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).
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 over the interval: displacement = ∫ 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.
Net displacement lives in Topic 8.2 (Connecting Position, Velocity, and Acceleration Using Integrals) in Unit 8: Applications of Integration. It directly supports learning objective AP Calc 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).
Keep studying AP® Calculus Unit 8
Visual cheatsheet
view galleryVelocity Function (Unit 8)
Velocity is the function 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)
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)
Integrating acceleration gives the change in velocity, the same way integrating velocity gives displacement. It's the identical idea one level up the derivative 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 of position. Particle motion is how the exam tests whether you really get that.
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.'
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.
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.
Net displacement is a particle's change in position over a time interval, calculated as ∫v(t)dt without absolute value. Per the CED (CHA-4.C.1), the integral of velocity gives displacement, while the integral of speed gives 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.
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.
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.
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.
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