Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
If you know how fast a particle moves and where it started, you know where it is. The rule is s(t)=s(0)+∫0tv(τ)dτ, and the same rule works between any two times: s(b)=s(a)+∫abv(τ)dτ. The integral adds up velocity, and velocity carries a sign, so the sum can go backward. That single fact decides most of the questions on this page. A particle moving left has negative velocity, and its position decreases even though it is moving. The released exams always supply the starting position, because without it the integral gives a change in position rather than a position. Read the question for the word “position” and for the initial value; those two pieces are the whole setup.
AP Calculus AB tests this skill through the representations and reasoning described below.
Figure 1 — Velocity with a sign, and the position it produces. The two positive lobes add and the negative lobe subtracts.
The statement “velocity is the derivative of position” runs backward as an integral:
s(b)−s(a)=∫abv(t)dt⟹s(b)=s(a)+∫abv(t)dt
This is the Fundamental Theorem read as a recipe. The left side is a change in position, which the released questions call displacement. The right side is the signed area under the velocity graph. Adding the starting position turns a displacement into a position.
Velocity is not speed. A particle moving left has v<0, and the integral counts that motion as negative.
So the position graph can rise, fall, and rise again while the velocity graph stays perfectly smooth. Every question on this page is asking you to turn one picture into the other.
When two particles move on the same line, the distance between them is the absolute difference of their positions:
distance at time t=∣sP(t)−sQ(t)∣
Each position needs its own starting value. A question that gives sP(0) but only the velocity of Q will usually hand you sQ at some later time instead, and you work backward from there.
When the velocity arrives as a picture, you never integrate symbolically. You count regions.
| Question | What to do on the graph |
|---|---|
| position at t=b | start at s(0), add areas above, subtract areas below |
| maximum position | the point where the accumulated signed area is largest |
| particle at rest | where the graph crosses the axis |
| position equal to the starting position | where the signed areas cancel to zero |
| displacement on [a,b] | net signed area between a and b |
The released questions state it in one of three ways. Sometimes it is a sentence: “at time t=0, the particle is at x=5.” Sometimes it is a table entry. Occasionally it is buried in a previous part, and the phrase “using the result from part (b)” is the signal.
If no starting value appears anywhere, the question is asking for a displacement. Answering with a position in that case is a units error as well as a conceptual one, because a displacement and a position are different quantities.
Borrowed idea: A position is an accumulation with a starting value attached. The integral of velocity gives a change, not a location. Add the anchor point first. Then let the signs of the velocity decide the direction of travel. A negative velocity lowers the position even though the particle moves. Reading a velocity graph is counting signed areas. Reading a velocity formula is integrating. Both routes end at the same number when the anchor is correct.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| “at t=0 the particle is at x=5” | an anchor point | s(0)=5 | add it to the integral |
| “find the position at t=3” | a location at a time | s(0)+∫03v | integrate, then add |
| “find the displacement on [0,3]” | a change, not a location | ∫03v | do not add s(0) |
| “the particle is at rest when …” | a zero of velocity | v(t)=0 | solve, then locate |
| “distance between the particles” | two positions | ∣sP−sQ∣ | build both positions first |
| a velocity graph with labelled areas | numbers, no formula | signed sum of areas | add and subtract regions |
① Locate the anchor point before touching the integral, and write it as an equation.
② Decide whether the question wants a position or a displacement. Only one of them includes the anchor.
③ Keep velocity and speed apart. Every sign in the answer descends from that distinction.
Core Archetype (Simplest Form): A particle moves with v(t)=3t2−8t+4 for 0≤t≤2.5, and s(0)=2. Find s(t) and s(2.5).
Standard Execution:
- Integrate. s(t)=2+∫0t(3τ2−8τ+4)dτ=2+t3−4t2+4t.
- Evaluate. s(2.5)=2+15.625−25+10=2.625.
- Check. The velocity is positive at t=0 and negative near t=1, so the position first rises and then falls back; a value near the starting value is plausible.
Exam Variation (Context / Parameter Shift): Two particles move along the same line. Particle P starts at x=2 and has vP(t)=4; particle Q starts at x=−3 and has vQ(t)=1+2t. Find the distance between them at t=5.
Solution: Particle P has constant velocity, so sP(5)=2+4⋅5=22. Particle Q satisfies sQ(5)=−3+∫05(1+2t)dt=−3+[t+t2]05=−3+30=27. The distance is ∣22−27∣=5. The order of subtraction does not matter, but the absolute value does.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
A particle moves along a line with v(t)=3t2−2 and s(0)=5. Its position at t=2 is
MCQ 2
A particle has v(t)=4−2t for 0≤t≤3 and starts at s(0)=1. Its position at t=3 is
MCQ 3
Particle P starts at x=2 with constant velocity 4. Particle Q starts at x=−3 with velocity vQ(t)=1+2t. The distance between the particles at t=5 is
FRQ 4
A particle moves along the x-axis with velocity v(t)=3t2−8t+4 for 0≤t≤2.5, measured in meters per second. At time t=0 the particle is at x=2.
(a) Find s(t), the position of the particle at time t, and the value of s(2.5).
(b) Find every time in (0,2.5) at which the particle is at rest, and give its position at each of those times.
(c) Explain why the particle is never farther to the right than it is at t=32 during the interval 0≤t≤2.5.
| Action | Failure if skipped | |
|---|---|---|
| 1 | locate the starting position and write s(0)=… | the answer is a displacement |
| 2 | write the accumulation ∫abv | the sign structure is lost |
| 3 | evaluate the integral | no amount of algebra fixes a missing area |
| 4 | add the anchor | the location is off by a constant |
| 5 | state position or displacement, matching the noun | the units and the meaning both break |
| 6 | for a graph, count signed areas | adding magnitudes inflates the position |
Strengthen this topic with AP-style practice, or continue to next topic.
Reviewed by ACEpath AP Math Editorial Team
Last Updated: September 2026
30-second summary
If you know how fast a particle moves and where it started, you know where it is. The rule is s(t)=s(0)+∫0tv(τ)dτ, and the same rule works between any two times: s(b)=s(a)+∫abv(τ)dτ. The integral adds up velocity, and velocity carries a sign, so the sum can go backward. That single fact decides most of the questions on this page. A particle moving left has negative velocity, and its position decreases even though it is moving. The released exams always supply the starting position, because without it the integral gives a change in position rather than a position. Read the question for the word “position” and for the initial value; those two pieces are the whole setup.
AP Calculus AB tests this skill through the representations and reasoning described below.
Figure 1 — Velocity with a sign, and the position it produces. The two positive lobes add and the negative lobe subtracts.
The statement “velocity is the derivative of position” runs backward as an integral:
s(b)−s(a)=∫abv(t)dt⟹s(b)=s(a)+∫abv(t)dt
This is the Fundamental Theorem read as a recipe. The left side is a change in position, which the released questions call displacement. The right side is the signed area under the velocity graph. Adding the starting position turns a displacement into a position.
Velocity is not speed. A particle moving left has v<0, and the integral counts that motion as negative.
So the position graph can rise, fall, and rise again while the velocity graph stays perfectly smooth. Every question on this page is asking you to turn one picture into the other.
When two particles move on the same line, the distance between them is the absolute difference of their positions:
distance at time t=∣sP(t)−sQ(t)∣
Each position needs its own starting value. A question that gives sP(0) but only the velocity of Q will usually hand you sQ at some later time instead, and you work backward from there.
When the velocity arrives as a picture, you never integrate symbolically. You count regions.
| Question | What to do on the graph |
|---|---|
| position at t=b | start at s(0), add areas above, subtract areas below |
| maximum position | the point where the accumulated signed area is largest |
| particle at rest | where the graph crosses the axis |
| position equal to the starting position | where the signed areas cancel to zero |
| displacement on [a,b] | net signed area between a and b |
The released questions state it in one of three ways. Sometimes it is a sentence: “at time t=0, the particle is at x=5.” Sometimes it is a table entry. Occasionally it is buried in a previous part, and the phrase “using the result from part (b)” is the signal.
If no starting value appears anywhere, the question is asking for a displacement. Answering with a position in that case is a units error as well as a conceptual one, because a displacement and a position are different quantities.
Borrowed idea: A position is an accumulation with a starting value attached. The integral of velocity gives a change, not a location. Add the anchor point first. Then let the signs of the velocity decide the direction of travel. A negative velocity lowers the position even though the particle moves. Reading a velocity graph is counting signed areas. Reading a velocity formula is integrating. Both routes end at the same number when the anchor is correct.
| Exam disguise | What you see on the page | What your brain must translate | Required execution |
|---|---|---|---|
| “at t=0 the particle is at x=5” | an anchor point | s(0)=5 | add it to the integral |
| “find the position at t=3” | a location at a time | s(0)+∫03v | integrate, then add |
| “find the displacement on [0,3]” | a change, not a location | ∫03v | do not add s(0) |
| “the particle is at rest when …” | a zero of velocity | v(t)=0 | solve, then locate |
| “distance between the particles” | two positions | ∣sP−sQ∣ | build both positions first |
| a velocity graph with labelled areas | numbers, no formula | signed sum of areas | add and subtract regions |
① Locate the anchor point before touching the integral, and write it as an equation.
② Decide whether the question wants a position or a displacement. Only one of them includes the anchor.
③ Keep velocity and speed apart. Every sign in the answer descends from that distinction.
Core Archetype (Simplest Form): A particle moves with v(t)=3t2−8t+4 for 0≤t≤2.5, and s(0)=2. Find s(t) and s(2.5).
Standard Execution:
- Integrate. s(t)=2+∫0t(3τ2−8τ+4)dτ=2+t3−4t2+4t.
- Evaluate. s(2.5)=2+15.625−25+10=2.625.
- Check. The velocity is positive at t=0 and negative near t=1, so the position first rises and then falls back; a value near the starting value is plausible.
Exam Variation (Context / Parameter Shift): Two particles move along the same line. Particle P starts at x=2 and has vP(t)=4; particle Q starts at x=−3 and has vQ(t)=1+2t. Find the distance between them at t=5.
Solution: Particle P has constant velocity, so sP(5)=2+4⋅5=22. Particle Q satisfies sQ(5)=−3+∫05(1+2t)dt=−3+[t+t2]05=−3+30=27. The distance is ∣22−27∣=5. The order of subtraction does not matter, but the absolute value does.
Common Traps & Failure Signals:
6D Cognitive Error Diagnosis (Where Your Thinking Broke):
MCQ 1
A particle moves along a line with v(t)=3t2−2 and s(0)=5. Its position at t=2 is
MCQ 2
A particle has v(t)=4−2t for 0≤t≤3 and starts at s(0)=1. Its position at t=3 is
MCQ 3
Particle P starts at x=2 with constant velocity 4. Particle Q starts at x=−3 with velocity vQ(t)=1+2t. The distance between the particles at t=5 is
FRQ 4
A particle moves along the x-axis with velocity v(t)=3t2−8t+4 for 0≤t≤2.5, measured in meters per second. At time t=0 the particle is at x=2.
(a) Find s(t), the position of the particle at time t, and the value of s(2.5).
(b) Find every time in (0,2.5) at which the particle is at rest, and give its position at each of those times.
(c) Explain why the particle is never farther to the right than it is at t=32 during the interval 0≤t≤2.5.
| Action | Failure if skipped | |
|---|---|---|
| 1 | locate the starting position and write s(0)=… | the answer is a displacement |
| 2 | write the accumulation ∫abv | the sign structure is lost |
| 3 | evaluate the integral | no amount of algebra fixes a missing area |
| 4 | add the anchor | the location is off by a constant |
| 5 | state position or displacement, matching the noun | the units and the meaning both break |
| 6 | for a graph, count signed areas | adding magnitudes inflates the position |
Strengthen this topic with AP-style practice, or continue to next topic.