Unit 8 · Topic 8.2 · about 30 minutes · 7 parts · 13 scored checks, 3 of them multiple choice

AP Calculus AB 8.2: Connecting Position, Velocity, and Acceleration of Functions Using Integrals

Q: Is displacement the same as total distance traveled?

Only when the particle never turns around. Displacement subtracts the motion back from the motion forward, while total distance adds them. A runner who does one lap of a track has a displacement of 0 and a total distance of one lap.

Q: Why is the integral of velocity a change in position?

Velocity is the derivative of position, so position is an antiderivative of velocity. The Fundamental Theorem of Calculus then says ab v(t)\,dt = x(b) - x(a).

Q: How do I find total distance without a calculator?

Find the times where v(t) changes sign, split the interval there, and integrate v(t) on each piece. Make every piece positive and add. On a graph, that is the total area between the velocity graph and the t-axis.

Q: Can displacement be negative?

Yes. A negative displacement means the particle ends up on the negative side of where it started. Total distance, an integral of speed, can never be negative.

Q: What units does the integral of velocity have?

The units of velocity times the units of time. Meters per second times seconds gives meters, and for acceleration, meters per second per second times seconds gives meters per second.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 8: 10 to 15% of the multiple-choice section

What you'll be able to do

Use definite integrals to find a moving particle's displacement, total distance traveled, position and velocity.

Key vocabulary

Term	What it means	The exam trap
Rectilinear motion	Motion along a straight line, where the position at each time is a single number such as $x (t)$ .	Read which direction the problem calls positive before you interpret the sign of any answer.
Displacement	The change in a particle's position over an interval of time, given by the definite integral of its velocity over that interval.	Displacement and position are different quantities, so check which one a question asks for.
Speed	The absolute value of the velocity, $\| v (t) \|$ .	Velocity carries a sign and speed does not, so read which of the two a question names.
Total distance traveled	The definite integral of speed over an interval of time, $\int_{a}^{b} \| v (t) \| d t$ .	Read whether a question asks for total distance or for displacement before you set up the integral.

Graph of the velocity v(t) = 6 - 2t meters per second for t from 0 to 5 seconds. The line crosses the t-axis at t = 3. The shaded triangle above the axis from t = 0 to t = 3 has area 9, and the shaded triangle below the axis from t = 3 to t = 5 has area 4.

Part 1 Think first Where did the robot end up?

Guess before you learn. A wrong guess still makes the answer stick.

Predict first

A delivery robot rolls along a straight hallway. For the first 40 seconds its velocity is 2 meters per second, toward the lobby. For the next 20 seconds its velocity is $- 3$ meters per second, so it rolls back the other way. After those 60 seconds, how far is the robot from where it started?

In short. For a particle moving along a line, the definite integral of its velocity over a time interval is its displacement, and the definite integral of its speed is the total distance it travels.

A later position is a known position plus the integral of velocity: $x (b) = x (a) + \int_{a}^{b} v (t) d t$ .
Total distance is $\int_{a}^{b} | v (t) | d t$ ; without a calculator, split the interval wherever $v (t)$ changes sign.
Velocity comes from acceleration the same way: $v (b) = v (a) + \int_{a}^{b} a (t) d t$ .

Part 2 Learn it Position, distance and velocity from integrals

2a. The integral of velocity is displacement

A particle in rectilinear motion moves along a line, so its position at time $t$ is a single number, $x (t)$ . Its velocity is $v (t) = x^{'} (t)$ : positive while it moves in the positive direction, negative while it moves back.

Because $x$ is an antiderivative of $v$ , the Fundamental Theorem of Calculus gives

\int_{a}^{b} v (t) d t = x (b) - x (a) .

That difference is the particle's displacement from $t = a$ to $t = b$ , its change in position. Motion in the positive direction adds to it and motion back subtracts from it. To say where the particle is, add the displacement to a position you already know:

x (b) = x (a) + \int_{a}^{b} v (t) d t .

The graph at the top shows $v (t) = 6 - 2 t$ meters per second. From $t = 0$ to $t = 3$ the velocity is positive and the area above the axis is 9. From $t = 3$ to $t = 5$ it is negative and the area below the axis is 4. So $\int_{0}^{5} v (t) d t = 9 - 4 = 5$ meters. If the particle is at $x = - 2$ at $t = 0$ , then at $t = 5$ it is at $- 2 + 5 = 3$ .

The units multiply: meters per second times seconds gives meters.

Quick check · Calculate

No calculator. A particle moves along the x-axis with velocity $v (t) = 3 t^{2} - 12$ for $t \geq 0$ . At time $t = 0$ the particle is at $x = 5$ . What is its position at time $t = 3$ ?

Your answer

2b. The integral of speed is total distance

Displacement can hide motion. In the graph at the top the particle went 9 meters forward and 4 back, and the displacement, 5 meters, is what is left after the 4 cancel part of the 9.

Speed is the size of the velocity, $| v (t) |$ , and it is never negative. The total distance traveled from $t = a$ to $t = b$ is the definite integral of speed:

\int_{a}^{b} | v (t) | d t .

Every stretch of motion now counts as positive, so for the graph at the top the total distance is $9 + 4 = 13$ meters.

Without a calculator, find where $v (t)$ changes sign, split the interval at those times, and put a minus sign in front of each piece where $v (t) < 0$ :

\int_{0}^{5} | 6 - 2 t | d t = \int_{0}^{3} (6 - 2 t) d t - \int_{3}^{5} (6 - 2 t) d t = 9 - (- 4) = 13.

With a graphing calculator, integrate $| v (t) |$ directly from $a$ to $b$ .

Total distance is never less than the size of the displacement, and the two are equal only when the velocity never changes sign on the interval.

Quick check · Calculate

No calculator. A particle moves along a line with velocity $v (t) = 3 t^{2} - 6 t$ for $0 \leq t \leq 3$ . What is the total distance the particle travels from $t = 0$ to $t = 3$ ?

Your answer

2c. The integral of acceleration is change in velocity

The same idea works one level down. Acceleration is the rate of change of velocity, $a (t) = v^{'} (t)$ , so $v$ is an antiderivative of $a$ , and the integral of acceleration over a time interval is the change in velocity:

v (b) = v (a) + \int_{a}^{b} a (t) d t .

The value you start from has to be at the lower limit. If you know the velocity at $t = 2$ , the integral starts at 2, not at 0.

Suppose a cyclist is riding at 2 meters per second at $t = 0$ , and her acceleration is $a (t) = 0.6 t$ meters per second per second. Then

v (4) = 2 + \int_{0}^{4} 0.6 t d t = 2 + {[0.3 t^{2}]}_{0}^{4} = 2 + 4.8 = 6.8

meters per second. The units work as before: meters per second per second times seconds gives meters per second.

Put the two levels together and a starting position, a starting velocity and the acceleration are enough to find where a particle is at any later time. Each step back is a definite integral added to a value you know.

Quick check · Multiple choice

No calculator. A cart moves along a straight track with acceleration $a (t)$ meters per second per second. At $t = 1$ second its velocity is $v (1) = 3$ meters per second. Which expression gives its velocity at $t = 5$ seconds?

Common mistakes

The mistake	Why it is wrong	Do this instead
Reporting $\int_{a}^{b} v (t) d t$ as the particle's position at $t = b$ .	The integral is the change in position. It says nothing about where the particle started.	Add the position you know: $x (b) = x (a) + \int_{a}^{b} v (t) d t$ .
Finding total distance as $\| \int_{a}^{b} v (t) d t \|$ , the size of the displacement.	Taking the absolute value at the end lets motion back cancel motion forward before anything is counted.	Integrate the speed, $\int_{a}^{b} \| v (t) \| d t$ : split where $v (t)$ changes sign, or integrate $\| v (t) \|$ on the calculator.
Starting the integral at $t = 0$ when the value you know is at a different time.	A known value already includes all the change up to its own time, so integrating from 0 counts some of the change twice.	Use the time of the known value as the lower limit of the integral.

Part 3 Watch it done Total distance when the particle turns around twice

Watch each step, then do the last one yourself.

A particle moves along the x-axis with velocity $v (t) = t^{2} - 4 t + 3$ for $0 \leq t \leq 4$ . Find the total distance the particle travels from $t = 0$ to $t = 4$ .

Step 1 · Find where the velocity changes sign
$v (t) = (t - 1) (t - 3)$ , so $v (t) = 0$ at $t = 1$ and $t = 3$ . $v (t) > 0$ on $0 < t < 1$ , $v (t) < 0$ on $1 < t < 3$ , and $v (t) > 0$ on $3 < t < 4$ .
Step 2 · Write the distance as an integral of speed, split at those times
Total distance $= \int_{0}^{4} | v (t) | d t = \int_{0}^{1} v (t) d t - \int_{1}^{3} v (t) d t + \int_{3}^{4} v (t) d t$ .
Why this step?
Why does the middle integral get a minus sign?
On $1 < t < 3$ the velocity is negative, so $| v (t) | = - v (t)$ there. The particle is moving back, and the minus sign turns that negative change in position into a positive distance.
Step 3 · Find an antiderivative
$F (t) = \frac{t^{3}}{3} - 2 t^{2} + 3 t$ , so $F (0) = 0$ , $F (1) = \frac{4}{3}$ , $F (3) = 0$ and $F (4) = \frac{4}{3}$ .
Step 4 · Evaluate each piece
$\int_{0}^{1} v (t) d t = \frac{4}{3} - 0 = \frac{4}{3}$ , $\int_{1}^{3} v (t) d t = 0 - \frac{4}{3} = - \frac{4}{3}$ , and $\int_{3}^{4} v (t) d t = \frac{4}{3} - 0 = \frac{4}{3}$ .
Your step · Calculate

No calculator. Finish it. For $v (t) = t^{2} - 4 t + 3$ , the integrals of $v$ over $[0, 1]$ , $[1, 3]$ and $[3, 4]$ are $\frac{4}{3}$ , $- \frac{4}{3}$ and $\frac{4}{3}$ . What is the total distance the particle travels from $t = 0$ to $t = 4$ ?

Your answer

Part 4 Try it A calculator integral and a claim to judge

Try it · Calculate

Graphing calculator required. A slider moves along a straight rail with velocity $v (t) = 3 \cos (0.9 t) + 0.5$ centimeters per second for $0 \leq t \leq 6$ seconds. In radian mode, find the total distance the slider travels from $t = 0$ to $t = 6$ . Give your answer to three decimal places.

Your answer

Try it · Decide and justify

No calculator. A particle moves along a line with a continuous velocity $v (t)$ meters per second. Its velocity at $t = 1$ is $v (1) = 2$ , and $\int_{0}^{6} v (t) d t = 0$ . A classmate says the particle traveled a total distance of 0 meters from $t = 0$ to $t = 6$ . Is the classmate right?

Which reason supports your answer?

Part 5 Explain it Write it the way the exam asks

On the exam free response is handwritten, so work on paper with your graphing calculator in radian mode. Nothing you type here leaves this page. Then compare with the model answer and score each point.

Free response is 50% of the exam score. This one is shaped like Question 1, from free-response Part A (graphing calculator required).

Free-response practice · Question 1 style

Graphing calculator required. A kayaker paddles along a straight canal. Her velocity is $v (t) = 25 \sin (0.5 t) + 5$ meters per minute for $0 \leq t \leq 15$ minutes, where a positive velocity means she is moving east. At time $t = 0$ she is 40 meters east of a bridge. Your calculator is in radian mode. Give each decimal answer to three decimal places.

(a) Find the kayaker's position relative to the bridge at time $t = 15$ . Show the setup that leads to your answer. Indicate units of measure.

(b) Find all times $t$ , for $0 < t < 15$ , at which the kayaker changes direction. Justify your answer.

(c) Find the total distance the kayaker paddles from $t = 0$ to $t = 15$ . Explain why this distance is greater than the distance between where she starts and where she ends. Indicate units of measure.

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $x (15) = 40 + \int_{0}^{15} (25 \sin (0.5 t) + 5) d t = 40 + 107.668 = 147.668$ . At $t = 15$ minutes the kayaker is 147.668 meters east of the bridge.

(b) Solving $25 \sin (0.5 t) + 5 = 0$ on the calculator for $0 < t < 15$ gives $t = 6.686$ and $t = 12.164$ minutes (truncated, 6.685 and 12.163). The velocity changes from positive to negative at $t = 6.686$ and from negative to positive at $t = 12.164$ , so she changes direction at both times.

(c) Total distance $= \int_{0}^{15} | 25 \sin (0.5 t) + 5 | d t = 248.850$ meters (truncated, 248.849). From $t = 6.686$ to $t = 12.164$ she paddles west. In the displacement, $\int_{0}^{15} v (t) d t = 107.668$ meters, that stretch is subtracted, but in the total distance it is added, so the total distance is greater than the 107.668 meters between her start and her end.

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

(a) Writes $40 + \int_{0}^{15} v (t) d t$ with the integrand, both limits and $d t$ , and gets 147.668 meters east of the bridge.
(b) Writes the equation $v (t) = 0$ and gets both times, $t = 6.686$ and $t = 12.164$ (or 6.685 and 12.163 truncated).
(b) Justifies each time by the change in the sign of $v (t)$ there.
(c) Writes $\int_{0}^{15} | v (t) | d t$ and gets 248.850 meters (or 248.849 truncated).
(c) Explains that the westward stretch adds to the total distance but is subtracted in the displacement.

Part 6 Check yourself A robot on a track

The three case questions use the same table. Answer them from it.

A small robot moves along a straight track. Its velocity $v (t)$ , in centimeters per second, is made of straight line segments joining the points in the table, for $0 \leq t \leq 10$ seconds. A positive velocity means the robot moves to the right.

Velocity of the robot, joined by straight segments
$t$ (seconds)	$v (t)$ (centimeters per second)
0	0
3	6
5	6
7	0
10	$- 4$

Case question 1 · Calculate

No calculator. What is the total distance, in centimeters, that the robot travels from $t = 0$ to $t = 10$ ?

Your answer

Case question 2 · Multiple choice

No calculator. At what time $t$ , for $0 \leq t \leq 10$ , is the robot farthest to the right?

Case question 3 · Calculate

No calculator. At $t = 0$ the robot is at $x = - 5$ centimeters. What is its position $x (10)$ , in centimeters?

Your answer

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A particle moves along a line with velocity $v (t) = 2 t - 6$ meters per second. A student finds the total distance it travels from $t = 0$ to $t = 4$ . Tap the first line with a mistake.

Check yourself · Match

No calculator. A particle moves along a line with position $x (t)$ , velocity $v (t)$ and acceleration $a (t)$ . Tap an expression, then what it gives.

Part 7 Keep it From earlier lessons

This brings back the previous topic's average value, and from Unit 4 what the signs of velocity and acceleration mean. Bringing them back is what makes them last.

Bring it back · Calculate

No calculator. What is the average value of $f (x) = 3 x^{2}$ on the interval $[0, 2]$ ?

Your answer

Bring it back · Multiple choice

No calculator. A particle moves along a line. At time $t = 3$ its velocity is $v (3) = - 4$ and its acceleration is $a (3) = 2$ . Which statement is true at $t = 3$ ?

In three lines

The integral of velocity over a time interval is the displacement, and adding it to a known position gives a later position.
The integral of speed, $| v (t) |$ , is the total distance traveled; split the interval where the velocity changes sign.
The integral of acceleration is the change in velocity, so $v (b) = v (a) + \int_{a}^{b} a (t) d t$ .

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

When a question asks where a particle is, write the known position plus the integral of velocity, with both limits and $d t$ , before any number. When it asks how far the particle travels, write $\int_{a}^{b} | v (t) | d t$ ; on a calculator part, evaluate that integral directly and give three decimal places with units.

Questions students ask

Is displacement the same as total distance traveled?

Only when the particle never turns around. Displacement subtracts the motion back from the motion forward, while total distance adds them. A runner who does one lap of a track has a displacement of 0 and a total distance of one lap.

Why is the integral of velocity a change in position?

Velocity is the derivative of position, so position is an antiderivative of velocity. The Fundamental Theorem of Calculus then says $\int_{a}^{b} v (t) d t = x (b) - x (a)$ .

How do I find total distance without a calculator?

Find the times where $v (t)$ changes sign, split the interval there, and integrate $v (t)$ on each piece. Make every piece positive and add. On a graph, that is the total area between the velocity graph and the $t$ -axis.

Can displacement be negative?

Yes. A negative displacement means the particle ends up on the negative side of where it started. Total distance, an integral of speed, can never be negative.

What units does the integral of velocity have?

The units of velocity times the units of time. Meters per second times seconds gives meters, and for acceleration, meters per second per second times seconds gives meters per second.

Course alignment, for teachers

AP Calculus AB topic 8.2, Unit 8: Applications of Integration.

Skill 1.D: Identify an appropriate mathematical rule or procedure based on the relationship between concepts (e.g., rate of change and accumulation) or processes (e.g., differentiation and its inverse process, anti-differentiation) to solve problems.

This lesson maps the one essential knowledge statement of the topic to the part that teaches each one: 1 is checked by a scored question. The map is in the lesson file.

What you'll be able to do

Key vocabulary

Part 1 Think first Where did the robot end up?

Part 2 Learn it Position, distance and velocity from integrals

2a. The integral of velocity is displacement

2b. The integral of speed is total distance

2c. The integral of acceleration is change in velocity

Part 3 Watch it done Total distance when the particle turns around twice

Part 4 Try it A calculator integral and a claim to judge

Part 5 Explain it Write it the way the exam asks

Part 6 Check yourself A robot on a track

Part 7 Keep it From earlier lessons

Exam strategy for this topic

Questions students ask

Past exam questions on this topic