What is the difference between velocity and speed?

Velocity has a sign that shows direction. Speed is the absolute value of velocity, so it's never negative.

How do I know if a particle is speeding up?

Compare the signs of velocity and acceleration at that moment. The same sign means speeding up, and opposite signs mean slowing down. Both need to be nonzero.

How do I find total distance traveled?

Integrate the absolute value of velocity over the interval. By hand, you can find where velocity changes sign and add the size of each leg of the trip.

Does a velocity of zero mean the particle changes direction?

Not by itself. The particle changes direction only if velocity changes sign at that time.

Particle motion for AP Calculus AB

Updated November 9, 2026

A particle moving along a line is one of the most common settings on the AP Calculus AB exam. The calculus is familiar, and most lost points come from mixing up related ideas: velocity and speed, or displacement and total distance.

This guide defines each quantity, then works through the questions you're likely to see from a formula, from a calculator, and from a graph of velocity.

The three functions

Position $s(t)$ tells you where the particle is.
Velocity $v(t) = s'(t)$ tells you how fast position is changing and in which direction.
Acceleration $a(t) = v'(t) = s''(t)$ tells you how velocity is changing.
Speed is $|v(t)|$ . It's never negative, and it has no direction.

In this guide, moving right means position is increasing. If a question uses up and down, or a different convention, follow the question.

Reading the motion

At rest: $v(t) = 0$ at that moment.
Moving right or left: $v(t) > 0$ means right, and $v(t) < 0$ means left.
Changes direction: $v$ changes sign at that time. It isn't enough for $v$ to equal zero. For example, $v(t) = (t - 3)^2$ is zero at $t = 3$ , but the particle keeps moving right on both sides.
Speeding up: $v$ and $a$ have the same sign.
Slowing down: $v$ and $a$ have opposite signs.

A full example from a formula

No calculator

A particle moves along the

x

-axis with position

s(t) = 2t^3 - 15t^2 + 24t + 3

on the interval

[0, 5]

. (a) When does the particle change direction? (b) Is the particle speeding up or slowing down at

t = 2

? At

t = 3

? (c) Find the displacement and the total distance on

[0, 5]

(a) $v(t) = 6t^2 - 30t + 24 = 6(t - 1)(t - 4)$ . Velocity is positive on $(0, 1)$ , negative on $(1, 4)$ , and positive on $(4, 5)$ . It changes sign at $t = 1$ and $t = 4$ , so the particle changes direction at both times.

(b) $a(t) = 12t - 30$ . At $t = 2$ , $v = -12$ and $a = -6$ . They have the same sign, so the particle is speeding up. At $t = 3$ , $v = -12$ and $a = 6$ . They have opposite signs, so it's slowing down.

s(t) = 2t^3 - 15t^2 + 24t + 3

For total distance, add the size of each leg of the trip. From $t = 0$ to $1$ , it moves $|14 - 3| = 11$ . From $t = 1$ to $4$ , it moves $|-13 - 14| = 27$ . From $t = 4$ to $5$ , it moves $|-2 - (-13)| = 11$ . The total is $49$ .

Answer: (a) $t = 1$ and $t = 4$ . (b) Speeding up at $t = 2$ , slowing down at $t = 3$ . (c) Displacement $-5$ , total distance $49$ .

Displacement vs total distance

Displacement from $t = a$ to $t = b$ is $\int_a^b v(t)\,dt$ . Motion left counts as negative, so trips can cancel.
Total distance from $t = a$ to $t = b$ is $\int_a^b |v(t)|\,dt$ . Nothing cancels.
Final position is the starting position plus the displacement: $s(b) = s(a) + \int_a^b v(t)\,dt$ .
Average velocity is displacement divided by elapsed time. Average acceleration is the change in velocity divided by elapsed time.

The two integrals match only when $v$ doesn't go below zero on the interval. These are the same accumulation ideas used in the fundamental theorem guide.

With a calculator

A particle moves along a line with velocity

v(t) = 5\cos(0.9t) + 0.4t - 1

on the interval

[0, 6]

. At

t = 0

its position is

3

. (a) Find its position at

t = 6

. (b) Find the total distance traveled on

[0, 6]

. (c) Is the particle speeding up or slowing down at

t = 4

v(t) = 5\cos(0.9t) + 0.4t - 1

(a) $s(6) = 3 + \int_0^6 v(t)\,dt$ . The integral is about $-3.09314$ , so the position is about $-0.093$ .

(b) Type $\int_0^6 |v(t)|\,dt$ directly into Desmos. It returns about $16.630$ . You don't have to split the interval on a calculator, but the two zeros in the graph show why the distance is larger than the size of the displacement.

Answer: (a) $-0.093$ . (b) $16.630$ . (c) Slowing down, because $v(4)$ and $a(4)$ have opposite signs.

For the steps in the calculator, see how to use Desmos on the AP Calculus AB exam.

Reading a velocity graph

On a graph of velocity, area above the axis is motion to the right, and area below the axis is motion to the left. The slope of the graph is the acceleration.

v(t)

The velocity of a particle for the example below, made of straight segments.

Areas on a velocity graph

The graph above shows the velocity of a particle on the interval

[0, 8]

. At

t = 0

its position is

-2

. (a) Find the position at

t = 8

. (b) Find the total distance traveled. (c) Is the particle speeding up or slowing down at

t = 5

The areas are $6$ on $[0, 2]$ , $4$ on $[2, 4]$ , $3$ below the axis on $[4, 6]$ , and $6$ below the axis on $[6, 8]$ .

(a) The displacement is $6 + 4 - 3 - 6 = 1$ , so the position is $-2 + 1 = -1$ .

(b) The total distance is $6 + 4 + 3 + 6 = 19$ .

(c) At $t = 5$ , $v(5) = -1.5$ , and the slope of the segment gives $a(5) = -1.5$ . They have the same sign, so the particle is speeding up.

Answer: (a) $-1$ . (b) $19$ . (c) Speeding up.

Keep going

Practice on exam-style questions

Intersect Prep has 50 multiple-choice questions for Unit 4, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $44 unlocks the course for 6 months.

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Questions

What is the difference between velocity and speed?: Velocity has a sign that shows direction. Speed is the absolute value of velocity, so it's never negative.
How do I know if a particle is speeding up?: Compare the signs of velocity and acceleration at that moment. The same sign means speeding up, and opposite signs mean slowing down. Both need to be nonzero.
How do I find total distance traveled?: Integrate the absolute value of velocity over the interval. By hand, you can find where velocity changes sign and add the size of each leg of the trip.
Does a velocity of zero mean the particle changes direction?: Not by itself. The particle changes direction only if velocity changes sign at that time.