# Position, velocity, acceleration via integration

> AP Calculus AB · Applications of Integration
> Source: https://www.owlsprep.com/study/ap-calculus-ab-u8-position-velocity-acceleration-via-integration/

This module connects linear motion variables to integration, covering recovering position/velocity from acceleration, solving for constants of integration with initial conditions, and distinguishing net displacement from total distance traveled, a frequent AP exam topic.

**Prerequisites:** Derivative relationship between position, velocity, and acceleration; Basic integration rules for polynomial, trigonometric, and exponential functions; Fundamental Theorem of Calculus for evaluating definite integrals

## Learning objectives

- Recover velocity and position functions from acceleration using integration and initial conditions
- Distinguish between net displacement and total distance traveled over an interval
- Calculate total distance traveled by integrating absolute value of velocity
- Use the Fundamental Theorem of Calculus to find velocity/position at a point without deriving the full function

## Recovering Velocity and Position with Initial Conditions

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.

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

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

**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 $a(t) = 6t - 2$ 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: $v(t) = 3t^2 - 2t + 3$. 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.

## Net Displacement vs. Total Distance Traveled

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.

**Net Displacement** — 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.

*Notation:* s(b) - s(a)

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.

## Finding Values at a Point with the Fundamental Theorem

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 $a(t) = 2e^{0.1t}$ 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 $2e^{0.1t}$ is $20e^{0.1t}$:

   $$20e^{1} - 20e^{0} = 20(e-1) \approx 34.36$$
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.

## Concept Check

**Check your understanding**

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

1. 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 $t = \pi$?

   - $3 + \pi$
   - $5 + \pi$
   - $5 - \pi$
   - $1 + 3\pi$

   *Answer:* $5 + \pi$

   *Why:* First integrate acceleration to get $v(t) = \cos t + 1$, then integrate to get $s(t) = \sin t + t + 5$. Evaluating at $t = \pi$ gives $0 + \pi + 5 = 5 + \pi$.

## Common pitfalls

- **Wrong:** Forgetting to flip the sign of $v(t)$ when integrating $|v(t)|$ on intervals where $v(t)$ is negative.
  - Why it fails: 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: 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:** Solving for only one constant of integration after integrating acceleration twice to get position.
  - Why it fails: 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: 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:** Calculating total distance by integrating velocity instead of absolute value of velocity.
  - Why it fails: 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: 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:** Including roots of $v(t)=0$ outside the interval of integration when splitting the integral for total distance.
  - Why it fails: 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: After finding all roots of $v(t)=0$, only keep roots that are strictly between $a$ and $b$ when splitting the interval.
- **Wrong:** Using initial position to find the constant for velocity, or initial velocity to find the constant for position.
  - Why it fails: Students mix up which initial condition goes with which integration step.
  - Correct: 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.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Velocity from Acceleration | $v(t) = \int a(t) dt + C$ | Solve for $C$ using initial velocity $v(t_0) = v_0$ |
| Position from Velocity | $s(t) = \int v(t) dt + D$ | Solve for $D$ using initial position $s(t_0) = s_0$ |
| Change in Velocity | $v(b) - v(a) = \int_a^b a(t) dt$ | No need to solve for $C$, works for any interval |
| Net Displacement | $s(b) - s(a) = \int_a^b v(t) dt$ | Signed, accounts for direction of motion |
| Total Distance Traveled | $\int_a^b \|v(t)\| dt$ | Always non-negative, all motion counts |
| Position at time $b$ | $s(b) = s(a) + \int_a^b v(t) dt$ | Uses FTC, avoids full function derivation |
| Splitting Absolute Value Integral | $\int_a^b \|v(t)\| dt = \sum \left\|\int_{c_i}^{c_{i+1}} v(t) dt\right\|$ | Split only at roots inside $[a,b]$ |

## 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.

- [Average value of a function](https://www.owlsprep.com/study/ap-calculus-ab-u8-average-value-of-a-function/)
- [Accumulation functions and definite integrals in applied contexts](https://www.owlsprep.com/study/ap-calculus-ab-u8-accumulation-functions-and-definite-integrals/)
- [Area between curves expressed as functions of x](https://www.owlsprep.com/study/ap-calculus-ab-u8-area-between-curves-expressed-as/)

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