Unit 4Relationships:

v(t)=s'(t)

Concept 4.268Built for May 2027 AP Exam

Relationships: $v(t)=s'(t)$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Velocity is the derivative of position, so every statement about motion can be read as a statement about a slope. Write $v(t) = s'(t) = \frac{ds}{dt}$ , and remember that the two notations describe the same number: change in position per unit of time. The sign of $v$ is a direction and never a speed, which is why the absolute value has to be taken last. Going the other way, position is recovered by integrating velocity and applying one initial condition, and the same integral answers two different questions depending on whether it is signed. A signed integral gives displacement, an unsigned one gives distance, and the gap between them is the part of the path that doubled back.

On the AP Exam

What you’ll see: A position function given as a formula, a velocity given as a formula, a table or a graph, or a particle described only by "moves along the $x$ -axis with velocity vector" when the problem lives in the plane.
What you’ll do: Differentiate once to get velocity or integrate once to get position, then read the sign of the result as a direction before doing anything else with it.
Memory hook: Slope first, sign second. The derivative gives a number, and only the sign of that number says which way the particle went.
Use it on the exam: For MCQs: Speed is $\lvert v \rvert$ , so an option offering a negative speed is wrong by construction. Velocity and speed appear together in the same question precisely to catch that slip. For FRQs: The sentence "since $v(t) < 0$ on $(1,3)$ , the particle moves in the negative direction on that interval" is a scored statement on its own. On 2024 Q2(d) the guideline wants the sign of $y'(t)$ and its meaning, in that order.

Concept Review

The Tangent Line Is the Velocity

Figure 1 — A position curve with two tangents. The steep tangent belongs to the faster instant, the flat tangent to the moment the particle stops, and the sign of each slope is the direction of travel.

The height of the curve says where the particle is, and the slope says how fast that position is changing. Those are different questions, and the exam asks them in the same sentence often enough that separating them has to become automatic. At the flat point the slope is zero, so velocity is zero, and the particle is momentarily at rest rather than permanently stopped.

A position curve with a steep tangent, a horizontal tangent at the peak and a falling section, beside two note cards separating height from slope

Three Symbols, One Number

Symbol	Reads as	Units	How it is obtained
$s(t)$	position	meters or feet	integrate $v$ and apply $s(0)$
$v(t) = s'(t) = \frac{ds}{dt}$	velocity	meters per second	differentiate $s$
$\lvert v(t) \rvert$	speed	meters per second	take the magnitude of $v$

The middle row is the definition this page is built on. The third row is not a new function; it is the middle row with the sign removed, which is why a speed can never come out negative and a velocity frequently does.

The Sign of Velocity Is a Direction

If the particle moves along a line, the sign of $v$ splits the timeline into intervals of travel:

$v(t) > 0 \;\Rightarrow\; \text{moving in the positive direction}, \qquad v(t) < 0 \;\Rightarrow\; \text{moving in the negative direction}$

A zero of $v$ is a candidate reversal, and it becomes an actual reversal only when the sign changes across it. This is the difference between the particle stopping and the particle turning around, and the exam tests it directly.

Recovering Position

Differentiation never needs a starting point, and integration always does:

$s(t) = s(0) + \int_0^t v(u)\,du$

The initial condition is usually buried in the stem as "at time $t = 0$ the particle is at $x = 2$ ". Reading it out before integrating keeps one constant from turning into two, and on the plane version of the same question the condition is a point rather than a number:

$y(t) = y(t_0) + \int_{t_0}^{t} y'(u)\,du$

Displacement and Distance

The signed integral measures net change, and the unsigned one measures ground covered:

$\text{displacement} = \int_a^b v(t)\,dt, \qquad \text{distance} = \int_a^b \lvert v(t) \rvert\,dt$

They agree exactly when $v$ keeps one sign. When $v$ changes sign, the distance integral must be split at every zero of $v$ , and the pieces added with their signs stripped. Writing the split before evaluating is what keeps a negative piece from cancelling a positive one and quietly shrinking the answer.

Common Misconception Check

Myth 1: "A negative velocity means the particle is slowing down." Truth: It means the particle moves in the negative direction. Slowing down is a statement about $\lvert v \rvert$ , which needs the sign of acceleration as well.
Myth 2: "Position and velocity are read from the same feature of the graph." Truth: Position is the height and velocity is the slope. A curve can be high and slow, or low and fast.
Myth 3: "Distance travelled equals displacement whenever the particle moves forward." Truth: That is exactly the case where they agree, and the exam usually asks the case where they do not. A single reversal is enough to separate them.

Topic-Specific Study Method

Read the Slope, Then the Sign

Borrowed idea: Read the slope. Every question on this page asks what the position graph is doing at one input, and the answer is never that graph's height. The velocity at an instant is the slope of the position curve at that instant, so a steeper curve means a faster particle and a flat tangent means the particle has stopped. When the position curve turns downward the velocity is negative, and that sign is a direction rather than a slowdown. Take the slope first, then decide what its sign means for the question.

What the stem gives	What is being asked	What to do
$s(t)$ as a formula	velocity, direction, turning points	differentiate, then solve $v = 0$
$v(t)$ as a formula	position, displacement, distance	integrate, split at zeros for distance
graph of $s$	where the particle is fastest	steepest slope, not highest point
graph of $v$	where the particle turns	zero crossing with a sign change
velocity components $x'(t)$ , $y'(t)$	speed, distance in the plane	$\sqrt{\left(x'\right)^2 + \left(y'\right)^2}$ , then integrate

① Take one position formula, differentiate it, and mark on the curve where the slope is zero. The marks must land where the particle is at rest, and if they do not, the derivative is wrong.

② For a velocity that changes sign, compute the displacement and the distance without simplifying either. The two answers differ by twice the area of the region that doubles back.

③ Read a graph of $v$ and write three sentences, one per sign interval, each beginning "the particle moves". The habit transfers to the FRQ sentence that carries a point by itself.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle moves along the $x$ -axis with position $s(t) = t^3 - 6t^2 + 9t$ meters for $t \ge 0$ . Find the intervals on which the particle moves in the negative direction, and the velocity at $t = 2$ .

Standard Execution:

Differentiate. $v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$ .

Locate the stops. $v = 0$ at $t = 1$ and $t = 3$ .

Sign test. Positive on $(0,1)$ , negative on $(1,3)$ , positive for $t > 3$ .

Answer in words. The particle moves in the negative direction on $1 < t < 3$ , and $v(2) = -3$ meters per second.

Exam Variation (Context / Parameter Shift): A particle moves in the plane so that $x'(t) = 4t$ and $y'(t) = 3t^2 - 12$ , with both measured in centimeters per second. At time $t = 2$ seconds the particle is at $(6, -4)$ . Find the speed at $t = 2$ and the position at $t = 0$ .

Solution: The speed is $\sqrt{(4\cdot 2)^2 + (3\cdot 4 - 12)^2} = \sqrt{64 + 0} = 8$ centimeters per second, so the particle is moving horizontally at that instant. Going backwards in time, $x(0) = 6 - \int_0^2 4u\,du = 6 - 8 = -2$ and $y(0) = -4 - \int_0^2 \left(3u^2 - 12\right)du = -4 - (8 - 24) = 12$ , so the position at $t = 0$ is $(-2, 12)$ .

Common Traps & Failure Signals:

The height read as the slope (Signal: velocity reported as a position value): the graph was read vertically when the question asked for a tangent.
The unsigned speed (Signal: a negative speed, or a direction dropped from a sentence): the magnitude was taken before the sign was interpreted.
The unsplit distance (Signal: distance equal to displacement on an interval with a reversal): the integral was evaluated in one piece across a zero of $v$ .

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Height for Slope: You read height where the slope was asked.
Sign Slipped: You found the correct velocity and then described the particle as slowing down, when the negative sign was telling you it had turned around and was heading back.
Integral Unsplit: You integrated the velocity across a zero without breaking the interval there, so the part of the trip that doubled back cancelled part of the trip that did not, and the distance came out smaller than the displacement it should have exceeded.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five of the nine released BC sets build a part around a velocity that is itself a derivative of a given position, and the plane version of the question has become the standard form
10-Year Total Count: 5 of 9 Exams · 13 FRQ Parts; Counted where a scored answer is a velocity, a speed, a direction of travel, a position recovered from a rate, or a distance read from $\lvert v \rvert$
FRQ Years Appeared: 2018 (Q2d), 2021 (Q2a, Q2b, Q2c), 2022 (Q2c, Q2d), 2023 (Q2b, Q2c, Q2d), 2024 (Q2a, Q2b, Q2c, Q2d)

Practice Examples

MCQ 1

Direction From a Position Formula

A particle moves along the $x$ -axis with position $s(t) = 2t^3 - 9t^2 + 12t$ for $t \ge 0$ . On which interval is the particle moving in the negative direction?

MCQ 2

Velocity Against Speed

A particle moves along the $x$ -axis with velocity $v(t) = 3 - t$ meters per second for $0 \le t \le 5$ . At $t = 4$ seconds, what are the velocity and the speed of the particle?

MCQ 3

Position From Velocity

A particle moves along the $x$ -axis with velocity $v(t) = 4t - 6$ meters per second, and its position at $t = 0$ is $s(0) = 5$ meters. What is the position at $t = 2$ seconds?

FRQ 1 • A Particle With Two Reversals

A particle moves along the $x$ -axis with velocity $v(t) = t^2 - 4t + 3$ meters per second for $0 \le t \le 5$ . At time $t = 0$ the particle is at position $x = 2$ meters.

(a) Find all times in $0 < t < 5$ at which the particle is at rest, and state whether the particle changes direction at each.

(b) Find the position of the particle at $t = 5$ .

(c) Find the total distance travelled by the particle on $0 \le t \le 5$ .

(d) Find the average velocity and the average speed of the particle on $0 \le t \le 5$ , and explain why the two values differ.

Cheatsheet