---
title: "Speed Increasing — AP Calc Definition & Exam Guide"
description: "Speed increases when velocity and acceleration share the same sign. Learn the sign-check rule for AP Calc straight-line motion FRQs and how to justify it."
canonical: "https://fiveable.me/ap-calc/key-terms/speed-increasing"
type: "key-term"
subject: "AP Calculus AB/BC"
unit: "Unit 4"
---

# Speed Increasing — AP Calc Definition & Exam Guide

## Definition

In AP Calculus, a particle's speed is increasing when its velocity v(t) and acceleration a(t) have the same sign (both positive or both negative); speed is the absolute value of velocity, so matching signs mean the particle is speeding up regardless of direction.

## What It Is

[Speed](/ap-calc/key-terms/speed "fv-autolink") is the absolute value of [velocity](/ap-calc/unit-4/straight-line-motion-connecting-position-velocity-acceleration/study-guide/2ZIESajDNiJ4ENTrnDT6 "fv-autolink"), |v(t)|. It measures how fast a particle moves without caring about direction. That last part is the whole trick. A particle moving left with velocity -8 is faster than one moving right with velocity +3.

So when is speed increasing? When velocity and acceleration point the same way. If v(t) > 0 and a(t) > 0, the particle moves right and pushes right, so it speeds up. If v(t) < 0 and a(t) < 0, it moves left and pushes left, so it also speeds up, even though velocity is getting *more negative*. Think of acceleration as a push. If the push is in the same direction the particle is already moving, the particle goes faster. If the push opposes the motion, the particle slows down. On the exam, you check this by finding the signs of v(t) and a(t) at the given time and stating whether they match.

## Why It Matters

This lives in Topic 4.2 (Straight-Line Motion) in [Unit 4](/ap-calc/unit-4 "fv-autolink"): Contextual Applications of Differentiation, supporting learning objective 4.2.A, which asks you to calculate [rates of change](/ap-calc/key-terms/rate-of-change "fv-autolink") in applied contexts. The essential knowledge here is that derivatives solve rectilinear motion problems involving position, speed, velocity, and acceleration. The speed-increasing question is the single most common way the College Board tests whether you truly understand that speed and velocity are different things. It's a near-guaranteed FRQ part: 'Is the speed of the particle increasing or decreasing at time t = c? Give a reason for your answer.' The answer is worth points only if your justification mentions the signs of both v(t) and a(t).

## Connections

### Velocity Function v(t) (Unit 4)

Velocity is the [derivative](/ap-calc/unit-10-infinite-sequences-and-series-bc-only/finding-taylor-polynomial-approximations-functions/study-guide/LszguYzKz0M6GdqTRSr6 "fv-autolink") of position, and its sign tells you direction of motion. The speed-increasing check starts here. You need the sign of v(t) at the given time before you can say anything about speed.

### Acceleration Function x''(t) (Unit 4)

Acceleration is the derivative of velocity (the [second derivative](/ap-calc/unit-3/calculating-higher-order-derivatives/study-guide/Mh7ZnLES3ycIzEEKWl95 "fv-autolink") of position). It's the 'push' in the speed-increasing rule. Compare its sign to the sign of v(t), and you have your answer.

### [Absolute Value (Units 1 & 4)](/ap-calc/key-terms/absolute-value)

Speed is |v(t)|, which is why a velocity going from -2 to -7 means speed went UP, from 2 to 7. Forgetting the [absolute value](/ap-calc/key-terms/absolute-value "fv-autolink") is the number one source of wrong answers on this question type.

### [Change Direction (Unit 4)](/ap-calc/key-terms/change-direction)

A particle changes direction when v(t) changes sign, and at that instant speed hits zero. Right before and right after a direction change, the velocity-acceleration sign comparison flips, so speed switches between [decreasing](/ap-calc/unit-5/determining-intervals-on-which-function-is-increasing-or-decreasing/study-guide/Y2fgjyl7H1dKPI2YsB4Y "fv-autolink") and increasing.

## On the AP Exam

This shows up as a justify-your-answer FRQ part almost every time particle motion appears. The 2024 FRQ Q2 gave a particle with velocity v(t) = ln(t² − 4t + 5) − 0.2t and asked about its motion. On questions like this, you evaluate v and a at the specified time (calculator allowed in the calculator section), then write a sentence comparing their signs. A full-credit justification looks like: 'Since v(2) < 0 and a(2) < 0, velocity and acceleration have the same sign, so the speed is increasing.' MCQs test the same idea conceptually, often handing you a velocity graph and asking on which interval speed is increasing. Read the graph for where v is below the x-axis AND decreasing (or above and increasing). Saying only 'acceleration is positive' earns nothing. The signs of both v and a must appear in your reasoning.

## speed increasing vs Velocity increasing

Velocity increasing just means a(t) > 0, since acceleration is the derivative of velocity. Speed increasing requires v(t) and a(t) to have the same sign. These can disagree. If v(t) = -5 and a(t) = +2, velocity is increasing (climbing toward zero) but speed is decreasing (the particle is slowing down). The exam loves this exact trap, so always ask which one the question wants.

## Key Takeaways

- Speed is the absolute value of velocity, |v(t)|, so it measures how fast, not which way.
- Speed is increasing when v(t) and a(t) have the same sign, and decreasing when they have opposite signs.
- A particle with negative velocity and negative acceleration is speeding up, even though its velocity is decreasing.
- An FRQ justification must state the signs of both velocity and acceleration; mentioning only one earns no credit.
- Speed increasing is not the same as velocity increasing, because velocity increasing only means acceleration is positive.
- At the moment a particle changes direction, v(t) = 0, so speed momentarily equals zero before increasing again.

## FAQs

### What does it mean for speed to be increasing in AP Calc?

Speed is increasing when velocity v(t) and acceleration a(t) have the same sign at that time. Both positive or both negative means the particle is speeding up, because the acceleration 'push' matches the direction of motion.

### Does positive acceleration mean speed is increasing?

No. Positive acceleration only guarantees speed is increasing if velocity is also positive. If v(t) = -5 and a(t) = +2, the particle is actually slowing down because the push opposes the motion.

### What's the difference between speed increasing and velocity increasing?

Velocity increasing means a(t) > 0, period. Speed increasing means v(t) and a(t) share the same sign. A particle with v = -3 and a = +1 has increasing velocity but decreasing speed.

### How do I justify that speed is increasing on an FRQ?

Compute or read off the signs of v(t) and a(t) at the given time, then write something like 'v(2) < 0 and a(2) < 0, so velocity and acceleration have the same sign and the speed is increasing.' Both signs must appear in your sentence for credit.

### Can speed increase while velocity is negative?

Yes. If velocity is negative and acceleration is also negative, the particle moves left faster and faster. Speed is |v(t)|, so velocity going from -2 to -7 means speed climbed from 2 to 7.

## Related Study Guides

- [4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration](/ap-calc/unit-4/straight-line-motion-connecting-position-velocity-acceleration/study-guide/2ZIESajDNiJ4ENTrnDT6)

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