Position, velocity, acceleration via integration

AP Calculus AB· AP Calculus AB CED — Applications of Integration· 14 min read

1. Recovering Velocity and Position with Initial Conditions★★☆☆☆⏱ 4 min

We start from the fundamental derivative relationships you already know: velocity is the rate of change of position, and acceleration is the rate of change of velocity. Integration reverses differentiation, so we can integrate acceleration to get velocity, then integrate velocity to get position, using initial conditions to solve for unknown constants of integration.

v(t) = \frac{ds}{dt}, \quad a(t) = \frac{dv}{dt}

v(t) = \int a(t) dt + C \\ s(t) = \int v(t) dt + D

📘 Definition

Constants of Integration

Unknown constants that arise from indefinite integration. We solve for each constant separately using the corresponding initial condition after each integration step, regardless of whether the starting time is $t = 0$ .

📐 Worked Example

A particle moving along the x-axis has acceleration given by $a (t) = 6 t - 2$ m/s². At time $t = 1$ , the velocity is $v (1) = 4$ m/s, and the position is $s (1) = - 2$ m. Find the position function $s (t)$ .

1
First integrate acceleration to get the general form of $v (t)$ :
$v(t) = \int (6t - 2) dt = 3t^2 - 2t + C$
2
Use the initial velocity condition to solve for $C$ :
$3(1)^2 - 2(1) + C = 4 \implies 1 + C = 4 \implies C = 3$
3
This gives the full velocity function: $v (t) = 3 t^{2} - 2 t + 3$ . Next, integrate velocity to get the general form of $s (t)$ :
$s(t) = \int (3t^2 - 2t + 3) dt = t^3 - t^2 + 3t + D$
4
Use the position condition at $t = 1$ to solve for $D$ :
$1^3 - 1^2 + 3(1) + D = -2 \implies 3 + D = -2 \implies D = -5$
5
Final position function:
$s(t) = t^3 - t^2 + 3t - 5$

Exam tip:

Always label your constants $C$ and $D$ separately when integrating twice, and solve for each constant as soon as you have the required initial condition. Don't wait to solve for both at the end, which often leads to substitution errors.

2. Net Displacement vs. Total Distance Traveled★★★☆☆⏱ 5 min

Once you have the velocity function $v (t)$ , you can calculate two distinct quantities over an interval $[a, b]$ that are frequently confused on the AP exam. Net displacement is the net change in position from start to end of the interval, accounting for direction, while total distance is the total length of the path traveled, regardless of direction.

📘 Definition

Net Displacement

s(b) - s(a)

Net change in position over an interval, equal to the signed integral of velocity. Negative velocity (motion in the negative direction) contributes negative displacement that cancels positive displacement.

To calculate total distance, all motion counts as positive, so we integrate the absolute value of velocity. First find all points in $[a, b]$ where $v (t) = 0$ (direction changes), split the integral at these points, adjust the sign for intervals where $v (t)$ is negative, then add the results.

\text{Net Displacement} = \int_a^b v(t) dt

\text{Total Distance Traveled} = \int_a^b |v(t)| dt

📐 Worked Example

A particle has velocity $v (t) = t^{2} - 4$ m/s for $t \in [0, 5]$ seconds. Find (a) net displacement, (b) total distance traveled over the interval.

1
Calculate net displacement directly:
$\int_0^5 (t^2 - 4) dt = \left[\frac{t^3}{3} - 4t\right]_0^5 = \frac{125}{3} - 20 = \frac{65}{3} \approx 21.67 \text{ m}$
2
For total distance, first find where $v (t) = 0$ on $[0, 5]$ :
$t^2 - 4 = 0 \implies t = 2 \text{ (discard } t=-2 \text{ for non-negative time)}$
3
Check sign: $v (t) < 0$ on $[0, 2]$ , so $∣ v (t) ∣ = - (t^{2} - 4)$ on this interval, and $∣ v (t) ∣ = v (t)$ on $[2, 5]$ . Split the integral:
$\begin{align*} \int_0^5 |t^2 - 4| dt &= \int_0^2 -(t^2 - 4) dt + \int_2^5 (t^2 - 4) dt \\ &= \left[-\frac{t^3}{3} + 4t\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^5 \\ &= \frac{16}{3} + 27 = \frac{97}{3} \approx 32.33 \text{ m} \end{align*}$
4
Final results: Net displacement = $\frac{65}{3}$ m, total distance = $\frac{97}{3}$ m.

Exam tip:

Always check that the total distance is greater than or equal to the absolute value of net displacement. If your calculation gives the opposite, you made a sign error when splitting the integral.

3. Finding Values at a Point with the Fundamental Theorem★★☆☆☆⏱ 3 min

A common AP exam problem asks for velocity or position at a specific time, given an initial value, but does not require the full function. In these cases, you can use the Fundamental Theorem of Calculus directly to avoid solving for constants of integration, saving time and reducing errors.

v(b) = v(a) + \int_a^b a(t) dt \\ s(b) = s(a) + \int_a^b v(t) dt

📐 Worked Example

A rocket moving straight upward has acceleration $a (t) = 2 e^{0.1 t}$ m/s² at time $t$ seconds. If the velocity at $t = 0$ is 100 m/s, what is the velocity at $t = 10$ seconds?

1
We only need $v (10)$ , not the full function, so use the FTC form directly:
$v(10) = v(0) + \int_0^{10} 2e^{0.1t} dt$
2
Compute the definite integral: the antiderivative of $2 e^{0.1 t}$ is $20 e^{0.1 t}$ :
$20e^{1} - 20e^{0} = 20(e-1) \approx 34.36$
3
Add the initial velocity to get the final result:
$v(10) \approx 100 + 34.36 = 134.36 \text{ m/s}$

Exam tip:

If the question only asks for a value of velocity/position at a specific time, skip finding the full function and use the Fundamental Theorem form directly to save 1–2 minutes on exam day.

4. Concept Check★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

A particle moving along the x-axis has acceleration given by $a (t) = - sin t$ for $t \geq 0$ . If the initial velocity is $v (0) = 2$ and the initial position is $s (0) = 5$ , what is the position at $t = π$ ?
- $3 + π$
- $5 + π$
- $5 - π$
- $1 + 3 π$
Reveal answer
1 —
First integrate acceleration to get $v (t) = cos t + 1$ , then integrate to get $s (t) = sin t + t + 5$ . Evaluating at $t = π$ gives $0 + π + 5 = 5 + π$ .

5. Common Pitfalls

Wrong move:

Forgetting to flip the sign of $v (t)$ when integrating $∣ v (t) ∣$ on intervals where $v (t)$ is negative.

Why:

Students remember to split the integral at zeros of $v (t)$ but just drop the absolute value without checking the sign, treating $∣ v (t) ∣ = v (t)$ everywhere.

Correct move:

Always test the sign of $v (t)$ in each subinterval after splitting, and add a negative sign to any interval where $v (t)$ is negative before integrating.

Wrong move:

Solving for only one constant of integration after integrating acceleration twice to get position.

Why:

Students confuse the constants, using the same $C$ for both integration steps and solving once instead of solving for a new constant after the second integration.

Correct move:

Label the constant after integrating acceleration $C_{1}$ and the constant after integrating velocity $C_{2}$ , and solve each separately using the corresponding initial condition.

Wrong move:

Calculating total distance by integrating velocity instead of absolute value of velocity.

Why:

Students confuse net displacement with total distance, especially when the question says 'how far did the particle travel' which can seem ambiguous to new learners.

Correct move:

Always read the question carefully: 'what is the change in position' or 'what is the displacement' means integrate $v (t)$ ; 'how far did the particle travel' or 'what is the total distance traveled' means integrate $∣ v (t) ∣$ .

Wrong move:

Including roots of $v (t) = 0$ outside the interval of integration when splitting the integral for total distance.

Why:

Students solve for all roots automatically and split at all roots, even ones outside the interval, leading to extra unnecessary subintervals and calculation errors.

Correct move:

After finding all roots of $v (t) = 0$ , only keep roots that are strictly between $a$ and $b$ when splitting the interval.

Wrong move:

Using initial position to find the constant for velocity, or initial velocity to find the constant for position.

Why:

Students mix up which initial condition goes with which integration step.

Correct move:

Match the derivative order: acceleration integrates to velocity, so use velocity's initial condition to find $C$ for velocity; velocity integrates to position, so use position's initial condition to find $D$ for position.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Velocity from Acceleration	$v (t) = \int a (t) d t + C$	Solve for $C$ using initial velocity $v (t_{0}) = v_{0}$
Position from Velocity	$s (t) = \int v (t) d t + D$	Solve for $D$ using initial position $s (t_{0}) = s_{0}$
Change in Velocity	$v (b) - v (a) = \int_{a}^{b} a (t) d t$	No need to solve for $C$ , works for any interval
Net Displacement	$s (b) - s (a) = \int_{a}^{b} v (t) d t$	Signed, accounts for direction of motion
Total Distance Traveled	$\int_{a}^{b} ∣ v (t) ∣ d t$	Always non-negative, all motion counts
Position at time $b$	$s (b) = s (a) + \int_{a}^{b} v (t) d t$	Uses FTC, avoids full function derivation
Splitting Absolute Value Integral	$\int_{a}^{b} ∣ v (t) ∣ d t = \sum \int_{c_{i}}^{c_{i + 1}} v (t) d t$	Split only at roots inside $[a, b]$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AB
FRQ linear motion problem
2022 · AB
MCQ distance calculation

What's Next

This topic is the foundational example for applying integration to all net change problems across AP Calculus AB, not just linear motion. Immediately next in Unit 8, you will extend the core reasoning of this topic—finding net change from a rate of change—to other applied contexts, like population growth, fluid flow, and accumulated cost. Without mastering this relationship, you will struggle to generalize this reasoning to other rate-of-change problems, which are common in both multiple-choice and free-response sections of the AP exam. This topic also reinforces the core inverse relationship between differentiation and integration that unites the entire course.