---
title: "Net Displacement — AP Calc Definition & Exam Guide"
description: "Net displacement is the change in position, found by ∫v(t)dt with no absolute value. Learn how it differs from total distance on AP Calc Unit 8 motion FRQs."
canonical: "https://fiveable.me/ap-calc/key-terms/net-displacement"
type: "key-term"
subject: "AP Calculus AB/BC"
unit: "Unit 8"
---

# Net Displacement — AP Calc Definition & Exam Guide

## Definition

Net displacement is a particle's change in position over a time interval, computed as the definite integral of velocity, ∫v(t)dt, without absolute value. It can be negative or zero, and it only equals total distance traveled when the particle never changes direction (CHA-4.C.1, AP Calc Topic 8.2).

## What It Is

Net displacement answers one question. Where did the particle end up compared to where it started? In AP Calculus, you find it by integrating [velocity](/ap-calc/unit-4/straight-line-motion-connecting-position-velocity-acceleration/study-guide/2ZIESajDNiJ4ENTrnDT6 "fv-autolink") over the interval: [displacement](/ap-calc/key-terms/displacement "fv-autolink") = ∫ from a to b of v(t) dt. That's it. No absolute value, no splitting the interval. The CED states this directly in CHA-4.C.1: the definite integral of velocity gives displacement, while the definite integral of speed, |v(t)|, gives total distance traveled.

The reason these two can differ is direction. When v(t) is negative, the particle moves backward, and that backward motion subtracts from the integral. So a particle can walk 4 meters right, walk 4 meters left, and have a net displacement of zero even though it traveled 8 meters total. Think of displacement as 'as the crow flies' along the line, and total distance as the odometer reading. If v(t) never changes sign on the interval, the two values match (up to sign). The moment velocity crosses zero, they split apart.

## Why It Matters

Net displacement lives in Topic 8.2 (Connecting Position, Velocity, and Acceleration Using Integrals) in [Unit 8](/ap-calc/unit-8 "fv-autolink"): Applications of Integration. It directly supports learning objective [AP Calc](/ap-calc "fv-autolink") 8.2.A, determining values for positions and rates of change using definite integrals in rectilinear motion problems. This is one of the most reliable contexts on the whole exam. A particle-motion FRQ shows up almost every year, and it nearly always asks you to distinguish displacement from total distance, or to use displacement to find a new position via s(b) = s(a) + ∫v(t)dt. It's also the cleanest real-world example of the Fundamental Theorem of Calculus in action: integrating a rate of change (velocity) recovers the net change in the quantity (position).

## Connections

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

Velocity is the [function](/ap-calc/unit-1/defining-continuity-at-point/study-guide/JbsR9iQfAzCznNOCG6JK "fv-autolink") you actually integrate. The sign of v(t) tells you direction, which is exactly why displacement can be negative or zero. Finding where v(t) = 0 tells you where displacement and total distance start to disagree.

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

Displacement is the change in position, s(b) − s(a). A classic exam move gives you an initial position and asks for position later, and the formula s(b) = s(a) + ∫ from a to b of v(t) dt is just 'start plus displacement.'

### [Acceleration Function (Unit 8)](/ap-calc/key-terms/acceleration-function)

Integrating acceleration gives the change in velocity, the same way integrating velocity gives displacement. It's the identical idea one level up the [derivative](/ap-calc/unit-10-infinite-sequences-and-series-bc-only/finding-taylor-polynomial-approximations-functions/study-guide/LszguYzKz0M6GdqTRSr6 "fv-autolink") chain, so if you understand displacement, you already understand ∫a(t)dt.

### Fundamental Theorem of Calculus (Unit 6)

Net displacement is the FTC wearing a physics costume. The net change theorem says integrating a rate gives net change in the original quantity, and v(t) is the [rate of change](/ap-calc/key-terms/rate-of-change "fv-autolink") of position. Particle motion is how the exam tests whether you really get that.

## On the AP Exam

Particle motion is a near-guaranteed FRQ context. The 2024 exam (FRQ 2) gave v(t) = ln(t² − 4t + 5) − 0.2t and asked questions built on this exact integral setup, usually calculator-active since the velocity is messy. You need to do three things: (1) compute ∫v(t)dt for displacement and ∫|v(t)|dt for total distance, and know which is which; (2) interpret the value in context with units, like explaining that ∫₀⁴ v(t)dt = 0 means the particle ended exactly where it started even though it moved; (3) use displacement to find position, s(b) = s(a) + ∫v(t)dt. Multiple-choice stems love the interpretation angle. A negative answer like ∫₀³ v(t)dt = −13.5 for a drone means its final height is 13.5 meters below its starting height, not that it traveled 13.5 meters. Watch the units and the word 'net' in the prompt, since graders award interpretation points for saying 'change in position' rather than 'distance.'

## net displacement vs Total distance traveled

Displacement is ∫v(t)dt and total distance is ∫|v(t)|dt. The absolute value is the whole difference. Displacement lets forward and backward motion cancel, so it can be zero or negative. Total distance counts every meter regardless of direction, so it's always ≥ |displacement|. With v(t) = 4t − 8 on [0, 4], displacement is 0 but total distance is 16, because the particle goes backward for 2 seconds and then retraces its path. If an exam question says 'how far did it travel,' that's total distance. If it says 'change in position' or 'displacement,' skip the absolute value.

## Key Takeaways

- Net displacement equals the definite integral of velocity, ∫v(t)dt, computed with no absolute value, and it represents the change in position from start to end (CHA-4.C.1).
- Displacement can be negative or zero; a negative value means the particle ended up behind (or below) where it started, not that it traveled a negative distance.
- Displacement equals total distance traveled only when v(t) never changes sign on the interval, so finding where v(t) = 0 tells you when the two values split.
- To find a particle's position at time b, add displacement to the starting position: s(b) = s(a) + ∫ from a to b of v(t) dt.
- On FRQs, interpretation points require saying 'change in position' with correct units; calling ∫v(t)dt a 'distance' costs you the point.
- Total distance is ∫|v(t)|dt, which is always greater than or equal to the absolute value of displacement.

## FAQs

### What is net displacement in AP Calculus?

Net displacement is a particle's change in position over a time interval, calculated as ∫v(t)dt without [absolute value](/ap-calc/key-terms/absolute-value "fv-autolink"). Per the CED (CHA-4.C.1), the integral of velocity gives displacement, while the integral of speed gives total distance traveled.

### Is net displacement the same as total distance traveled?

No. Displacement is ∫v(t)dt and total distance is ∫|v(t)|dt. With v(t) = 4t − 8 on [0, 4], displacement is 0 meters but total distance is 16 meters, because the particle moves backward and then retraces its path. They only match when velocity never changes sign.

### Can net displacement be negative or zero?

Yes to both. A negative displacement like ∫₀³ v(t)dt = −13.5 means the object's final position is 13.5 meters in the negative direction from its start (a drone 13.5 m lower, for example). A displacement of zero means the particle ended exactly where it began, even if it moved the whole time.

### How do I find a particle's position from its velocity on the AP exam?

Use s(b) = s(a) + ∫ from a to b of v(t) dt, which is just starting position plus displacement. This setup shows up on nearly every particle-motion FRQ, including the 2024 FRQ 2 with v(t) = ln(t² − 4t + 5) − 0.2t.

### When do I use the absolute value of velocity in the integral?

Only when the question asks for total distance traveled or how far the object went. Use ∫|v(t)|dt for distance and plain ∫v(t)dt for displacement or change in position. Reading which one the prompt wants is half the battle on these problems.

## Related Study Guides

- [8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals](/ap-calc/unit-8/connecting-position-velocity-acceleration-functions-using-integrals/study-guide/k9tY28YXs7YDVu1uqFuw)

## Structured Data

```json
{"@context":"https://schema.org","@graph":[{"@type":"LearningResource","@id":"https://fiveable.me/ap-calc/key-terms/net-displacement#resource","name":"Net Displacement — AP Calc Definition & Exam Guide","url":"https://fiveable.me/ap-calc/key-terms/net-displacement","learningResourceType":"Concept explainer","educationalLevel":"AP® / High School","about":{"@id":"https://fiveable.me/ap-calc/key-terms/net-displacement#term"},"audience":{"@type":"EducationalAudience","educationalRole":"student"},"dateModified":"2026-06-11T05:27:22.730Z","isPartOf":{"@type":"Collection","name":"AP Calculus AB/BC Key Terms","url":"https://fiveable.me/ap-calc/key-terms"},"publisher":{"@type":"Organization","name":"Fiveable","url":"https://fiveable.me"}},{"@type":"DefinedTerm","@id":"https://fiveable.me/ap-calc/key-terms/net-displacement#term","name":"net displacement","description":"Net displacement is a particle's change in position over a time interval, computed as the definite integral of velocity, ∫v(t)dt, without absolute value. It can be negative or zero, and it only equals total distance traveled when the particle never changes direction (CHA-4.C.1, AP Calc Topic 8.2).","url":"https://fiveable.me/ap-calc/key-terms/net-displacement","inDefinedTermSet":{"@type":"DefinedTermSet","name":"AP Calculus AB/BC Key Terms","url":"https://fiveable.me/ap-calc/key-terms"}},{"@type":"FAQPage","mainEntity":[{"@type":"Question","name":"What is net displacement in AP Calculus?","acceptedAnswer":{"@type":"Answer","text":"Net displacement is a particle's change in position over a time interval, calculated as ∫v(t)dt without [absolute value](/ap-calc/key-terms/absolute-value \"fv-autolink\"). Per the CED (CHA-4.C.1), the integral of velocity gives displacement, while the integral of speed gives total distance traveled."}},{"@type":"Question","name":"Is net displacement the same as total distance traveled?","acceptedAnswer":{"@type":"Answer","text":"No. Displacement is ∫v(t)dt and total distance is ∫|v(t)|dt. With v(t) = 4t − 8 on [0, 4], displacement is 0 meters but total distance is 16 meters, because the particle moves backward and then retraces its path. They only match when velocity never changes sign."}},{"@type":"Question","name":"Can net displacement be negative or zero?","acceptedAnswer":{"@type":"Answer","text":"Yes to both. A negative displacement like ∫₀³ v(t)dt = −13.5 means the object's final position is 13.5 meters in the negative direction from its start (a drone 13.5 m lower, for example). A displacement of zero means the particle ended exactly where it began, even if it moved the whole time."}},{"@type":"Question","name":"How do I find a particle's position from its velocity on the AP exam?","acceptedAnswer":{"@type":"Answer","text":"Use s(b) = s(a) + ∫ from a to b of v(t) dt, which is just starting position plus displacement. This setup shows up on nearly every particle-motion FRQ, including the 2024 FRQ 2 with v(t) = ln(t² − 4t + 5) − 0.2t."}},{"@type":"Question","name":"When do I use the absolute value of velocity in the integral?","acceptedAnswer":{"@type":"Answer","text":"Only when the question asks for total distance traveled or how far the object went. Use ∫|v(t)|dt for distance and plain ∫v(t)dt for displacement or change in position. Reading which one the prompt wants is half the battle on these problems."}}]},{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"AP Calculus AB/BC","item":"https://fiveable.me/ap-calc"},{"@type":"ListItem","position":2,"name":"Key Terms","item":"https://fiveable.me/ap-calc/key-terms"},{"@type":"ListItem","position":3,"name":"Unit 8","item":"https://fiveable.me/ap-calc/unit-8"},{"@type":"ListItem","position":4,"name":"net displacement"}]}]}
```