2.3 Nonuniform Acceleration and Integral Relationships

When the Formulas Stop Working

[Video: Two side-by-side clips. On the left, a sports car launches from a stop. At first the acceleration is fierce --- the driver's head snaps back. But as the speedometer climbs past 100, 150, 200 km/h, the acceleration visibly fades. The car creeps toward its top speed, barely gaining. On the right, a rocket lifts off. At ignition it barely moves --- the thrust fights the full weight of fuel. But as fuel burns away, the rocket gets lighter and lighter. The acceleration increases. The rocket climbs faster and faster, punching through the atmosphere. A split-screen graph shows $a(t)$ for each: the car's acceleration curve slopes downward; the rocket's slopes upward. Neither is a horizontal line.]

Watch those two clips carefully. In Section 2.2, we derived a beautiful set of kinematic equations --- and every one of them rested on a single assumption: constant acceleration. The acceleration was a flat horizontal line on the $a$-vs-$t$ graph, and that made the integrals simple enough to produce clean formulas.

But the car's acceleration fades. The rocket's acceleration grows. Neither one is constant. If you tried to use $v = v_0 + at$ on either of these situations, you would get a prediction that is not just inaccurate --- it is qualitatively wrong.

So now what?

Before you read on: If a rocket's acceleration is increasing with time, will its speed grow faster than, slower than, or at the same rate as a rocket with constant acceleration equal to the initial value?

Commit to your answer. We will come back to it.

Quick Recall: Concepts from Chapter 1

Before we push forward, let's make sure the foundations are solid. Answer these from memory --- no scrolling back.

Recall prompt: In Chapter 1, we discussed the difference between displacement and path length. Can you state the distinction in one sentence? And what does the derivative of a position function $x(t)$ give you, physically?

If those felt shaky, revisit Sections 1.2 and 1.4. Everything in this section builds on the derivative-integral chain you met there.

The Problem with "Plugging In"

Let's make the failure of the constant-acceleration formulas concrete with a specific example.

Suppose a particle starts from rest and has an acceleration that increases linearly with time:

$$a(t) = 2t \quad (\text{in m/s}^2)$$

At $t = 0$, the acceleration is zero. At $t = 1$ s, it is $2 \text{ m/s}^2$. At $t = 3$ s, it is $6 \text{ m/s}^2$. The acceleration itself is changing.

What happens if you try the constant-acceleration formula $v = v_0 + at$ anyway? You would need to pick a single value for $a$. Which one? The value at $t = 0$? That gives $v = 0$ for all time --- clearly wrong, because the acceleration is not zero for $t > 0$. The value at $t = 3$? That gives $v = 18 \text{ m/s}$ at $t = 3$ s --- but it assumes the acceleration was $6 \text{ m/s}^2$ the entire time, which it was not.

There is no right answer because the question is wrong. The formula $v = v_0 + at$ was derived under the assumption that $a$ is a constant. When $a$ varies, the derivation does not apply. It is not a matter of choosing the right $a$ to plug in. The formula itself does not fit the situation.

Pause and think: Why can't you just use the average acceleration over the interval and plug that into the constant-acceleration formula? What would go wrong?

Exploration: Watching Integration in Action

[Interactive: Acceleration Function Explorer. A text input field accepts a mathematical expression for $a(t)$ (pre-loaded with $a(t) = 2t$). Below, three graphs update in real time: $a(t)$, $v(t)$, and $x(t)$. The system numerically integrates $a(t)$ to produce $v(t)$, then integrates $v(t)$ to produce $x(t)$, both starting from zero initial conditions. A dashed line shows what the constant-acceleration prediction would be using $a(0)$ as the constant value. The difference between the two is shaded.]

Try the following guided explorations:

Step 1: Start with the default, $a(t) = 2t$.

Look at the velocity graph. With constant acceleration, $v(t)$ would be a straight line. What shape is it instead? Notice how the velocity curve bends upward --- it grows faster and faster, because the acceleration itself is growing.

Step 2: Now change the acceleration to $a(t) = 6$ (a constant).

The velocity should become a straight line: $v(t) = 6t$. The position should become a parabola: $x(t) = 3t^2$. This is Section 2.2 --- the constant-acceleration case. Notice how the dashed prediction line and the solid curve overlap perfectly. The formulas work because the assumption holds.

Step 3: Try $a(t) = 10 - 2t$.

This is an acceleration that starts at $10 \text{ m/s}^2$ and decreases. What happens to the velocity? It still increases at first, but the rate of increase slows down. At $t = 5$ s, the acceleration reaches zero. What happens to the velocity at that moment? Does the object stop?

Before you read on: Based on your exploration, how would you describe the relationship between $a(t)$ and $v(t)$ in words? What operation connects them?

The General Integral Relationships

Here is the key insight, and if you have been following the exploration, it should feel familiar rather than surprising.

In Section 2.2, we started with constant acceleration and integrated:

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

The integral was trivial because $a$ was constant --- it came outside the integral sign, and we were left with $a \cdot t$.

But the integral itself does not require $a$ to be constant. The general relationship is:

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

This says: the velocity at time $t$ equals the initial velocity plus the accumulated effect of all the acceleration between time 0 and time $t$. When the acceleration varies, you cannot shortcut the accumulation. You must actually perform the integral.

Similarly, position comes from integrating velocity:

$$\boxed{x(t) = x_0 + \int_0^t v(t') \, dt'}$$

These two equations are the general kinematic relationships for one-dimensional motion. They always work. The constant-acceleration formulas from Section 2.2 are the special case where the integrals happen to evaluate to simple algebraic expressions.

A note on notation: The variable $t'$ (read "t-prime") inside the integral is a dummy variable --- a placeholder for the integration variable. We use $t'$ instead of $t$ to distinguish the integration variable (which runs from $0$ to $t$) from the upper limit $t$ (which is the specific time we are asking about). You could equally well write $\int_0^t a(\tau) \, d\tau$ or $\int_0^t a(s) \, ds$. The letter does not matter; the structure does.

Connection: Generalization, Not Replacement

It is worth pausing to see how the pieces fit together.

In Section 2.2, you learned:

Quantity	Constant acceleration formula
Velocity	$v(t) = v_0 + at$
Position	$x(t) = x_0 + v_0 t + \frac{1}{2}at^2$

These are not wrong. They are special cases. Watch what happens when you set $a(t) = a$ (a constant) in the general integral:

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

That is exactly the formula from Section 2.2. And then:

$$x(t) = x_0 + \int_0^t (v_0 + at') \, dt' = x_0 + v_0 t + \frac{1}{2}at^2$$

Again, exactly the Section 2.2 result. The kinematic equations did not disappear. They became the easy version of something more general.

This is how physics grows. You do not throw away old tools --- you discover that they were a special case of a bigger framework all along.

Worked Example: $a(t) = 3t^2$

Let's work through a complete example with all steps visible.

Problem: A particle starts from rest at the origin. Its acceleration is $a(t) = 3t^2$ (in m/s$^2$). Find $v(t)$ and $x(t)$.

Step 1: Find $v(t)$ by integrating $a(t)$.

Starting from rest means $v_0 = 0$. The general relationship gives:

$$v(t) = v_0 + \int_0^t a(t') \, dt' = 0 + \int_0^t 3t'^2 \, dt'$$

Evaluate the integral:

$$v(t) = 3 \cdot \frac{t'^3}{3} \Big|_0^t = t'^3 \Big|_0^t = t^3 - 0 = t^3$$

So $v(t) = t^3$ m/s.

Step 2: Find $x(t)$ by integrating $v(t)$.

Starting at the origin means $x_0 = 0$:

$$x(t) = x_0 + \int_0^t v(t') \, dt' = 0 + \int_0^t t'^3 \, dt'$$

Evaluate:

$$x(t) = \frac{t'^4}{4} \Big|_0^t = \frac{t^4}{4}$$

So $x(t) = \frac{t^4}{4}$ m.

Step 3: Sanity check.

Let's verify the chain works in reverse. Take the derivative of $x(t) = \frac{t^4}{4}$:

$$\frac{dx}{dt} = t^3 = v(t) \quad \checkmark$$

Take the derivative of $v(t) = t^3$:

$$\frac{dv}{dt} = 3t^2 = a(t) \quad \checkmark$$

Everything is consistent. The integral-derivative chain from Chapter 1 runs in both directions.

Faded Example: $a(t) = \sin(t)$

Now try one where some steps are left for you to fill in.

Problem: A particle starts from rest at $x_0 = 0$. Its acceleration is $a(t) = \sin(t)$ (in m/s$^2$). Find $v(t)$ and $x(t)$.

Step 1: Find $v(t)$.

$$v(t) = v_0 + \int_0^t a(t') \, dt' = 0 + \int_0^t \sin(t') \, dt'$$

Your turn: Evaluate $\int_0^t \sin(t') \, dt'$. What is the antiderivative of $\sin(t')$?

Check your answer

The antiderivative of $\sin(t')$ is $-\cos(t')$. Evaluating the definite integral: $$v(t) = \left[-\cos(t')\right]_0^t = -\cos(t) - (-\cos(0)) = -\cos(t) + 1 = 1 - \cos(t)$$ So $v(t) = 1 - \cos(t)$ m/s.

Step 2: Find $x(t)$.

$$x(t) = 0 + \int_0^t v(t') \, dt' = \int_0^t \bigl(1 - \cos(t')\bigr) \, dt'$$

Your turn: Evaluate this integral. Split it into two terms if it helps.

Check your answer

$$x(t) = \int_0^t 1 \, dt' - \int_0^t \cos(t') \, dt' = t - \sin(t)$$ So $x(t) = t - \sin(t)$ m.

Step 3: Sanity check.

Your turn: Verify the result by differentiating. Does $dx/dt$ give you $v(t)$? Does $dv/dt$ give you $a(t)$?

Check your answer

$$\frac{dx}{dt} = \frac{d}{dt}\bigl(t - \sin(t)\bigr) = 1 - \cos(t) = v(t) \quad \checkmark$$ $$\frac{dv}{dt} = \frac{d}{dt}\bigl(1 - \cos(t)\bigr) = \sin(t) = a(t) \quad \checkmark$$ Everything checks out.

Notice something interesting about this example. The acceleration $a(t) = \sin(t)$ oscillates --- it is sometimes positive, sometimes negative. Yet the velocity $v(t) = 1 - \cos(t)$ is always greater than or equal to zero. The object never moves backward, even though the acceleration periodically reverses. Can you see why? The velocity oscillates but never dips below zero because $\cos(t)$ never exceeds 1.

Why Closed-Form Answers Are Not Always Available

In the two examples above, we were able to find antiderivatives and write $v(t)$ and $x(t)$ as explicit formulas. That is not always possible.

Consider $a(t) = e^{-t^2}$. This is a perfectly reasonable acceleration function --- it describes an acceleration that starts at $1 \text{ m/s}^2$ and fades to zero as a bell-shaped curve. But the integral

$$v(t) = \int_0^t e^{-t'^2} \, dt'$$

has no elementary closed form. There is no combination of polynomials, trig functions, exponentials, and logarithms that equals this integral. It is related to the "error function" from probability and statistics, but it cannot be written in terms of the functions you learn in a standard calculus course.

Does this mean the physics is broken? Not at all. The velocity $v(t)$ is perfectly well-defined --- it is the area under the $e^{-t^2}$ curve from $0$ to $t$. We can compute it to any desired accuracy using numerical integration. We can graph it. We can reason about its behavior (it increases, it has a horizontal asymptote). We just cannot write it as a tidy formula.

This is an important lesson: the integral relationships always work, but they do not always produce simple formulas. The constant-acceleration case was special precisely because the integrals were easy. In the general case, computing the answer may require numerical methods, series approximations, or simply leaving the result in integral form.

Returning to the Prediction

At the start of this section, you were asked: if a rocket's acceleration is increasing with time, will its speed grow faster or slower than a constant-acceleration case?

You now have the tools to answer this precisely. Suppose the constant-acceleration case uses $a_0$ (the rocket's initial acceleration) for all time. Then:

• Constant case: $v(t) = a_0 t$ --- speed grows linearly.
• Increasing acceleration: $v(t) = \int_0^t a(t') \, dt'$ where $a(t') > a_0$ for all $t' > 0$.

Since the acceleration is always at least as large as the constant value (and usually larger), the integral accumulates more area. The speed grows faster than the constant-acceleration prediction.

[Video: Animation showing two velocity-time graphs overlaid. One is the straight line $v = a_0 t$. The other curves upward above it, pulling further and further away. The shaded area between them represents the extra velocity gained from the increasing acceleration. The narration emphasizes: "The integral accumulates the total effect. When the acceleration is larger, the area is larger, and the velocity is greater."]

Recall Prompt: Reference Frames

Quick check from Chapter 1: Suppose two observers are watching the same rocket. Observer A is on the ground. Observer B is in a car moving at constant velocity $V$ to the right.

Do both observers measure the same acceleration for the rocket? Do they measure the same velocity?

Think about this for a moment before checking.

Check your answer

Both observers measure the **same acceleration**. Acceleration is the same in all inertial reference frames because the constant relative velocity between the frames cancels out when you take derivatives. They measure **different velocities**, however. Observer B sees the rocket's velocity shifted by $-V$ relative to what Observer A measures. But when they compute $dv/dt$, the constant $V$ disappears, and they agree on $a(t)$. This is why the integral relationships work the same way in any inertial frame: the acceleration function $a(t)$ does not depend on the observer's constant velocity.

Practice

Layer 1: Concrete

A particle starts from rest at the origin with acceleration $a(t) = 4t$ (in m/s$^2$).

(a) Find $v(t)$.

(b) Find $x(t)$.

(c) What is the particle's velocity and position at $t = 3$ s?

Check your answer

**(a)** $v(t) = \int_0^t 4t' \, dt' = 4 \cdot \frac{t'^2}{2}\Big|_0^t = 2t^2$ m/s. **(b)** $x(t) = \int_0^t 2t'^2 \, dt' = 2 \cdot \frac{t'^3}{3}\Big|_0^t = \frac{2t^3}{3}$ m. **(c)** At $t = 3$ s: $v(3) = 2(9) = 18$ m/s. $x(3) = \frac{2(27)}{3} = 18$ m. **Sanity check:** $dv/dt = 4t = a(t)$ and $dx/dt = 2t^2 = v(t)$. Consistent.

Layer 2: Pattern

For each acceleration function below, determine $v(t)$ (starting from rest) and classify whether the resulting speed grows without bound as $t \to \infty$ or approaches a finite limit.

(a) $a(t) = 5$

(b) $a(t) = 3t$

(c) $a(t) = \frac{6}{(1+t)^2}$

(d) $a(t) = 10 e^{-t}$

Check your answer

**(a)** $v(t) = 5t$. This grows without bound --- speed increases forever. (This is the constant-acceleration case.) **(b)** $v(t) = \frac{3}{2}t^2$. This also grows without bound, and in fact grows *faster* than the constant case --- the speed is a parabola, not a line. **(c)** $v(t) = \int_0^t \frac{6}{(1+t')^2} \, dt' = -\frac{6}{1+t'}\Big|_0^t = 6 - \frac{6}{1+t} = \frac{6t}{1+t}$. As $t \to \infty$, $v(t) \to 6$ m/s. The speed **approaches a finite limit**. Even though the acceleration is always positive (the object is always speeding up), the acceleration dies off fast enough that the total accumulated velocity is bounded. **(d)** $v(t) = \int_0^t 10 e^{-t'} \, dt' = -10 e^{-t'}\Big|_0^t = 10(1 - e^{-t})$. As $t \to \infty$, $v(t) \to 10$ m/s. Again, **bounded speed**. The exponentially decaying acceleration adds less and less velocity over time, and the total converges. **The pattern:** Whether speed is bounded depends on whether the integral $\int_0^\infty a(t') \, dt'$ converges. If the total area under the $a(t)$ curve is finite, the speed approaches a limit. If the area is infinite, the speed grows without bound.

Layer 3: Structure

A student in your study group says: "When the acceleration is not constant, I just use the average acceleration and plug it into $v = v_0 + \bar{a} \cdot t$. It should give the same answer."

Is this correct? If not, explain precisely what goes wrong and under what special condition it would give the correct result.

Check your answer

This is **not correct in general**. Here is why. The true velocity is $v(t) = v_0 + \int_0^t a(t') \, dt'$. The average acceleration over the interval $[0, t]$ is $\bar{a} = \frac{1}{t}\int_0^t a(t') \, dt'$. Plugging the average into the constant-acceleration formula gives: $$v_0 + \bar{a} \cdot t = v_0 + \frac{1}{t}\int_0^t a(t') \, dt' \cdot t = v_0 + \int_0^t a(t') \, dt'$$ Wait --- that is actually the same! So the student's approach gives the correct **velocity at time $t$**. But here is the catch: the student's method gives the right velocity at the *end* of the interval, but it does **not** give the correct position. The position formula $x = x_0 + v_0 t + \frac{1}{2}\bar{a}t^2$ would be wrong because it assumes the velocity grew linearly from $v_0$ to $v_0 + \bar{a}t$, when in reality the velocity followed a curved path. The area under the actual $v(t)$ curve is generally not the same as the area under the straight line connecting its endpoints. So: using the average acceleration to find the *final velocity* happens to work (by definition of the average), but using it to find the *position* does not.

Layer 4: Transfer

A population of bacteria starts at $N_0 = 1000$. The growth rate decreases over time due to resource depletion: $r(t) = r_0 e^{-t/\tau}$, where $r_0 = 0.5$ per hour and $\tau = 3$ hours. The rate of population change is $dN/dt = r(t) \cdot N_0$ (assuming the population stays close to $N_0$ for the calculation).

(a) Find the population $N(T)$ at time $T$.

(b) What is the maximum population as $T \to \infty$?

(c) How is this problem structurally identical to a kinematics problem?

Check your answer

**(a)** This has the same structure as integrating a time-varying acceleration: $$N(T) = N_0 + \int_0^T r_0 e^{-t/\tau} \cdot N_0 \, dt = N_0 + N_0 r_0 \int_0^T e^{-t/\tau} \, dt$$ $$= N_0 + N_0 r_0 \left[-\tau e^{-t/\tau}\right]_0^T = N_0 + N_0 r_0 \tau \left(1 - e^{-T/\tau}\right)$$ Plugging in: $N(T) = 1000 + 1000(0.5)(3)(1 - e^{-T/3}) = 1000 + 1500(1 - e^{-T/3})$. **(b)** As $T \to \infty$: $N \to 1000 + 1500 = 2500$ bacteria. The population approaches a finite limit because the growth rate decays to zero fast enough --- the integral converges. **(c)** The structural mapping is: | Kinematics | Population | |:---|:---| | $a(t)$ (acceleration) | $r(t) \cdot N_0$ (population growth rate) | | $v(t)$ (velocity) | $N(t)$ (population) | | $v_0$ (initial velocity) | $N_0$ (initial population) | | $\int_0^T a(t') \, dt'$ | $\int_0^T r(t') \cdot N_0 \, dt'$ | The math is identical. Integrating a time-varying rate to find a cumulative quantity is the same operation whether the rate is acceleration, population growth, flow rate, or anything else. The integral is the universal tool for accumulating a changing quantity over time.

Reflection

Think about what you have learned in this section, then consider these questions:

When would you use the constant-acceleration formulas from Section 2.2, and when would you need the integral approach from this section?

The constant-acceleration formulas are not obsolete --- they are efficient tools for the situations where they apply. The integral relationships are the general framework. Knowing which to reach for in a given problem is part of the skill.

You might also think about: In the faded example with $a(t) = \sin(t)$, the acceleration oscillated but the velocity never went negative. Did that surprise you? Can you state a general condition for when velocity stays non-negative?

Looking Ahead

You now have the general machinery for connecting acceleration, velocity, and position through integration. This works for any $a(t)$ --- polynomial, exponential, sinusoidal, or even functions with no closed-form antiderivative.

But real motion rarely obeys a single rule from start to finish. A rocket burns at full thrust for 10 seconds, then the engine cuts off and gravity alone takes over. A car accelerates, then cruises, then brakes. The acceleration changes abruptly from one expression to another.

In the next section, we tackle this head-on: motion with piecewise-defined acceleration. The integral tools you just learned still apply within each stage. The new challenge is stitching the stages together --- making sure that the velocity and position do not suddenly "jump" at the boundaries where the acceleration rule changes. The physics demands continuity, and enforcing it will become the central skill of Section 2.4.