration law says $a = -v$. Since $a = dv/dt$, we can write:

$$\frac{dv}{dt} = -v$$

Read this aloud: "The rate of change of velocity equals the negative of the velocity itself." At every instant, this equation must be true. If $v$ is large, the rate of change is large and negative --- the velocity is decreasing rapidly. If $v$ is small, the rate of change is small --- the velocity is barely changing. If $v$ is zero, the rate of change is zero --- the velocity has stopped changing.

This is a differential equation: an equation that involves an unknown function and its derivative. The unknown is $v(t)$ --- a function of time that we do not yet know. The equation constrains how that function can behave: its rate of change must always equal the negative of its current value.

This is worth saying plainly: a differential equation is not a formula you evaluate. It is a condition that a function must satisfy. The function $v(t)$ is not given to you. You have to find it --- or, as we will see, let the equation generate it.

Exploration: Watching the Rule Build a Motion

Before we solve anything algebraically, let's watch the differential equation in action. The following interactive implements the acceleration law as a step-by-step recipe: start with an initial velocity, compute the acceleration from the current velocity, update the velocity by a small amount, and repeat.

[Interactive: Motion Builder. The student selects an acceleration rule from a menu: $a = -v$, $a = -2v$, $a = +v$, or $a = -0.5v$. They also set an initial velocity $v_0$ using a slider (range: $-10$ to $10$ m/s).

A table builds row by row. Each row shows: time $t$, current velocity $v$, current acceleration $a$ (computed from the rule), and the updated velocity after a small time step. Next to the table, a velocity-versus-time graph plots the motion as it unfolds.

Controls: a "Step" button advances one row at a time, and a "Run" button builds the entire table automatically. A time-step slider lets the student choose $\Delta t = 0.5$, $0.1$, or $0.01$ s.

Guided prompts appear as the student explores:]

Step 1: Set the rule to $a = -v$ and the initial velocity to $v_0 = 10$ m/s. Click "Step" several times and watch the table fill in.

What happens to the velocity? Does it ever reach zero? Does it ever become negative?

Step 2: Now change the rule to $a = +v$ (positive sign). Keep $v_0 = 10$ m/s. Run the simulation.

What happens now? How is this motion qualitatively different from the previous one?

Step 3: Go back to $a = -v$ but try $v_0 = -5$ m/s (a negative initial velocity). What does the equation predict?

Step 4: Compare $a = -v$ and $a = -2v$ with the same initial velocity. What effect does doubling the constant have on how quickly the velocity changes?

If you worked through those explorations, you discovered several things:

With $a = -v$ and positive $v_0$: the velocity decreases, approaching zero but never reaching it. The deceleration itself slows down as $v$ shrinks. The motion dies away gradually, like a volume knob being turned down smoothly.
With $a = +v$ and positive $v_0$: the velocity increases without bound. The acceleration grows as $v$ grows, producing runaway growth. This is qualitatively different from anything in Chapter 2.
With negative $v_0$ and $a = -v$: the velocity is negative, so the acceleration is positive. The velocity increases (becomes less negative), approaching zero from below. The object slows down regardless of the sign of $v_0$.
Doubling the constant from $-v$ to $-2v$ makes the velocity approach zero faster, but the qualitative story is the same.

Return to the prediction. The answer is (c): the object slows down forever but never quite stops. The acceleration is always pushing the velocity toward zero, but as the velocity shrinks, the acceleration shrinks too. The velocity gets arbitrarily close to zero without ever arriving. If you guessed (a), the idea of "stops in finite time" feels natural, but it is wrong here --- the braking force weakens as the object slows. If you guessed (b), nothing in the equation ever reverses the velocity. The acceleration law $a = -v$ always opposes the current velocity, pushing it toward zero, not past it.

The Concept: A Motion-Generating Rule

Here is the key idea of this section, and it is more than just a mathematical formality.

A differential equation like $\frac{dv}{dt} = -v$ is a recipe that generates an entire motion from a single starting condition. It does not tell you what the velocity is at any particular time. Instead, it tells you what happens next, given where you are now:

If $v = 10$, then $dv/dt = -10$: the velocity is decreasing at $10$ m/s per second.
If $v = 3$, then $dv/dt = -3$: the velocity is decreasing at $3$ m/s per second.
If $v = 0.01$, then $dv/dt = -0.01$: the velocity is barely changing.

At each instant, the equation looks at the current state and prescribes the next infinitesimal change. Link all those infinitesimal changes together, and you get a complete function $v(t)$ stretching from the initial moment into the indefinite future.

This is fundamentally different from the equations you worked with in Chapter 2. There, $a(t) = 6 - 2t$ told you the acceleration at every moment, like a script that the object follows. The differential equation $dv/dt = -v$ is more like a rule of conduct: "at each moment, adjust your velocity in proportion to its current value." The motion is not scripted in advance. It emerges from the interaction between the current state and the rule.

An analogy. Think of the difference between a recipe that says "add 200 grams of flour" and one that says "keep adding flour until the dough stops sticking to your hands." The first is a fixed instruction. The second is a rule that depends on the current state of the dough --- and the amount of flour you end up adding depends on how the dough responds. Differential equations are like the second kind of recipe.

Translating Physics into Differential Equations

Now that you understand what a differential equation is, the practical skill is translating a physical situation into one. The procedure is the same every time:

Identify the acceleration law. This comes from the physics --- the forces acting on the object determine its acceleration.
Replace $a$ with $dv/dt$ (or $d^2x/dt^2$). This converts the acceleration statement into a differential equation.
Identify the unknown function. What are you solving for? Usually $v(t)$ or $x(t)$.
Identify the initial conditions. What do you know about the state of the system at the starting moment?

Let's work through several examples.

Example 1: Linear drag.

A raindrop falls through air. The drag force is proportional to its speed, so the net acceleration is:

$$a = g - bv$$

where $g$ is gravitational acceleration (downward is positive) and $b$ is a positive constant that depends on the raindrop's size and the air's density. The term $-bv$ represents drag opposing the motion.

Replacing $a$ with $dv/dt$:

$$\frac{dv}{dt} = g - bv$$

Unknown function: $v(t)$
Initial condition: $v(0) = 0$ (released from rest)
What the equation says: "The rate of change of velocity equals the gravitational pull minus a braking term proportional to the current speed."

Notice something interesting. At $t = 0$, when $v = 0$, the equation gives $dv/dt = g$ --- the raindrop accelerates at full gravity, as if there were no air resistance. As $v$ increases, the drag term $bv$ grows, and the acceleration decreases. Eventually, when $v$ reaches $g/b$, the acceleration is zero. The raindrop has reached terminal velocity.

Pause and think: The equation $dv/dt = g - bv$ contains the complete story of the raindrop's motion --- from the initial free-fall acceleration to the eventual constant-speed drift. All of that is encoded in a single line.

Example 2: Restoring force (a spring).

A mass on a spring experiences a force proportional to its displacement from equilibrium:

$$a = -kx$$

where $k$ is a positive constant and $x$ is the displacement. Replacing $a$ with $d^2x/dt^2$:

$$\frac{d^2x}{dt^2} = -kx$$

Unknown function: $x(t)$
Initial conditions: $x(0) = x_0$ (initial displacement) and $v(0) = v_0$ (initial velocity). Two conditions are needed because this is a second-order differential equation --- it involves the second derivative.
What the equation says: "The second derivative of position is proportional to the negative of the position itself." Displaced to the right ($x > 0$), the acceleration points left. Displaced to the left ($x < 0$), the acceleration points right. The equation always pushes back toward the center.

You might already suspect that this produces oscillation. We will confirm that in Section 4.3.

Example 3: Gravitational attraction.

A ball is launched straight up from Earth's surface. If we account for the fact that gravity weakens with distance, the acceleration is:

$$a = -\frac{GM}{r^2}$$

where $r$ is the distance from Earth's center, $G$ is the gravitational constant, and $M$ is Earth's mass. Since $r$ changes as the ball moves, the acceleration depends on position. Writing $a = d^2r/dt^2$:

$$\frac{d^2r}{dt^2} = -\frac{GM}{r^2}$$

Unknown function: $r(t)$
Initial conditions: $r(0) = R_E$ (Earth's radius) and $v(0) = v_0$ (launch speed).
What the equation says: "The acceleration is inversely proportional to the square of the distance." Close to Earth, the pull is strong. Far away, the pull is weak. The motion depends on whether the launch speed is large enough to escape this weakening pull.

What changed? What stayed the same? In all three examples, the procedure was identical: write down the acceleration law, replace $a$ with a derivative, identify the unknown and the initial conditions. But the equations look very different, and they will produce very different motions --- exponential approach in Example 1, oscillation in Example 2, and (depending on initial speed) either escape or return in Example 3. The form of the acceleration law determines the character of the motion.

Three Types of Acceleration Laws

The examples above illustrate a useful classification. Acceleration laws in mechanics generally fall into three categories, depending on what the acceleration depends on:

Type	Form	Example	Chapter 2 method?
Time-dependent	$a = f(t)$	$a = 6 - 2t$	Yes --- integrate directly
Velocity-dependent	$a = f(v)$	$a = g - bv$	No --- $v$ is the unknown
Position-dependent	$a = f(x)$	$a = -kx$	No --- $x$ is the unknown

Time-dependent acceleration is the only type you can handle with straight integration, because $t$ is the independent variable --- the clock ticks forward regardless of what the object does. When $a$ depends on $v$ or $x$, the unknown has become entangled with the equation, and you need the differential equation framework.

Some situations involve combinations. The drag-plus-gravity example ($a = g - bv$) is velocity-dependent. A damped spring ($a = -kx - bv$) depends on both position and velocity. The classification is not rigid, but it helps you recognize when Chapter 2's methods apply and when they do not.

Before you read on: For each of the following, identify whether the acceleration law is time-dependent, velocity-dependent, position-dependent, or some combination.

A rocket with a steadily decreasing thrust: $a = a_0 - ct$

Quadratic air drag on a falling ball: $a = g - cv^2$

A pendulum (small angle): $a = -\frac{g}{L}\theta$

An object decelerating at a rate that depends on both speed and position: $a = -bv - kx$

Check your answer

1. **Time-dependent.** The acceleration depends only on $t$. You can integrate this directly using Chapter 2's methods. 2. **Velocity-dependent.** The acceleration depends on $v$. This requires a differential equation. 3. **Position-dependent** (where $\theta$ plays the role of position). The acceleration depends on the angular displacement. This requires a differential equation. 4. **Combination** --- both velocity-dependent and position-dependent. The acceleration depends on $v$ and $x$ simultaneously. This requires a differential equation, and the equation is more complex than either type alone.

Why Position-Dependent Acceleration Is Fundamentally Different

There is a structural distinction that is worth dwelling on. Compare these two differential equations:

$$\frac{dv}{dt} = f(t) \qquad \text{vs.} \qquad \frac{dv}{dt} = -kx$$

The first is time-dependent acceleration. You can integrate it directly. One integration gives $v(t)$, another gives $x(t)$. Done.

The second says $dv/dt = -kx$. But $v = dx/dt$, so this is really:

$$\frac{d^2x}{dt^2} = -kx$$

The second derivative of the unknown function is proportional to the negative of the function itself. This is a second-order differential equation. It requires two initial conditions ($x_0$ and $v_0$), and its solution involves oscillation --- behavior that is qualitatively absent from anything in Chapter 2.

Why oscillation? Think about it physically. If $x > 0$, the acceleration is negative --- it pushes the object back toward the origin. The object decelerates, stops, and then the negative position creates a positive acceleration that pushes it back the other way. The object overshoots the origin, and the cycle repeats.

No time-dependent acceleration law can do this. If $a = f(t)$, the acceleration follows a predetermined script regardless of where the object is. It cannot "know" that the object has passed the origin and needs to reverse. Position-dependent acceleration creates feedback --- the acceleration responds to the object's state, which creates the possibility of sustained, self-correcting motion.

This is one of the deepest structural differences in mechanics, and it is entirely visible from the form of the differential equation.

A Historical Interlude

Newton's Principia Mathematica (1687) is often described as a book about gravity and planetary motion. But at its core, it is a book about exactly the idea we have just developed: if the force law is known, the motion follows from solving the differential equation.

Newton's Second Law says $F = ma$, which means $a = F/m$. If the force depends on position (like gravity: $F \propto 1/r^2$) or on velocity (like drag: $F \propto v$), then the resulting equation $d^2x/dt^2 = F(x, v)/m$ is a differential equation. The entire future of the object's motion is encoded in that equation, once you specify where the object starts and how fast it is moving.

This was a staggering insight. Before Newton, the motion of planets was described by Kepler's Laws --- empirical rules extracted from observational data. Newton showed that all three of Kepler's Laws follow from a single differential equation: $d^2r/dt^2 = -GM/r^2$. The universe was not following three separate rules. It was following one rule, and the three laws were consequences.

In a very real sense, the universe is a collection of initial-value problems. Specify the positions and velocities of all the particles, write down the force laws, and the differential equations determine everything that happens next. This is what physicists mean by determinism in classical mechanics.

Worked Example: Setting Up a Differential Equation

A boat cuts its engine and coasts through water. The water resistance is proportional to the boat's speed: $F_{\text{drag}} = -\beta v$, where $\beta$ is a positive constant. The boat has mass $m$.

Step 1: Identify the acceleration law.

Newton's Second Law gives $ma = F_{\text{drag}}$, so:

$$a = -\frac{\beta}{m} v$$

Let $b = \beta/m$ to simplify notation. Then $a = -bv$.

Step 2: Write the differential equation.

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

Step 3: Identify the unknown and initial conditions.

Unknown: $v(t)$.
Initial condition: $v(0) = v_0$ (the boat's speed when the engine is cut).

Step 4: Interpret the equation before solving it.

At the moment the engine is cut, $v = v_0$, so $dv/dt = -bv_0$. The boat is decelerating.
As $v$ decreases, $dv/dt$ decreases in magnitude. The deceleration slows down.
The equation predicts that $v$ approaches zero asymptotically --- the boat drifts to a stop gradually.
The larger $b$ is (stronger drag relative to mass), the faster the velocity decays.

We will learn how to solve this equation explicitly in Section 4.2. For now, the point is the translation: a physical statement ("drag proportional to speed") became a mathematical object ($dv/dt = -bv$) that encodes the entire motion.

Faded Example: Your Turn to Set It Up

A ball bearing sinks through honey. Gravity pulls it down with acceleration $g$. The honey exerts a drag force proportional to speed, with drag coefficient $c$. The ball bearing has mass $m$ and starts from rest.

Step 1: Write down the net acceleration of the ball bearing. (Take downward as positive.)

Check your answer

The net acceleration is gravity minus drag: $$a = g - \frac{c}{m}v$$ or, writing $b = c/m$: $$a = g - bv$$

Step 2: Write the differential equation.

Check your answer

$$\frac{dv}{dt} = g - bv$$

Step 3: What is the unknown function? What is the initial condition?

Check your answer

- Unknown: $v(t)$. - Initial condition: $v(0) = 0$ (released from rest).

Step 4: What does the equation predict about the long-time behavior? (Hint: what value of $v$ makes $dv/dt = 0$?)

Check your answer

Setting $dv/dt = 0$ gives $0 = g - bv$, so $v = g/b$. At this speed, the gravitational acceleration and the drag deceleration exactly balance. The ball bearing reaches a **terminal velocity** of $v_{\text{term}} = g/b$ and then sinks at constant speed. Before reaching terminal velocity, $dv/dt > 0$ (gravity wins over drag), so the ball is accelerating downward. As $v$ approaches $g/b$, the acceleration shrinks toward zero. The ball bearing asymptotically approaches terminal velocity from below --- it never overshoots.

The Unknown Function and Its Initial Conditions

Let's collect a pattern that has appeared in every example. A differential equation involves:

An unknown function --- the thing you are solving for. In mechanics, this is usually $v(t)$ or $x(t)$.
A derivative of the unknown function --- $dv/dt$ or $d^2x/dt^2$ --- which relates the function to its own rate of change.
Initial conditions --- the values of the unknown function (and its derivatives, if the equation is second-order) at the starting time.

The order of the differential equation tells you how many initial conditions you need:

Order	Example	Initial conditions needed
First-order	$\frac{dv}{dt} = -bv$	One: $v(0)$
Second-order	$\frac{d^2x}{dt^2} = -kx$	Two: $x(0)$ and $v(0) = \frac{dx}{dt}(0)$

The reason is intuitive. A first-order equation tells you the rate of change of $v$, so if you know the starting value, you can build forward step by step. A second-order equation tells you the rate of change of the rate of change of $x$. You need to know both the starting position and the starting velocity to get started.

Together, the differential equation and the initial conditions form an initial-value problem (IVP) --- the complete mathematical specification of a particular motion. Different initial conditions produce different motions, but the differential equation is the same for every motion governed by the same physical law.

Before you read on: A mass on a spring obeys $\frac{d^2x}{dt^2} = -9x$. Two experiments are done:

Experiment A: $x(0) = 2$ m, $v(0) = 0$.

Experiment B: $x(0) = 0$, $v(0) = 6$ m/s.

Will the motions look the same? Why or why not?

[Interactive: Predict-Then-Reveal. The student writes a brief response, then clicks to see two animated oscillations side by side --- same frequency, but Experiment A starts at maximum displacement while Experiment B starts at the center with maximum speed. The motions have the same period but are shifted in phase.]

Multiple Representations: The Same DE Four Ways

A good way to deepen your understanding is to see the same differential equation from multiple angles. Let's take $\frac{dv}{dt} = -v$ with $v(0) = 10$ m/s and look at it four ways.

Algebraically: $\frac{dv}{dt} = -v$. The rate of change of $v$ is always equal to $-v$.

Numerically (step by step):

$t$ (s)	$v$ (m/s)	$a = -v$ (m/s$^2$)	$\Delta v \approx a \cdot \Delta t$
0.0	10.00	$-10.00$	$-5.00$
0.5	5.00	$-5.00$	$-2.50$
1.0	2.50	$-2.50$	$-1.25$
1.5	1.25	$-1.25$	$-0.63$
2.0	0.62	$-0.62$	$-0.31$

(Using $\Delta t = 0.5$ s. Each row: compute $a$ from current $v$, then update $v \gets v + a \Delta t$. The values are approximate because the step size is coarse, but the pattern is clear.)

Graphically: The $v$-vs-$t$ curve starts at $10$ and decays toward zero, bending downward but flattening as it approaches the axis. The curve never crosses zero. It is the shape of exponential decay.

Physically: An object losing speed in proportion to its current speed. The faster it goes, the harder it brakes. As it slows, the braking eases. The object coasts to a halt that it never quite reaches.

These four views --- algebraic, numerical, graphical, physical --- are not four different topics. They are four windows onto one idea. Fluency with differential equations means being able to move between these views freely.

Practice

Layer 1: Concrete

Problem 1. A particle moves along a line. Its acceleration is proportional to the square of its velocity: $a = -cv^2$, where $c > 0$. Write the differential equation for $v(t)$.

Check your answer

Replacing $a$ with $dv/dt$: $$\frac{dv}{dt} = -cv^2$$ The unknown function is $v(t)$. You would need one initial condition --- the velocity at $t = 0$ --- to specify a particular motion. This is the differential equation for quadratic drag (relevant at high speeds, unlike the linear drag we considered earlier).

Problem 2. A spring-mass system has a restoring acceleration of $a = -\omega^2 x$, where $\omega$ is a positive constant and $x$ is the displacement from equilibrium. Write the differential equation for $x(t)$ and state the required initial conditions.

Check your answer

Since $a = d^2x/dt^2$: $$\frac{d^2x}{dt^2} = -\omega^2 x$$ This is a second-order differential equation. Two initial conditions are needed: $x(0)$ (the initial displacement) and $\frac{dx}{dt}(0) = v(0)$ (the initial velocity). Together with the equation, these form a complete initial-value problem.

Problem 3. Translate each verbal statement into a differential equation.

(a) "The velocity of the object decreases at a rate of 3 m/s every second."

(b) "The acceleration is always twice the current velocity."

Check your answer

(a) "Decreases at a rate of 3 m/s every second" means $dv/dt = -3$. This is constant acceleration --- actually just a Chapter 2 problem in disguise. The acceleration does not depend on $v$ or $x$, so you can integrate directly. (b) $\frac{dv}{dt} = 2v$. This is a first-order DE with velocity-dependent acceleration. Note the positive sign: the acceleration reinforces the velocity, leading to exponential growth rather than decay. (c) "Proportional to the distance from the origin" means $a \propto x$. "Directed toward the origin" means the acceleration is in the opposite direction from the displacement: when $x > 0$, $a < 0$, and vice versa. So $a = -kx$ for some positive constant $k$, giving: $$\frac{d^2x}{dt^2} = -kx$$ This is the spring equation again --- the language is different, but the mathematics is the same.

Layer 2: Pattern

Problem 4. Classify each of the following acceleration laws. For each, state whether it is time-dependent, velocity-dependent, position-dependent, or a combination, and whether you could solve it using the methods of Chapter 2.

	Acceleration law	Type	Chapter 2?
(a)	$a = 5\sin(2t)$
(b)	$a = -3v + 2$
(c)	$a = -\frac{g R^2}{(R+x)^2}$
(d)	$a = -kx - bv$
(e)	$a = t^2 - 4t + 3$

Check your answer

| | Acceleration law | Type | Chapter 2? | |:---|:---|:---|:---| | (a) | $a = 5\sin(2t)$ | Time-dependent | Yes --- integrate $\int 5\sin(2t')\,dt'$ | | (b) | $a = -3v + 2$ | Velocity-dependent | No --- $v$ is the unknown | | (c) | $a = -\frac{g R^2}{(R+x)^2}$ | Position-dependent | No --- $x$ is the unknown | | (d) | $a = -kx - bv$ | Both velocity- and position-dependent | No --- both unknowns are involved | | (e) | $a = t^2 - 4t + 3$ | Time-dependent | Yes --- integrate the polynomial | The dividing line is clear: if $a$ depends only on $t$, you can integrate. If it involves $v$ or $x$, you cannot --- you need a differential equation.

Layer 3: Structure

Problem 5. Why is $a = -kx$ fundamentally different from $a = f(t)$?

Your answer should address at least two of the following: (i) What kind of motion each produces. (ii) How many initial conditions each requires. (iii) Whether the acceleration "knows" about the state of the object.

Check your answer

The differences run deep: **(i) Type of motion.** When $a = f(t)$, the acceleration follows a predetermined schedule. The object's velocity and position change in response, but the acceleration does not adjust. This produces motions that are "open-ended" --- the object does not inherently return to where it started. When $a = -kx$, the acceleration depends on where the object is. If the object moves to the right, the acceleration pushes it left. If it moves to the left, the acceleration pushes it right. This creates *feedback*: the motion corrects itself, producing oscillation. The object repeatedly overshoots equilibrium and gets pushed back. No time-dependent acceleration can produce this kind of self-correcting behavior. **(ii) Order and initial conditions.** The equation $dv/dt = f(t)$ is first-order in $v$. You need one initial condition: $v(0)$. The equation $d^2x/dt^2 = -kx$ is second-order in $x$. You need two initial conditions: $x(0)$ and $v(0)$. This reflects the richer physics: the future of the oscillation depends on both where you start and how fast you are moving. **(iii) State-awareness.** When $a = f(t)$, the acceleration is "blind" to the object --- it does not respond to the object's position or velocity. When $a = -kx$, the acceleration is "aware" of the object's state. This awareness is what makes feedback and oscillation possible. It is also what makes the equation a genuine differential equation rather than a direct integration problem.

Layer 4: Debug

Problem 6. A student is asked to write the differential equation for an object experiencing drag proportional to its speed. The student writes:

$$v = -kv$$

"The velocity equals negative $k$ times the velocity," the student explains. "So $v(1 + k) = 0$, which means $v = 0$. The object stops immediately."

What is wrong with the student's equation? What should it be, and why does the distinction matter?

Check your answer

The student wrote $v = -kv$, which is an algebraic equation. It says "the velocity equals negative $k$ times itself," which (for $k \neq -1$) implies $v = 0$. This is not a differential equation --- it is a statement about a single instant, and it has only the trivial solution $v = 0$. The correct equation is: $$\frac{dv}{dt} = -kv$$ This says "the *rate of change* of velocity equals negative $k$ times the velocity." It does not force $v$ to be zero. Instead, it constrains how $v$ evolves over time. The solution is an exponential decay: $v(t) = v_0 e^{-kt}$, which starts at $v_0$ and gradually approaches zero. The distinction matters because a differential equation relates a function to its *derivative*, not to itself. Leaving out the $d/dt$ changes the equation from a motion-generating rule to a static algebraic constraint. The student's equation collapses all the dynamics into a single instant and loses the entire motion. This is perhaps the most common error when students first encounter differential equations: confusing $v$ (the function) with $dv/dt$ (its rate of change).

Problem 7. Another student sets up the equation for a falling object with linear drag as:

$$\frac{dv}{dt} = g + bv$$

where $g$ and $b$ are both positive. The student says this represents "gravity pulling down and drag opposing the motion."

Is the sign correct? If not, what is the right equation, and what physical error did the student make?

Check your answer

The sign on the drag term is wrong. If we take downward as positive, then: - Gravity acts downward: $+g$. This is correct. - Drag opposes the motion. For a falling object, the velocity is positive (downward), so drag acts upward: $-bv$. The student wrote $+bv$, which means drag is *accelerating* the object in the direction of motion --- making it fall faster. The correct equation is: $$\frac{dv}{dt} = g - bv$$ With $+bv$, the student's equation predicts runaway acceleration: the faster the object goes, the more it speeds up. This is physically nonsensical for drag. With the correct sign $-bv$, the equation predicts a terminal velocity at $v = g/b$, which is what actually happens. Sign errors in differential equations are not minor typos. They change the qualitative behavior of the solution --- from decay to growth, from stable to unstable, from oscillating to exploding. Always check: does the equation predict behavior that makes physical sense?

Reflection

Think about what you have read and explored in this section.

What is the difference between an equation that describes an instant and one that generates an entire motion?

Consider the equation $v = 5$ m/s. It tells you the velocity at a single moment. Now consider the equation $dv/dt = -v$, along with $v(0) = 5$. It tells you the velocity at every moment --- an entire function $v(t)$, stretching from $t = 0$ to infinity.

The first is a snapshot. The second is a movie, encoded as a rule. That rule --- the differential equation --- is the mathematical heart of all of mechanics. Every force law in physics ultimately produces a differential equation, and solving that equation produces a prediction of motion.

Looking Ahead

You have just crossed a conceptual threshold. In Chapters 1--3, motion was described --- you were given $a(t)$, $v(t)$, or $x(t)$ and asked to analyze it. Now you have seen that the real starting point in physics is not a description of motion, but a rule for how motion unfolds: a differential equation.

But writing down the equation is only half the problem. You still need to solve it --- to find the function $v(t)$ or $x(t)$ that satisfies both the equation and the initial conditions. In the next section, you will learn how to do this for the simplest and most important cases. You will see that a differential equation defines a whole family of motions, and the initial conditions select one particular motion from that family. Together, the equation and the conditions form an initial-value problem --- the complete mathematical model of a specific physical motion.

The machinery you are building here is not a detour. It is the central tool of classical mechanics. Chapter 5 will hand you force laws. This chapter teaches you what to do with them.