2026 AP Calculus AB FRQ 3: Answer, Rubric Breakdown, and Point Maximization
Particle motion with a polynomial velocity function: every part answered, every rubric point mapped, and the exact sentences that earn the justification points.
- Part (a). a(4) = 2 units/sec². The particle is slowing down at t = 4, because v(4) = −3 and a(4) = 2 have opposite signs.
- Part (b). Total distance = 46/3 ≈ 15.333 units.
- Part (c). x(6) = −4 units.
- Part (d). The particle is farthest left at t = 5, where x(5) = −19/3 ≈ −6.333 units.
Why does this question matter on the AP Calculus AB exam?
Particle motion is the single most reliable free-response topic in AP Calculus AB. A velocity function plus an initial position is the setup the exam uses to test derivatives, definite integrals, the accumulation function, and justification language all in one question — which is why it is worth over-practicing even if you are short on study time.
| Fact | Detail |
|---|---|
| CED unit | Unit 4: Contextual Applications of Differentiation (with Unit 8 accumulation) |
| Unit exam weight | Units 4 and 8 together are roughly 27–35% of the exam |
| Points available | 9 of the 54 free-response points |
| Calculator | Calculator-active — store values, do not round mid-problem |
| How often this pattern appears | A motion or rate-in/rate-out question has appeared on essentially every released AP Calculus AB exam |
What does each part ask for?
| Part | What it asks | Skill tested | Points |
|---|---|---|---|
| (a) | Acceleration at t = 4, and whether the particle is speeding up or slowing down | Derivative of velocity; sign comparison | 2 |
| (b) | Total distance traveled on 0 ≤ t ≤ 6 | Integral of |v(t)|; splitting at sign changes | 3 |
| (c) | Position at t = 6 | Accumulation from an initial condition | 2 |
| (d) | When the particle is farthest left, with justification | Candidates test on a closed interval | 2 |
The question setup, in plain language
A particle moves along the x-axis for 0 ≤ t ≤ 6, where t is measured in seconds. Its velocity is given by v(t) = t² − 6t + 5, in units per second. At time t = 0 the particle is at position x(0) = 2.
Note that v(t) factors as (t − 1)(t − 5), so the velocity is zero at t = 1 and t = 5: positive on [0, 1), negative on (1, 5), and positive again on (5, 6]. Finding those sign changes first is what makes parts (b) and (d) fast.
Read the official released question and scoring guidelines on AP Central
What do you need to know to answer this?
If you cannot start this question, these are the two or three things to read first. Each one maps directly to a point on the rubric below.
- CED Topic 4.2 — Straight-line motion — Part (a) turns on the rule that a particle speeds up when velocity and acceleration share a sign, and slows down when they do not. Learn this before anything else on this page.
- CED Topic 8.2 — Connecting position, velocity, and acceleration — Part (b) asks for distance, not displacement. Distance is the integral of |v(t)|, which means splitting the interval at every zero of v where the sign changes.
- CED Topic 8.3 — Accumulation of change — Part (c) uses x(6) = x(0) + ∫₀⁶ v(t) dt. If you write the initial condition into the equation, you have already earned a point.
- CED Topic 5.4 — Extreme values — Part (d) needs the endpoints t = 0 and t = 6 checked alongside the interior critical time t = 5. Skipping the endpoints is the most common way to lose the justification point.
- Cramapple's FRQ strategy guide — The general template for writing justifications that readers can score.
Part (a): what is the acceleration at t = 4, and is the particle speeding up?
Short answer. a(4) = 2 units/sec². The particle is slowing down, because v(4) = −3 is negative while a(4) = 2 is positive.
Acceleration is the derivative of velocity: a(t) = v′(t) = 2t − 6. So a(4) = 2(4) − 6 = 2 units per second squared.
Evaluate velocity at the same time: v(4) = 16 − 24 + 5 = −3 units per second.
Velocity is negative and acceleration is positive, so they have opposite signs. Opposite signs means the speed is decreasing: the particle is slowing down at t = 4.
How the points are earned
- 1 point: a(4) = 2, with units of units/sec².
- 1 point: the correct conclusion (slowing down) supported by the signs of both v(4) and a(4). Stating the conclusion without both values earns nothing.
Part (b): what is the total distance traveled from t = 0 to t = 6?
Short answer. 46/3 ≈ 15.333 units.
Total distance is ∫₀⁶ |v(t)| dt. Because v(t) = (t − 1)(t − 5), the sign changes at t = 1 and t = 5, so split the integral there.
An antiderivative is F(t) = t³/3 − 3t² + 5t. Then F(0) = 0, F(1) = 7/3, F(5) = −25/3, and F(6) = −6.
Distance = |F(1) − F(0)| + |F(5) − F(1)| + |F(6) − F(5)| = 7/3 + 32/3 + 7/3 = 46/3 ≈ 15.333 units.
How the points are earned
- 1 point: setting up total distance as an integral of |v(t)| (or as a sum of integrals split at t = 1 and t = 5).
- 1 point: correct split points, identified from where v(t) changes sign.
- 1 point: the correct value, 46/3 ≈ 15.333, to at least three decimal places.
Part (c): where is the particle at t = 6?
Short answer. x(6) = −4 units.
Position is the initial position plus accumulated displacement: x(6) = x(0) + ∫₀⁶ v(t) dt.
∫₀⁶ v(t) dt = F(6) − F(0) = −6 − 0 = −6.
So x(6) = 2 + (−6) = −4 units. Note this is displacement, not distance — the −6 is much smaller in magnitude than the 46/3 from part (b) because the particle doubled back.
How the points are earned
- 1 point: writing x(6) = x(0) + ∫₀⁶ v(t) dt, with the initial condition included.
- 1 point: the correct answer, x(6) = −4.
Part (d): when is the particle farthest to the left, and how do you justify it?
Short answer. At t = 5, where x(5) = −19/3 ≈ −6.333 units — the smallest of the candidate values.
Position changes direction only where v(t) = 0, so the candidates are the interior critical times t = 1 and t = 5, plus the endpoints t = 0 and t = 6.
x(0) = 2, x(1) = 2 + 7/3 = 13/3 ≈ 4.333, x(5) = 2 − 25/3 = −19/3 ≈ −6.333, and x(6) = −4.
The smallest position value is −19/3 at t = 5, so the particle is farthest left at t = 5. The structure of the argument matters as much as the number: velocity changes from negative to positive at t = 5, so position has a minimum there, and the endpoint values confirm it is the absolute minimum.
How the points are earned
- 1 point: identifying t = 5 as the answer.
- 1 point: a complete candidates-test justification that compares x at t = 5 against both endpoints. Naming t = 5 without checking the endpoints does not earn the justification point.
How do you get the most points on this question?
Nine points are available, and most students who miss them do not miss the calculus — they miss the writing. The order you work in, the sentences you use, and what you write down when you get stuck are each worth more than one point on this question.
Score in this order if you are short on time
- Factor v(t) and write down where it is zero. That one line feeds parts (b) and (d) and takes fifteen seconds.
- Do part (a) next — it is two points of pure arithmetic and the fastest points on the page.
- Write the setup line for part (c), x(6) = x(0) + ∫₀⁶ v(t) dt, before evaluating anything. The setup is a point on its own.
- Write the setup line for part (b), ∫₀⁶ |v(t)| dt split at t = 1 and t = 5, even if the arithmetic is not done yet.
- Finish part (d) last; it reuses numbers you already computed in (b) and (c).
Sentence templates that earn the justification points
- Part (a) speeding-up/slowing-down point: “Because v(4) = −3 < 0 and a(4) = 2 > 0 have opposite signs, the particle is slowing down at t = 4.”
- Part (b) setup point: “Total distance = ∫₀¹ v(t) dt − ∫₁⁵ v(t) dt + ∫₅⁶ v(t) dt, splitting where v(t) changes sign at t = 1 and t = 5.”
- Part (c) accumulation point: “x(6) = x(0) + ∫₀⁶ v(t) dt = 2 + ∫₀⁶ (t² − 6t + 5) dt.”
- Part (d) justification point: “The only interior candidates are t = 1 and t = 5 because v(t) = 0 there. Comparing x(0) = 2, x(1) = 13/3, x(5) = −19/3, and x(6) = −4, the minimum is x(5) = −19/3, so the particle is farthest left at t = 5.”
What still scores when you cannot finish
- A correct integral setup with no evaluation still earns the setup point in parts (b) and (c). Never leave the setup blank because you ran out of time.
- If your value in part (b) is wrong but your split points and setup are right, you keep two of three points.
- In part (d), an incorrect x-value list still earns the justification point if the candidates test is applied correctly to your own numbers.
- In part (a), the acceleration point and the reasoning point are independent: a wrong a(4) with the correct sign-comparison sentence about your value can still earn the second point.
Where students lose points on this question
- Answering part (b) with displacement (−6) instead of total distance (46/3). Distance means the integral of the absolute value.
- Omitting units. Acceleration in part (a) needs units/sec², and readers do check.
- Rounding mid-problem on a calculator-active question. Store values; report at least three decimal places.
- Saying "the particle is slowing down because velocity is negative." Negative velocity alone means moving left, not slowing. You must compare it with the sign of acceleration.
- Naming t = 5 in part (d) without evaluating the endpoints. The candidates test is incomplete without them.
- Restating the prompt as a justification. "Because it is the farthest left" earns nothing.
The shortest full-credit response
- Part (a): "a(t) = 2t − 6, a(4) = 2 units/sec². v(4) = −3. Opposite signs, so the particle is slowing down."
- Part (b): "v = 0 at t = 1, 5. Distance = ∫₀¹v − ∫₁⁵v + ∫₅⁶v = 7/3 + 32/3 + 7/3 = 46/3 ≈ 15.333 units."
- Part (c): "x(6) = 2 + ∫₀⁶ v dt = 2 + (−6) = −4 units."
- Part (d): "Candidates t = 0, 1, 5, 6: x = 2, 13/3, −19/3, −4. Minimum is −19/3 at t = 5, so the particle is farthest left at t = 5."
- That is roughly eight written lines for nine points. Length is not what readers reward — completeness is.
Where else does this pattern show up?
Once you can write those eight lines, you can answer every motion FRQ the exam has used, plus the rate-in/rate-out questions that share the same structure.
- Rate-in/rate-out problems (water flowing into a tank) use the identical accumulation setup as part (c), with a different context.
- Any "is the object speeding up or slowing down" prompt is part (a) with different numbers.
- Any "at what time is the quantity greatest/least" prompt is part (d): interior critical points plus both endpoints.
- Graph-based motion questions replace the formula for v(t) with a graph, but the sign analysis and the split points work exactly the same way.
FAQ
What is the answer to 2026 AP Calculus AB FRQ 3?
Is FRQ 3 calculator-active?
How many points is this question worth?
Do I need units on the acceleration answer?
What is the difference between total distance and displacement here?
How do I justify that the particle is farthest left?
Practice this exact pattern against a real rubric.
Cramapple grades AP Calculus AB free response at the criterion level — setup, computation, units, and justification — and tells you the smallest repair that would have earned the next point.