Unit 4Relationships:

a(t)=v'(t)=s''(t)

Concept 4.161Built for May 2027 AP Exam

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

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Acceleration is the derivative of velocity and the second derivative of position, so it is one step further down the same ladder that produced velocity. Write $a(t) = v'(t) = s''(t) = \frac{d^2s}{dt^2}$ , and note that the units change once more: meters per second squared rather than meters per second. What acceleration answers is a comparison, not a value. Velocity says which way the particle moves, acceleration says which way that motion is changing, and only the two signs together say whether the speed is growing or shrinking. That comparison is the most tested sentence in the unit, and on a plane motion problem the same idea arrives as a vector with one component per axis.

On the AP Exam

What you’ll see: A position or velocity formula differentiated twice, a velocity vector whose acceleration vector is requested, or a table of velocities where the sign of the change decides whether the particle speeds up.
What you’ll do: Differentiate once more than the previous step, keep the units attached, and compare the sign of $a$ with the sign of $v$ before saying anything about speed.
Memory hook: Same sign speeds up, opposite sign slows down. Nothing else determines speed, and neither sign alone does.
Use it on the exam: For MCQs: Options usually offer "speeding up" and "slowing down" for a particle whose velocity is negative, which is exactly the case where the naive reading fails. For FRQs: On 2022 Q2(b) the acceleration vector earns a point per component, so an unlabelled ordered pair or a reversed pair loses both. Write the vector with its components in the order $x$ then $y$ .

Concept Review

A Comparison, Not a Number

Figure 2 — The four sign combinations of velocity and acceleration. Only the two rows where the signs agree describe a particle that is speeding up, and the direction of travel is set by velocity in every row.

Each cell is a state, not a value. A particle moving left with negative acceleration is speeding up, because both signs point the same way, and a particle moving right with negative acceleration is slowing down. The table is the fastest way to settle a speed question, and it costs one glance once the signs are written.

A two by two grid of velocity and acceleration signs with the direction of travel and whether speed rises or falls in each cell

Notation and Units

Quantity	Symbol	Units	One more step from
position	$s(t)$	meters	—
velocity	$v(t) = s'(t)$	meters per second	differentiate $s$
acceleration	$a(t) = v'(t) = s''(t)$	meters per second squared	differentiate $v$

In Leibniz notation the same statement is $a = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ . The superscript counts operations rather than multiplications, which is why $\frac{d^2s}{dt^2}$ is not a square of anything.

Reading Acceleration Off a Velocity Graph

On a graph of $v$ , acceleration is the slope. A rising velocity graph means positive acceleration regardless of whether the velocity itself is positive or negative, and a horizontal tangent means the acceleration is zero at that instant even though the particle may be moving quickly. This is the same slope-to-value translation that produced velocity from position, applied one level down.

Two Integrations, Two Constants

Going up the ladder needs one constant per step:

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

The exam supplies both numbers when both are needed. A missing $v(0)$ produces a velocity that is wrong by a constant, and the error then propagates into position, which is why the constant is worth writing down before the first integral is evaluated.

Acceleration in the Plane

When position is a pair of functions, everything happens component by component:

$\vec{a}(t) = \left\langle x''(t),\, y''(t)\right\rangle$

The magnitude of the velocity vector is speed, and the magnitude of the acceleration vector is not a quantity the exam asks for. What it asks for is the vector itself, sometimes as an ordered pair and sometimes with the components labelled, and on 2023 Q2(a) the setup is required before the value.

Common Misconception Check

Myth 1: "Acceleration tells you the direction of motion." Truth: Velocity does. Acceleration tells you how the velocity is changing, and the two can point in opposite directions for the whole of a slowing phase.
Myth 2: "Zero acceleration means the particle has stopped." Truth: Zero acceleration means constant velocity. The particle can be moving at full speed with the acceleration momentarily zero at a turning point of the velocity graph.
Myth 3: "The second derivative is the square of the first." Truth: It is the derivative of the first. For $s(t) = t^3$ the velocity is $3t^2$ and the acceleration is $6t$ , which is not $\left(3t^2\right)^2$ under any reading.

Topic-Specific Study Method

Ask What Changes the Change

Borrowed idea: Ask what changes the change. Velocity already describes how fast position moves, so acceleration has to describe how fast that description itself moves. When the two share a sign the speed grows, and when they disagree the speed shrinks, so the comparison is the whole content of this page. Which of those two cases holds at the instant the question names? Read both signs before computing anything. Is a negative acceleration a slowdown? Only when velocity is positive. The signs decide, and the numbers only fill in the blanks.

What the stem gives	What is asked	What to do
$s(t)$	acceleration	differentiate twice, keep units
$v(t)$	acceleration at a time	differentiate once, substitute
$v(t)$ and $a(t)$	speed increasing or decreasing	multiply the signs
$a(t)$ with $v(0)$	velocity, then position	integrate twice, two constants
velocity vector	acceleration vector	differentiate each component
graph of $v$	sign of acceleration	slope of the graph

① Take one velocity formula, find where it is zero, and mark the sign of $v$ on each side. Then compute $a$ at the same three times and fill in the table of Figure 2 from memory. If a cell surprises you, the sign convention is not yet automatic.

② For a particle that moves left, compute the acceleration and state in one sentence whether the speed is rising. The sentence has to name both signs, because the point is earned by the comparison rather than by the number.

③ Take a velocity vector, differentiate both components, and write the answer as an ordered pair with $x$ first. Doing that once under no pressure removes the reversed-pair error, which costs two points in a single stroke.

Training & Error Diagnosis

Core Archetype (Simplest Form): A particle moves along the $x$ -axis with velocity $v(t) = 12 - 3t^2$ meters per second for $0 \le t \le 4$ . Find $a(1)$ and state whether the speed is increasing or decreasing at $t = 1$ .

Standard Execution:

Differentiate. $a(t) = -6t$ , so $a(1) = -6$ meters per second squared.

Read velocity. $v(1) = 9$ meters per second, so the particle is moving in the positive direction.

Compare signs. Velocity is positive and acceleration is negative, so the signs disagree.

Conclude. The speed is decreasing at $t = 1$ .

Exam Variation (Context / Parameter Shift): A particle moves in the plane with velocity vector $\vec{v}(t) = \left\langle 2t,\, 3t^2 - 6\right\rangle$ centimeters per second. Find the acceleration vector at $t = 2$ and determine whether the speed is increasing or decreasing at that instant.

Solution: Differentiating each component gives $\vec{a}(t) = \left\langle 2,\, 6t\right\rangle$ , so $\vec{a}(2) = \langle 2, 12\rangle$ centimeters per second squared. At $t = 2$ the velocity is $\langle 4, 6\rangle$ and the acceleration is $\langle 2, 12\rangle$ . Both components agree in sign, so every part of the velocity is growing and the speed is increasing.

Common Traps & Failure Signals:

The sign read alone (Signal: "acceleration is negative, so the particle is slowing down"): the comparison with velocity was skipped.
The one-step derivative (Signal: an answer that is really $v$ when $a$ was asked): one differentiation was performed where two were needed.
The reversed pair (Signal: the $y$ -component listed first): the vector was assembled in the wrong order, which costs both component points.

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

Sign Alone: You judged the speed by the acceleration alone.
Level Missed: You differentiated once and reported the velocity, which means the ladder was walked one rung short of the question and every unit was wrong by a factor of seconds.
Units Dropped: You wrote the acceleration as a bare number, so the reader cannot tell whether the answer is a velocity, an acceleration, or a position, and on a contextual part that ambiguity costs the interpretation point even when the arithmetic is right.

Past 10-Year AP Exam Appearance

Testing Frequency: Rare in FRQs; Three released BC sets score an acceleration, and all three do it inside a plane motion question where the acceleration is a vector with two independently scored components
10-Year Total Count: 3 of 9 Exams · 3 FRQ Parts; Counted where the scored answer is an acceleration vector or an acceleration value obtained by differentiating a given velocity
FRQ Years Appeared: 2021 (Q2a), 2022 (Q2b), 2023 (Q2a)

Practice Examples

MCQ 1

Two Steps Down

A particle moves along the $x$ -axis with position $s(t) = t^3 - 6t^2 + 5$ meters. What is the acceleration of the particle at $t = 1$ second?

MCQ 2

The Sign Comparison

A particle moves along the $x$ -axis with velocity $v(t) = 6 - 2t$ meters per second for $0 \le t \le 5$ . At $t = 4$ seconds, is the speed of the particle increasing or decreasing?

MCQ 3

An Acceleration Vector

A particle moves in the $xy$ -plane with velocity vector $\vec{v}(t) = \left\langle 3t^2,\, 2t\right\rangle$ centimeters per second. What is the acceleration vector of the particle at $t = 1$ second?

FRQ 1 • Speed, Position, and a Reversal

A particle moves along the $x$ -axis with velocity $v(t) = 12 - 3t^2$ meters per second for $0 \le t \le 4$ . At time $t = 0$ the particle is at position $x = 5$ meters.

(a) Find $a(1)$ and explain what it says about the velocity at that instant.

(b) Find every time in $0 < t < 4$ at which the speed of the particle is increasing. Justify your answer.

(c) Find the position of the particle at $t = 2$ .

(d) Find the total distance travelled by the particle on $0 \le t \le 4$ .

Cheatsheet