---
title: "Acceleration Function — AP Calc Definition & Exam Guide"
description: "The acceleration function a(t) is the derivative of velocity and the second derivative of position. Master it for straight-line motion in AP Calc Units 4 and 8."
canonical: "https://fiveable.me/ap-calc/key-terms/acceleration-function"
type: "key-term"
subject: "AP Calculus AB/BC"
---

# Acceleration Function — AP Calc Definition & Exam Guide

## Definition

In AP Calculus, the acceleration function a(t) is the derivative of the velocity function and the second derivative of the position function, so a(t) = v'(t) = s''(t). It measures how fast an object's velocity is changing at any moment in rectilinear (straight-line) motion problems.

## What It Is

The acceleration function tells you how an object's [velocity](/ap-calc/unit-4/straight-line-motion-connecting-position-velocity-acceleration/study-guide/2ZIESajDNiJ4ENTrnDT6 "fv-autolink") is changing at every instant. If velocity answers "how fast and which direction?", acceleration answers "is that [speed](/ap-calc/key-terms/speed "fv-autolink") and direction changing, and how quickly?" Mathematically, a(t) = v'(t), and since velocity is already the derivative of position, a(t) = s''(t). That makes acceleration the second derivative of position, which is why you'll also see it written as x''(t).

Here's the part that trips people up. A negative a(t) does NOT automatically mean the object is slowing down. An object speeds up when velocity and acceleration have the **same sign** (both positive or both negative) and slows down when they have **opposite signs**. Think of acceleration as a push. If the push points the same way the object is already moving, it speeds up. If the push fights the motion, it slows down. This sign comparison is the single most tested idea attached to this term.

## Why It Matters

The acceleration function lives in two places in the CED. In Topic 4.2 ([Unit 4](/ap-calc/unit-4 "fv-autolink"): Contextual Applications of Differentiation), learning objective [AP Calc](/ap-calc "fv-autolink") 4.2.A has you calculate rates of change in applied contexts, and rectilinear motion is the classic setup. You differentiate position to get velocity, then differentiate again to get acceleration. In Topic 8.2 (Unit 8: Applications of Integration), learning objective AP Calc 8.2.A flips the direction. Now you integrate acceleration to recover velocity, and integrate velocity to recover position or displacement. The acceleration function is essentially the same motion story told twice, once with derivatives and once with antiderivatives, which is why it's one of the best concepts for seeing how Units 4 and 8 mirror each other.

## Connections

### [Velocity Function (Units 4, 8)](/ap-calc/key-terms/velocity-function)

Velocity is the bridge between position and acceleration. Differentiate v(t) and you get a(t); integrate a(t) and you get the change in velocity. Almost every motion problem makes you move along this chain in one direction or the other.

### [Position Function (Units 4, 8)](/ap-calc/key-terms/position-function)

Acceleration is the second derivative of position, s''(t) or x''(t). On the position graph, acceleration shows up as concavity. [Concave up](/ap-calc/key-terms/concave-up "fv-autolink") means positive acceleration, concave down means negative acceleration.

### Speeding Up vs. Slowing Down (Unit 4)

A particle speeds up when v(t) and a(t) share a sign and slows down when they don't. The exam loves this because the obvious answer (negative acceleration = slowing down) is wrong half the time.

### Definite Integrals of Velocity (Unit 8)

Per CHA-4.C.1, the definite integral of velocity gives [displacement](/ap-calc/key-terms/displacement "fv-autolink"), and the integral of speed (|v(t)|) gives total distance. Acceleration sits one derivative above this chain, so a(t) problems often start by integrating down to v(t) first.

## On the AP Exam

Multiple-choice questions test this in a few predictable ways. The most direct version hands you a velocity function like v(t) = 4 - 6t and asks for a(t), which is just one derivative. A step up gives you position, like s(t) = t² - 3t + 1, and asks for acceleration, requiring two derivatives. The conceptual version asks when a particle is "speeding up" or "decelerating," which means comparing the signs of v(t) and a(t) on intervals, not just looking at a(t) alone. On the free-response side, particle motion is a perennial FRQ context. You're often given v(t) or a(t) and asked to find velocity at a specific time (integrate a(t) and use an initial condition), determine when the particle changes direction, or compute displacement versus total distance. Show the calculus setup. Writing the integral or derivative you're using is what earns the points.

## Acceleration Function vs Velocity Function

Velocity v(t) tells you how fast position is changing; acceleration a(t) tells you how fast velocity is changing. They're one derivative apart, and the exam exploits the gap. A particle can have positive velocity with negative acceleration (moving forward but slowing down) or negative velocity with negative acceleration (moving backward and speeding up). Never judge "speeding up" from a(t) alone. You need both signs.

## Key Takeaways

- The acceleration function is the derivative of velocity and the second derivative of position, so a(t) = v'(t) = s''(t).
- A particle is speeding up when velocity and acceleration have the same sign, and slowing down when they have opposite signs.
- Negative acceleration does not automatically mean slowing down; if velocity is also negative, the particle is actually speeding up.
- In Unit 4 you differentiate down the chain (position to velocity to acceleration); in Unit 8 you integrate back up it.
- To find velocity from a(t), integrate and use an initial condition to solve for the constant; v(t) = v(0) + the integral of a(t) from 0 to t.
- On the graph of position, acceleration shows up as concavity, so concave up means positive acceleration.

## FAQs

### What is the acceleration function in calculus?

It's the function a(t) that gives an object's [rate of change](/ap-calc/key-terms/rate-of-change "fv-autolink") of velocity at time t. You get it by taking the derivative of the velocity function, or the second derivative of the position function: a(t) = v'(t) = s''(t).

### Does negative acceleration mean an object is slowing down?

No, not necessarily. An object slows down only when velocity and acceleration have opposite signs. If both v(t) and a(t) are negative, the object is moving in the negative direction and [speeding up](/ap-calc/key-terms/speeding-up "fv-autolink"). This is one of the most common trap answers on AP Calc motion questions.

### How is the acceleration function different from the velocity function?

Velocity measures how fast position changes; acceleration measures how fast velocity changes. They sit one derivative apart: differentiate v(t) to get a(t), or integrate a(t) to get the change in v(t). A question giving you position s(t) = t² - 3t + 1 needs two derivatives to reach a(t) = 2.

### How do you find velocity from the acceleration function?

Integrate a(t) and use an initial condition to pin down the constant of integration. For example, v(t) = v(0) + ∫₀ᵗ a(x) dx. This is the Unit 8 version of motion problems, where you work backward from acceleration instead of differentiating down from position.

### How do you tell when a particle is speeding up using a(t)?

Find where v(t) and a(t) are both positive or both negative; those are the speeding-up intervals. For v(t) = 4 - t, acceleration is the constant -1, so the particle speeds up only when v(t) is also negative, which happens for t > 4.

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