In AP Calculus, a particle is speeding up when its velocity v(t) and acceleration a(t) have the same sign (both positive or both negative), which means its speed |v(t)| is increasing. This is the standard test in straight-line motion problems (Topic 4.2).
Speeding up means the object's speed is increasing, where speed is the absolute value of velocity, |v(t)|. Here's the part that trips people up. Speeding up does NOT mean velocity is increasing, and it does NOT mean acceleration is positive. A particle moving in the negative direction (v < 0) with negative acceleration (a < 0) is speeding up, because it's moving backward faster and faster.
The rule you actually use: a particle is speeding up when v(t) and a(t) have the same sign, and slowing down when they have opposite signs. Think of acceleration as a push. If the push points the same way the particle is already moving, the particle goes faster. If the push fights the motion, the particle slows down. Since v(t) = x'(t) and a(t) = v'(t) = x''(t), you find these signs by taking derivatives of the position function and doing a sign analysis on each.
This lives in Topic 4.2 (Straight-Line Motion) in Unit 4: Contextual Applications of Differentiation, under learning objective 4.2.A: calculate rates of change in applied contexts. The essential knowledge here is that the derivative solves rectilinear motion problems involving position, speed, velocity, and acceleration. "Is the particle speeding up or slowing down?" is the classic way the exam checks whether you understand the difference between velocity (signed) and speed (unsigned). It also tests whether you can chain derivatives correctly, since you need both v(t) and a(t) from a single position or velocity function. Particle motion shows up on the AP exam almost every year in some form, so this sign-comparison move is one of the highest-value skills in Unit 4.
Keep studying AP Calculus Unit 4
Visual cheatsheet
view galleryAcceleration Function, x''(t) (Unit 4)
You can't answer a speeding-up question without acceleration. Differentiate v(t) to get a(t), then compare its sign to the sign of v(t). Acceleration alone never tells you the answer; it's the combination that matters.
Velocity Function, v(t) (Unit 4)
Velocity carries direction in its sign, and that sign is half of the speeding-up test. Most exam problems hand you v(t) directly and expect you to find where it's positive, negative, and zero before doing anything else.
Absolute Value and Speed (Unit 4)
Speed is |v(t)|, so 'speeding up' literally means |v(t)| is increasing. This is why a particle with v = -10 ft/s and a = -3 ft/s² is speeding up. Its velocity is dropping, but its speed is climbing from 10 toward 13.
Change Direction (Unit 4)
Direction changes happen where v(t) crosses zero, and those same zeros split the timeline into the intervals you analyze for speeding up versus slowing down. One sign chart for v(t) and a(t) answers both questions at once.
Multiple-choice questions hand you a velocity function and ask for the interval where the particle is speeding up or slowing down. For example, given v(t) = 4 - t, you find a(t) = -1, so the particle speeds up only when v is also negative, which is t > 4. Given v(t) = ln(t), a(t) = 1/t is always positive for t > 0, so the particle slows down on (0, 1) where v is negative and speeds up after t = 1.
On FRQs, this appears as a part of a larger particle-motion problem. The 2023 AB exam (FRQ 2, the swimmer with v(t) = 2.38e^(-0.02t)sin(πt/56)) and the 2026 AB exam (FRQ 5, the remote-controlled toy car) both center on velocity models for back-and-forth straight-line motion. The standard ask is "Is the speed increasing or decreasing at t = c? Justify your answer." Full credit requires you to state the sign of v(c) AND the sign of a(c), then conclude they match (speeding up) or differ (slowing down). Writing "acceleration is positive, so it's speeding up" loses the justification point.
Positive acceleration means velocity is increasing, not that the object is speeding up. If a car is rolling backward (v < 0) and you hit the gas forward (a > 0), the positive acceleration is actually slowing it down. Speeding up is about |v(t)| growing, which happens exactly when v and a point the same way. Match the signs, don't just check acceleration.
A particle is speeding up when v(t) and a(t) have the same sign, and slowing down when they have opposite signs.
Speeding up means speed |v(t)| is increasing, which is not the same as velocity increasing or acceleration being positive.
A particle moving in the negative direction with negative acceleration is speeding up, because it's moving backward faster.
To answer a speeding-up question, build a sign chart, find the zeros of v(t) and a(t), and check the sign of each function on every interval.
On an FRQ, justifying 'speeding up' requires stating the signs of both v and a at that time; citing only one of them won't earn the point.
This is a Topic 4.2 skill (LO 4.2.A) where the derivative connects position, velocity, acceleration, and speed in rectilinear motion.
Speeding up means the particle's speed, |v(t)|, is increasing. The test you use on the exam is that velocity v(t) and acceleration a(t) have the same sign, both positive or both negative.
No. Positive acceleration only means velocity is increasing. If velocity is negative while acceleration is positive (like v(t) = ln(t) on the interval 0 < t < 1), the particle is actually slowing down. Speeding up requires v and a to share the same sign.
Deceleration (slowing down) happens when v(t) and a(t) have opposite signs, so the acceleration fights the direction of motion and speed |v(t)| decreases. Speeding up is the mirror case where the signs match. Neither one is determined by the sign of acceleration alone.
Compute v(t) and a(t) at the given time, state the sign of each, and compare. Same sign means speeding up; opposite signs mean slowing down. Released FRQs like 2023 AB Q2 (the swimmer) expect you to cite both signs in your justification to earn the point.
Yes. If v(t) < 0 and a(t) < 0, the particle moves in the negative direction faster and faster, so it's speeding up. For example, with v(t) = 4 - t, the particle speeds up for t > 4 because both v and a (which equals -1) are negative there.