2.5 Turning Points, Extrema, and Qualitative Analysis of Motion

The Ball at the Top

A ball is thrown straight up. It rises, slows, reaches some highest point, and falls back down. You have seen this a thousand times.

Here is the question: at the exact instant the ball reaches its highest point, what do you know about its velocity? What do you know about its acceleration?

Most people answer the first part easily: the velocity is zero at the top. The ball is momentarily at rest.

But here is a harder question: without computing anything --- without knowing the initial speed, without writing down a single equation --- can you determine the exact moment the ball reaches its peak? What information would you need?

Think about what you actually need. Not formulas. Not numbers. Just the right idea.

If you know the position as a function of time, you are looking for the moment when the position reaches a maximum. And you already know a tool for finding maxima: set the derivative equal to zero. The velocity is the derivative of position. So the peak happens when $v(t) = 0$.

You just solved a physics problem using pure reasoning about derivatives --- no algebra, no plugging in numbers, no kinematic equations. That is what this section is about.

Before you read on: Qualitative reasoning --- reading the behavior of motion from graphs, signs, and calculus structure --- often gives you the answer faster than any formula. The goal of this section is to sharpen that kind of thinking.

A Graph with Secrets

Look at this position-time graph.

[Image: A position-time graph from $t = 0$ to $t = 10\,\text{s}$. The curve starts at $x = 0$, rises to a local maximum near $t = 3\,\text{s}$ (labeled point A, at approximately $x = 5\,\text{m}$), then descends to a local minimum near $t = 7\,\text{s}$ (labeled point B, at approximately $x = 1\,\text{m}$), then rises again through $t = 10\,\text{s}$. Between points A and B, the curve has a roughly symmetric shape with an inflection point near $t = 5\,\text{s}$ (labeled point C). The curve is smooth everywhere.]

Before you read on: Study the graph. Then answer these questions before scrolling further.

At which labeled point(s) is the velocity zero?

At which labeled point(s) is the acceleration zero?

At which point is the object moving fastest?

Commit to your answers. Write them down or say them out loud.

[Interactive: Predict-Then-Reveal. The student sees the position-time graph described above with labeled points A, B, and C. For each of the three questions, they select from the available points (A, B, C, or "none of these"). After submitting all three answers, the system reveals: - Question 1: Velocity is zero at A and B (local maximum and minimum --- the slope of the curve is zero at both). - Question 2: Acceleration is zero at C (the inflection point --- the curve changes from concave down to concave up, so the second derivative crosses zero). - Question 3: The object is moving fastest near C (the slope of $x(t)$ is steepest in magnitude at the inflection point between A and B). Each answer includes a brief explanation connecting slope to velocity and curvature to acceleration.]

If you got the first question right, good --- you are reading slopes correctly. The second and third questions are where most students stumble. The inflection point on a position graph does not look dramatic. Nothing "happens" visually at that point. But physically, it is where the acceleration vanishes and the speed reaches a peak. The curve is steepest there, even though the eye is drawn to the peaks and valleys instead.

The Guiding Question

How can we predict key events in motion --- turning points, maximum speed, changes in direction --- without solving every detail exactly?

The answer is that calculus gives you a toolkit for reading qualitative behavior directly from graphs and equations. You do not always need to find $x(t)$ at every instant. Sometimes you just need to know where the interesting things happen.

What Is a Turning Point?

When an object reverses direction, we call that moment a turning point. The ball thrown upward has a turning point at its highest position. A pendulum has turning points at the ends of its swing. A car that backs up and then drives forward has a turning point when it stops and reverses.

At a turning point, the velocity is zero. The object is momentarily at rest. But "momentarily at rest" is not the same as "stopped." Something important is happening at that instant --- the object is in the process of reversing. What drives the reversal is the acceleration, which is not zero at a turning point.

This distinction matters. Let's make it precise.

A turning point occurs where $v = 0$ and $a \neq 0$.

The velocity vanishes because the object has stopped moving in one direction. The nonzero acceleration is what pushes it back the other way. If both $v$ and $a$ were zero, there would be nothing to make the object start moving again --- it would simply stay put (at least momentarily).

Consider the ball thrown upward. At the peak, $v = 0$ (it has stopped rising) and $a = -g$ (gravity is still pulling it down). The nonzero acceleration is exactly what makes it fall back down. The turning point is not a moment of rest --- it is a moment of transition.

Pause and think: A car sits at a red light. Its velocity is zero and its acceleration is zero (it is not speeding up, slowing down, or turning). Is the car at a turning point? Why or why not?

No. The car is simply at rest. A turning point requires the object to have been moving, to momentarily stop, and to then reverse. The defining signature is $v = 0$ with $a \neq 0$. The car at the light has $v = 0$ and $a = 0$ --- it is stationary, not turning around.

Reading Motion from the Position Graph

The connection between the mathematics and the physics is direct:

Feature of $x(t)$ graph	What it means physically
Local maximum (peak)	Turning point --- object reverses from positive to negative velocity
Local minimum (valley)	Turning point --- object reverses from negative to positive velocity
Inflection point	Acceleration changes sign; speed is at a local extremum
Steepest slope	Maximum speed (in that region)
Zero slope (horizontal tangent)	Velocity is zero --- object is momentarily at rest
Concave up (curves upward like a bowl)	Positive acceleration --- velocity is increasing
Concave down (curves downward like a hill)	Negative acceleration --- velocity is decreasing

Every row in this table is a consequence of the derivative chain $x(t) \to v(t) \to a(t)$. Velocity is the slope of the position graph. Acceleration is the curvature. You learned these translations in Chapter 1. Here, they become tools for answering physical questions.

Extrema in Speed: Where Is the Object Moving Fastest?

This is a subtler question than it first appears. The speed of an object is $|v(t)|$ --- the magnitude of the velocity. Speed is always nonnegative. The object is moving fastest where the speed has a maximum.

When does speed reach a maximum? Think about what would have to be true. If the speed is at a maximum, then it is not increasing and not decreasing at that instant. The rate of change of speed is (momentarily) zero. And what controls the rate of change of velocity? Acceleration.

More precisely: extrema in speed correspond to moments where the acceleration is zero (or where the velocity itself is zero, which is a minimum of speed). This is because $a = 0$ means $v$ has stopped changing, which means $|v|$ has stopped changing too --- speed has reached a local peak or plateau.

On the position graph, this corresponds to the inflection point --- the place where the curvature changes sign. The slope is steepest at the inflection point, and the slope is the velocity. So the object is moving fastest right where the graph changes from curving one way to curving the other.

Before you read on: Go back to the graph at the beginning of this section. Verify that point C --- the inflection point --- is indeed where the slope of the curve has the greatest magnitude. Compare the steepness of the curve at C to the steepness near A or B. Does the visual impression match the mathematical claim?

Exploration: Finding Features in Graphs

[Interactive: Graph Detective. Students are presented with a series of $x(t)$ graphs of increasing complexity, one at a time. For each graph, they must: 1. Click to place red markers at all turning points (where $v = 0$ and the object reverses direction). 2. Click to place blue markers at all inflection points (where acceleration changes sign). 3. Click to place green markers at locations of maximum speed (where $|v|$ is greatest).

After the student places their markers, the system reveals the correct locations with brief annotations: - At each turning point: "The slope is zero here and changes sign --- the object reverses." - At each inflection point: "$a = 0$ here --- the curvature flips from concave up to concave down (or vice versa)." - At each maximum-speed point: "The slope has maximum magnitude here --- this is where the object moves fastest."

The system also highlights any markers the student placed incorrectly and explains why. For instance, if a student places a turning-point marker at an inflection point: "The slope is not zero here --- the object is not at rest. It is actually at maximum speed."

Five graphs are provided, progressing from simple (a single peak) to complex (multiple turning points, plateaus, and inflection points). The final graph includes a section that is flat --- a plateau where $v = 0$ for a finite interval, which is not a turning point because the object does not reverse.]

Spend time with this tool. The goal is to train your eye so that when you see a position graph, you can instantly read off the physically important moments without computing anything.

From Equations to Qualitative Features

Graphs are powerful for building intuition, but you will not always have a graph in front of you. Sometimes you start with an equation for $x(t)$ and need to extract the same qualitative information by calculation.

The procedure is a direct application of calculus:

Find turning points. Compute $v(t) = \frac{dx}{dt}$. Set $v(t) = 0$ and solve for $t$. These are your candidates. Check that $a(t) \neq 0$ at each candidate to confirm it is a genuine turning point (not a plateau or point of inflection with zero slope).
Find inflection points. Compute $a(t) = \frac{d^2x}{dt^2}$. Set $a(t) = 0$ and solve for $t$. At each such time, verify that $a(t)$ changes sign (by checking values on either side).
Determine direction of motion. The object moves in the positive direction when $v(t) > 0$ and in the negative direction when $v(t) < 0$. The sign of $v(t)$ in each interval between turning points tells you the direction.
Determine whether the object is speeding up or slowing down. The object speeds up when $v$ and $a$ have the same sign (both positive or both negative) and slows down when they have opposite signs.

Let's work through an example.

Worked Example

Consider the position function

$$x(t) = t^3 - 6t^2 + 9t + 2$$

where $x$ is in meters and $t$ is in seconds, for $t \geq 0$.

Step 1: Velocity.

$$v(t) = \frac{dx}{dt} = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3)$$

Setting $v(t) = 0$: the velocity vanishes at $t = 1\,\text{s}$ and $t = 3\,\text{s}$.

Step 2: Acceleration.

$$a(t) = \frac{dv}{dt} = 6t - 12 = 6(t - 2)$$

At $t = 1$: $a(1) = 6(1 - 2) = -6\,\text{m/s}^2$ (not zero --- so $t = 1$ is a genuine turning point).

At $t = 3$: $a(3) = 6(3 - 2) = +6\,\text{m/s}^2$ (not zero --- so $t = 3$ is also a genuine turning point).

Step 3: Direction of motion.

The velocity $v(t) = 3(t-1)(t-3)$ is a quadratic opening upward. Testing signs:

For $0 < t < 1$: both factors are negative, so $v > 0$. The object moves in the positive direction.
For $1 < t < 3$: $(t-1) > 0$ but $(t-3) < 0$, so $v < 0$. The object moves in the negative direction.
For $t > 3$: both factors are positive, so $v > 0$. The object moves in the positive direction again.

The object starts moving to the right, reverses at $t = 1$, moves to the left, and reverses again at $t = 3$.

Step 4: Speeding up or slowing down?

The acceleration $a(t) = 6(t - 2)$ is negative for $t < 2$ and positive for $t > 2$. Comparing signs of $v$ and $a$:

$0 < t < 1$: $v > 0$, $a < 0$. Opposite signs --- slowing down.
$1 < t < 2$: $v < 0$, $a < 0$. Same sign --- speeding up (in the negative direction).
$2 < t < 3$: $v < 0$, $a > 0$. Opposite signs --- slowing down.
$t > 3$: $v > 0$, $a > 0$. Same sign --- speeding up.

Step 5: Inflection point and maximum speed.

$a(t) = 0$ at $t = 2\,\text{s}$. The acceleration changes sign from negative to positive at this time. At $t = 2$, the velocity is $v(2) = 3(2-1)(2-3) = 3(1)(-1) = -3\,\text{m/s}$. The speed is $|v(2)| = 3\,\text{m/s}$, which is the maximum speed in the interval $1 < t < 3$ (between the two turning points). The object is moving fastest at the inflection point.

[Video: Animation of the motion described above. A dot moves along a number line. Its position, velocity, and acceleration are displayed as updating values. Below the number line, three synchronized graphs ($x(t)$, $v(t)$, $a(t)$) trace out in real time as the dot moves. The turning points at $t = 1$ and $t = 3$ are highlighted with brief pauses and labels. The inflection point at $t = 2$ is highlighted with a label reading "maximum speed." The animation emphasizes that at the turning points the dot stops and reverses, while at the inflection point it is moving fastest.]

Connection: Calculus You Already Know

If you have taken a calculus course, the procedure above should feel familiar. Finding turning points by setting the first derivative to zero, checking the second derivative to classify them, identifying inflection points --- this is exactly the same analysis you performed on abstract functions in a math class.

The difference is that now the function is not abstract. The function is $x(t)$, the first derivative is velocity, the second derivative is acceleration, and the "critical points" are physical events: a ball reaching its peak, a pendulum reversing direction, a car hitting maximum speed before braking.

The derivative and second derivative tests from calculus --- the same tools you used for function analysis --- now tell you about physical events in motion.

This is one of the deep rewards of learning calculus before physics. The mathematical machinery was built for analyzing functions. Motion is a function. The machinery fits perfectly.

Spaced Retrieval

Before moving to practice, recall some earlier ideas from memory.

Recall prompt 1: In Section 2.1, you learned about sign conventions. If a ball is thrown upward and you choose "up" as the positive direction, what is the sign of the acceleration at the top of the flight?

Recall prompt 2: In Section 2.2, you derived the constant-acceleration kinematic equations. What assumption must be true for those equations to apply? Is that assumption satisfied for the motion $x(t) = t^3 - 6t^2 + 9t + 2$?

Recall prompt 3: In Section 2.4, you studied piecewise-defined acceleration. At a boundary between two stages of motion, which quantities must be continuous?

Practice

Layer 1: Concrete

Problem 1. An object moves along the $x$-axis with position given by

$$x(t) = 2t^3 - 9t^2 + 12t$$

where $x$ is in meters and $t$ is in seconds.

(a) Find all turning points. At each one, state whether the object changes from moving right to moving left, or from moving left to moving right.

(b) Find the position of the object at each turning point.

Check your answer

(a) Compute the velocity: $$v(t) = \frac{dx}{dt} = 6t^2 - 18t + 12 = 6(t^2 - 3t + 2) = 6(t - 1)(t - 2)$$ Setting $v(t) = 0$: the turning points are at $t = 1\,\text{s}$ and $t = 2\,\text{s}$. Check the acceleration $a(t) = 12t - 18$: - At $t = 1$: $a(1) = -6\,\text{m/s}^2 \neq 0$. Confirmed turning point. - At $t = 2$: $a(2) = +6\,\text{m/s}^2 \neq 0$. Confirmed turning point. Checking the sign of $v$ in each interval: - For $0 < t < 1$: $v > 0$ (moving right). - For $1 < t < 2$: $v < 0$ (moving left). - For $t > 2$: $v > 0$ (moving right). At $t = 1$, the object changes from right to left. At $t = 2$, it changes from left to right. (b) $x(1) = 2 - 9 + 12 = 5\,\text{m}$. $x(2) = 16 - 36 + 24 = 4\,\text{m}$. (c) $a(t) = 12t - 18 = 0$ gives $t = 1.5\,\text{s}$. At that time, $v(1.5) = 6(0.5)(-0.5) = -1.5\,\text{m/s}$. The speed is $|v(1.5)| = 1.5\,\text{m/s}$.

Layer 2: Pattern

Problem 2. The velocity of an object is given by

$$v(t) = 4t^3 - 12t$$

where $v$ is in m/s and $t \geq 0$.

(a) Determine all time intervals where the object moves in the positive direction ($v > 0$).

(b) Determine all time intervals where the object is speeding up.

Check your answer

(a) Factor: $v(t) = 4t(t^2 - 3) = 4t(t - \sqrt{3})(t + \sqrt{3})$. Since $t \geq 0$, we ignore the factor $(t + \sqrt{3})$, which is always positive. The sign of $v$ for $t \geq 0$: - $0 < t < \sqrt{3}$: $v(t) = 4t \cdot (\text{negative}) = \text{negative}$ (moving in the negative direction). - $t > \sqrt{3}$: $v(t) = 4t \cdot (\text{positive}) = \text{positive}$ (moving in the positive direction). So the object moves in the positive direction for $t > \sqrt{3} \approx 1.73\,\text{s}$. (b) Compute $a(t) = \frac{dv}{dt} = 12t^2 - 12 = 12(t^2 - 1) = 12(t-1)(t+1)$. For $t \geq 0$: $a < 0$ for $0 < t < 1$ and $a > 0$ for $t > 1$. The object speeds up when $v$ and $a$ have the same sign: - $0 < t < 1$: $v < 0$, $a < 0$ --- same sign --- **speeding up**. - $1 < t < \sqrt{3}$: $v < 0$, $a > 0$ --- opposite signs --- **slowing down**. - $t > \sqrt{3}$: $v > 0$, $a > 0$ --- same sign --- **speeding up**. The object is speeding up during $0 < t < 1$ and $t > \sqrt{3}$. (c) Turning points occur where $v(t) = 0$ (with the object actually changing direction). At $t = 0$: $v = 0$, but the object starts from rest here --- this is a starting point, not a reversal (the velocity goes from zero to negative). At $t = \sqrt{3}$: $v = 0$, and the velocity changes from negative to positive --- this is a genuine turning point. Check: $a(\sqrt{3}) = 12(3 - 1) = 24\,\text{m/s}^2 \neq 0$. Confirmed.

Layer 3: Structure

Problem 3. Is it possible for an object to have a turning point where the acceleration is also zero? If so, give an example. If not, explain why it is impossible.

Before you read the answer: This is a genuine mathematical question, not a trick. Think about what happens at a point where both $v = 0$ and $a = 0$. Does the object necessarily stay put? Or could it still reverse direction?

Check your answer

It is possible, but only in special circumstances. The definition of a turning point requires the object to reverse direction --- the velocity must change sign. If both $v = 0$ and $a = 0$ at the same instant, we need to look at higher derivatives to determine what happens. **Example:** Consider $x(t) = t^5$ near $t = 0$. Then: - $v(t) = 5t^4$, so $v(0) = 0$. - $a(t) = 20t^3$, so $a(0) = 0$. But this is *not* a turning point. The velocity $v(t) = 5t^4$ is nonnegative for all $t$, so the object never reverses. It merely "pauses" at $t = 0$ (with zero velocity and zero acceleration) and then continues in the same direction. **Counterexample:** Now consider $x(t) = -t^4$ near $t = 0$. Then: - $v(t) = -4t^3$, so $v(0) = 0$. - $a(t) = -12t^2$, so $a(0) = 0$. Here the velocity $v(t) = -4t^3$ *does* change sign: it is positive for $t < 0$ and negative for $t > 0$. So the object does reverse direction at $t = 0$. This is a turning point where $a = 0$. The condition $v = 0$ and $a \neq 0$ is *sufficient* for a turning point, but it is not *necessary*. When $a = 0$ as well, you need to investigate further. In the standard problems of this course --- polynomials of degree 2 or 3, trig functions, or piecewise constant acceleration --- turning points almost always have $a \neq 0$. The edge cases arise in more exotic functions.

Layer 4: Debug

Problem 4. A student is analyzing the motion $x(t) = t^3 - 3t$ and writes the following:

"At $t = 1$, the velocity is zero, so the object is not moving. It is stuck at that point."

What is misleading or incorrect about this statement?

Check your answer

The student is correct that $v(1) = 0$: the velocity is indeed zero at $t = 1$. (You can verify: $v(t) = 3t^2 - 3$, so $v(1) = 0$.) But saying the object is "not moving" and "stuck at that point" is misleading in two ways. **First, "not moving" conflates a single instant with an interval.** The velocity is zero *at the instant* $t = 1$, but the object is not stationary --- it does not remain at rest. It was moving before $t = 1$ and it will be moving after $t = 1$. Saying "not moving" implies the object has stopped for a period of time. In reality, it is only instantaneously at rest. This is like saying a ball thrown upward "is not moving" at the peak --- true for an instant, but deeply misleading as a description of the situation. **Second, "stuck at that point" implies nothing is happening.** In fact, at $t = 1$, the acceleration is $a(1) = 6(1) = 6\,\text{m/s}^2$ --- far from zero. The object is in the middle of reversing direction. Forces are acting. Something is very much happening. The object is not stuck; it is transitioning. A better statement: "At $t = 1$, the object is at a turning point. It is momentarily at rest but is accelerating, so it will immediately begin moving in the opposite direction."

Putting It All Together: Reading the Full Story

Here is a position-time graph for a more complex motion.

[Image: A position-time graph from $t = 0$ to $t = 12\,\text{s}$. The curve starts at $x = 3\,\text{m}$, rises to a local maximum near $x = 8\,\text{m}$ at $t = 2\,\text{s}$, descends to a local minimum near $x = 1\,\text{m}$ at $t = 6\,\text{s}$, rises steeply to another local maximum near $x = 10\,\text{m}$ at $t = 9\,\text{s}$, then gently descends to about $x = 7\,\text{m}$ at $t = 12\,\text{s}$. The curve has two clear inflection points: one between $t = 2$ and $t = 6$ (approximately at $t = 4$), and another between $t = 6$ and $t = 9$ (approximately at $t = 7.5$).]

Before you read on: Without computing anything, answer these questions just by reading the graph:

How many turning points are there? At approximately what times?

During which interval(s) is the object moving in the negative direction?

At approximately what times is the acceleration zero?

Where is the object moving fastest?

Is the object speeding up or slowing down at $t = 5\,\text{s}$?

[Interactive: Full Motion Analysis. The complex position graph described above is displayed. Students answer each of the five questions using click-to-select or text entry. After all five answers are submitted, the system provides a complete motion narrative:

"The object has three turning points at approximately $t = 2$, $t = 6$, and $t = 9\,\text{s}$. It moves in the negative direction from $t = 2$ to $t = 6$ and from $t = 9$ to $t = 12$. The acceleration is zero near $t = 4$ and $t = 7.5$ (inflection points). The object moves fastest near $t = 7.5$, where the slope of the curve has the greatest magnitude. At $t = 5$, the object is moving in the negative direction (slope is negative) and the curve is concave up (curving upward) --- the velocity is negative and becoming less negative, so the object is slowing down."

Each answer is accompanied by visual annotations on the graph: tangent lines at key points, curvature indicators, and velocity arrows.]

This kind of qualitative analysis --- reading the complete motion story from a graph --- is a skill that takes practice but pays off enormously. When you encounter a new motion problem, sketching the graph and reading its qualitative features gives you a roadmap before you do any computation. You already know where the turning points are, where the object speeds up, where it slows down. The algebra just fills in the exact numbers.

Reflection

Think about what you have done in this section. You identified turning points, classified motion behavior, and read the entire story of a complex motion --- all from the shape of a graph and the logic of derivatives.

How much of the motion story can you read from a graph without ever writing an equation?

The answer, as you have seen, is: almost all of it. Direction of motion, turning points, maximum speed, speeding up versus slowing down. All of these are encoded in the slope and curvature of the position graph. The equations give you precision --- exact times, exact positions. But the qualitative structure is already there in the shape of the curve, waiting to be read by someone who knows where to look.

Looking Ahead

You have been working with smooth, clean functions --- equations and graphs that behave perfectly. But real motion data is not like that. Real data comes from sensors and measurements, and it comes with noise, gaps, and uncertainty.

In the next section, you will confront real experimental data: messy position measurements from a motion sensor tracking a cart on a track. You will try to extract velocity and acceleration from that data, and you will discover something troubling --- differentiation, the tool you have been using so confidently, amplifies every bit of noise in the data. The same derivative that reveals turning points in a clean equation creates chaos when applied to imperfect measurements.

That tension --- between the mathematical ideal and the experimental reality --- is one of the central themes of physics. And it starts in Section 2.6.