8.2 Connecting Position, Velocity, and Acceleration of Functions Using Integrals

For a particle moving along a line, the definite integral of velocity gives displacement, and the definite integral of speed (the absolute value of velocity) gives total distance traveled. To rebuild a function from a rate, integrate and then use an initial condition to solve for the constant: integrate acceleration to get velocity, and integrate velocity to get position. For AP Calculus, keep displacement and total distance separate because one is signed and the other is not.

Position, Velocity, and Acceleration Integrals Summary

For AP Calculus Topic 8.2, integration reverses the derivative chain from position to velocity to acceleration. If you integrate velocity on an interval, you get displacement. If you integrate speed, $|v(t)|$, you get total distance traveled. If you integrate acceleration, you get the change in velocity.

The exam often tests whether you can choose the correct integral and explain it in context. Displacement is signed, total distance is not, and an initial condition is needed when you want an actual position or velocity function instead of just a net change.

Why This Matters for the AP Calculus Exam

Rectilinear motion shows up across the AP Calculus exam because it ties together antiderivatives, the Fundamental Theorem of Calculus, and interpreting integrals in context. You will be asked to set up and evaluate definite integrals that represent displacement or total distance, work from velocity or acceleration data given as equations, graphs, or tables, and explain what your result means with correct units. Getting comfortable with this topic also builds the habit of choosing between differentiation and integration based on what a problem gives you and what it asks for.

Key Takeaways

• Displacement over $[a,b]$ is $\int_a^b v(t)\,dt$. It can be positive, negative, or zero.
• Total distance traveled over $[a,b]$ is $\int_a^b |v(t)|\,dt$. Distance is never negative.
• To use the distance formula, find where $v(t)=0$, split the interval where velocity changes sign, and integrate each piece.
• Integrate acceleration to recover velocity, and integrate velocity to recover position. Each integration adds a constant you solve for using an initial condition.
• Speed is $|v(t)|$, the magnitude of velocity. Velocity carries direction; speed does not.
• Always interpret results in context and attach units.

Position, Velocity, and Acceleration: The Setup

This topic is the integration version of straight-line motion from Unit 4. There, you used derivatives to go from position to velocity to acceleration. Here you reverse the chain with integration. Review the 4.2 derivative version here if you need it.

The three functions are connected by derivatives in one direction and integrals in the other:

• Position $s(t)$ is the location of the object at time $t$.
• Velocity $v(t) = s'(t)$ is the rate of change of position. Its sign tells you direction.
• Acceleration $a(t) = v'(t) = s''(t)$ is the rate of change of velocity.

Speed is the absolute value of velocity, $|v(t)|$. An object moving in the negative direction can still have a large speed.

Using Integration to Go Backward

When a problem gives you a rate (velocity or acceleration) and an initial value, integration rebuilds the original function.

Integrating Velocity to Find Position

Integrating velocity gives the change in position, called displacement:

$$\text{displacement} = s_f - s_i = \Delta s = \int_a^b v(t)\,dt$$

This is the Fundamental Theorem of Calculus applied to motion: the definite integral of a rate of change gives the net change. To get an actual position function, include the constant and solve for it with an initial position:

$$s(t) = \int v(t)\,dt + C$$

Displacement is signed. If the object moves forward then back, the negative part cancels some of the positive part.

Distance Traveled vs. Displacement

Total distance traveled measures the full length of the path, regardless of direction:

$$\text{distance traveled} = \int_a^b |v(t)|\,dt$$

Displacement only compares start and end positions. Distance counts every bit of movement. They are equal only when velocity never changes sign on the interval.

Integrating Acceleration to Find Velocity

The same idea works one level up. Integrate acceleration to find the change in velocity, and add the initial velocity:

$$v(t) = \int a(t)\,dt + C$$

So if you are given acceleration, you can integrate once to reach velocity and again to reach position. Each step needs its own initial condition to pin down the constant.

How to Use This on the AP Calculus Exam

Free Response

Motion problems often give velocity as an equation, a graph, or a table.

• For displacement, set up $\int_a^b v(t)\,dt$ and evaluate.
• For total distance, set up $\int_a^b |v(t)|\,dt$. Write a correct expression using proper notation, since clear setup matters for clear exam work.
• Interpret your answer in words with units, for example "the particle moved 4.5 meters to the right."

Problem Solving

To compute total distance by hand:

1. Solve $v(t) = 0$ to find when the particle changes direction.
2. Split the interval at those times.
3. On each piece, decide whether $v(t)$ is positive or negative.
4. Integrate $v(t)$ where it is positive and $-v(t)$ where it is negative, then add the pieces.

If a graphing calculator is allowed for that part, you can often evaluate $\int_a^b |v(t)|\,dt$ directly.

Common Trap

Watch for the difference between average value and average rate of change, and between displacement and distance. Reading the question carefully tells you which integral to set up.

Worked Examples

Example 1: Recover the Position Function

An object has velocity $v(t) = 2t + 5$ with initial position $s(0) = 10$. Find $s(t)$.

Integrate velocity and add the initial position:

$$s(t) = \int (2t + 5)\,dt + C = t^2 + 5t + C$$

Use $s(0) = 10$ to solve: $C = 10$. So:

$$s(t) = t^2 + 5t + 10$$

Example 2: Displacement and Distance

Given $v(t) = 3t - 2$ on $[1, 4]$ with $s(1) = 5$, find the displacement and distance traveled.

Displacement is the integral of velocity:

$$\text{displacement} = \int_{1}^{4} (3t - 2)\,dt = \left[\frac{3}{2}t^2 - 2t\right]_{1}^{4} = \frac{27}{2}$$

For distance, check whether $v(t)$ changes sign on $[1,4]$. Solving $3t - 2 = 0$ gives $t = \tfrac{2}{3}$, which is outside $[1,4]$, so $v(t) > 0$ on the whole interval. The absolute value does nothing here:

$$\text{distance traveled} = \int_{1}^{4} |3t - 2|\,dt = \int_{1}^{4} (3t - 2)\,dt = \frac{27}{2}$$

Because velocity never changes sign on this interval, displacement and distance are equal. If $v(t)$ had crossed zero inside the interval, you would split there and the two answers would differ.

Common Misconceptions

• Integrating velocity gives displacement, not the final position. You must add the initial position to get $s(t)$.
• Displacement and distance are not the same. They match only when velocity keeps one sign across the whole interval.
• $\int_a^b |v(t)|\,dt$ is not the same as $\left|\int_a^b v(t)\,dt\right|$. Take the absolute value inside the integral, after splitting where velocity changes sign.
• Speed is $|v(t)|$, the size of velocity. Negative velocity still means positive speed.
• A horizontal segment on a velocity vs. time graph means constant velocity, not that the object is stopped. The object is at rest only when velocity equals zero.
• Each integration introduces a new constant. You need a separate initial condition for velocity and for position when working from acceleration.

Quick Reference

To go forward through the chain, differentiate. To go backward, integrate and apply an initial condition.

Related AP Calculus Guides

• Unit 8 Overview: Applications of Integration
• 8.1 Finding the Average Value of a Function on an Interval
• 8.7 Volumes with Cross Sections: Squares and Rectangles
• 8.4 Finding the Area Between Curves Expressed as Functions of x
• 8.3 Using Accumulation Functions and Definite Integrals in Applied Contexts
• 8.6 Finding the Area Between Curves That Intersect at More Than Two Points