Position, velocity, acceleration via integration

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

1. Core Definitions and Relationships★☆☆☆☆⏱ 2 min

This topic applies integration to reverse the derivative relationships between motion quantities learned in differential calculus, for rectilinear (straight-line) motion, the primary context tested on the AP exam.

📘 Definition

Standard motion notation

$s (t)$ (or $x (t)$ ), $v (t)$ , $a (t)$

For $t \geq 0$ , $s (t)$ = position of a particle relative to a fixed origin, $v (t) = s^{'} (t)$ = velocity, $a (t) = v^{'} (t) = s^{''} (t)$ = acceleration.

When given acceleration or velocity and asked to find velocity or position, integration is the required tool. A key skill you will develop is distinguishing between net displacement (net change in position) and total distance traveled (total path length) over an interval. According to the AP Calculus CED, Applications of Integration make up 10-15% of the BC exam, with this subtopic accounting for roughly 2-4% of total exam points, appearing in both multiple-choice and free-response sections.

2. Finding Velocity and Position with Initial Conditions★★☆☆☆⏱ 4 min

Since acceleration is the rate of change of velocity, and velocity is the rate of change of position, we reverse these relationships via integration:

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

We can also write this in definite integral form to avoid carrying an unknown constant, where $τ$ is a dummy variable of integration:

v(t) = v(0) + \int_0^t a(\tau) d\tau \\ s(t) = s(0) + \int_0^t v(\tau) d\tau

The constant of integration $C$ is solved using an initial condition: a known value of velocity or position at a given starting time. Intuition: Integration accumulates all small changes in acceleration over time to get the total change in velocity, then adding the starting velocity gives the current velocity, with the same logic applying to position from velocity.

📐 Worked Example

A particle moving along the x-axis has acceleration $a (t) = 6 t - sin t$ for $t \geq 0$ . If the initial velocity is $v (0) = 2$ and initial position is $s (0) = - 3$ , find the position function $s (t)$ .

1
Integrate $a (t)$ to get the general form of $v (t)$ :
$v(t) = \int (6t - \sin t) dt = 3t^2 + \cos t + C$
2
Apply the initial velocity condition to solve for $C$ : $v (0) = 3 (0)^{2} + cos (0) + C = 1 + C = 2 ⟹ C = 1$ , so $v (t) = 3 t^{2} + cos t + 1$ .
3
Integrate $v (t)$ to get the general form of $s (t)$ :
$s(t) = \int (3t^2 + \cos t + 1) dt = t^3 + \sin t + t + D$
4
Apply the initial position condition to solve for $D$ : $s (0) = 0^{3} + sin (0) + 0 + D = D = - 3$ , so $s (t) = t^{3} + sin t + t - 3$ .

Exam tip:

Always solve for the constant of integration immediately after integrating each quantity, don’t wait until the end to solve for multiple constants. This avoids arithmetic errors from carrying unknown values through multiple steps.

3. Calculating Net Displacement Over a Time Interval★★☆☆☆⏱ 3 min

Net displacement is the net change in a particle’s position between two times $t = a$ and $t = b$ , or $Δ s = s (b) - s (a)$ . By the Fundamental Theorem of Calculus Part 2, since $s (t)$ is the antiderivative of $v (t)$ , we get:

\Delta s = s(b) - s(a) = \int_a^b v(t) \, dt

Displacement is a net quantity: positive velocity (movement in the positive direction) adds to displacement, while negative velocity (movement in the negative direction) subtracts from displacement. Displacement can be positive, negative, or zero, depending on where the particle ends up relative to its starting position. You do not need an initial condition to calculate displacement, because the constant of integration cancels out when evaluating the definite integral.

📐 Worked Example

A particle has velocity $v (t) = t^{2} - 4 t + 3$ for $0 \leq t \leq 4$ . Find the net displacement of the particle over the interval.

1
Recall displacement is the definite integral of velocity over the interval, so $Δ s = \int_{0}^{4} (t^{2} - 4 t + 3) d t$ .
2
Find the antiderivative of $v (t)$ :
$\int (t^2 - 4t + 3) dt = \frac{t^3}{3} - 2t^2 + 3t + C$
3
Evaluate the definite integral using the bounds $[0, 4]$ :
$\left[\frac{t^3}{3} - 2t^2 + 3t\right]_0^4 = \left(\frac{64}{3} - 32 + 12\right) - 0 = \frac{4}{3}$
4
Interpretation: The particle ends $\frac{4}{3}$ units in the positive direction from its starting position at $t = 4$ .

Exam tip:

If a question only asks for displacement, you do not need to solve for the position function or find the constant of integration — just evaluate the definite integral of velocity directly.

4. Calculating Total Distance Traveled Over a Time Interval★★★☆☆⏱ 5 min

Total distance traveled is the total length of the path a particle takes over $[a, b]$ , regardless of direction of movement. Unlike displacement, it is always non-negative. Any movement (forward or backward) adds to total distance, so we have to account for negative velocity by making it positive before integrating. The formula for total distance is:

D = \int_a^b |v(t)| \, dt

Find all times $t$ in $(a, b)$ where $v (t) = 0$ (these are turning points, where the particle changes direction).
Split the original interval into subintervals where $v (t)$ is entirely positive or entirely negative.
On subintervals where $v (t) < 0$ , replace $∣ v (t) ∣$ with $- v (t)$ to make it positive; leave $v (t)$ as-is on intervals where $v (t) > 0$ .
Integrate each subinterval and add the results.

📐 Worked Example

For the same particle with $v (t) = t^{2} - 4 t + 3$ on $0 \leq t \leq 4$ , find the total distance traveled over the interval.

1
Find turning points by solving $v (t) = 0$ : $t^{2} - 4 t + 3 = (t - 1) (t - 3) = 0 ⟹ t = 1, t = 3$ , both inside $[0, 4]$ .
2
Test the sign of $v (t)$ on each subinterval: $v (t) > 0$ on $(0, 1)$ and $(3, 4)$ , $v (t) < 0$ on $(1, 3)$ .
3
Set up the split integral for $∣ v (t) ∣$ :
$D = \int_0^1 (t^2 - 4t + 3) dt + \int_1^3 -(t^2 - 4t + 3) dt + \int_3^4 (t^2 - 4t + 3) dt$
4
Evaluate using the antiderivative $F (t) = \frac{t ^{3}}{3} - 2 t^{2} + 3 t$ :
- $\int_{0}^{1} v (t) d t = F (1) - F (0) = \frac{4}{3}$
- $\int_{1}^{3} - v (t) d t = - (F (3) - F (1)) = \frac{4}{3}$
- $\int_{3}^{4} v (t) d t = F (4) - F (3) = \frac{4}{3}$
5
Add the results: $D = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4$ units.

✓ Quick check

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

A particle moving along a straight line has acceleration $a (t) = 2 e^{2 t}$ for $t \geq 0$ . If the initial velocity is $v (0) = 3$ , what is $v (1)$ ?
- $e^{2}$
- $e^{2} + 2$
- $2 e^{2} + 1$
- $3 e^{2}$
Reveal answer
1 —
To find velocity from acceleration, integrate $a (t)$ to get $v (t) = e^{2 t} + C$ , then apply $v (0) = 1 + C = 3 ⟹ C = 2$ , so $v (1) = e^{2} + 2$ .

Exam tip:

If velocity is never zero on the interval, the particle never changes direction, so total distance equals displacement. Always confirm whether turning points exist inside the interval before splitting.

5. Common Pitfalls

Wrong move:

Ignoring the absolute value when asked for total distance, just integrating $v (t)$ directly like you would for displacement.

Why:

Students mix up the definitions of displacement and total distance, especially on rushed FRQs.

Correct move:

Always circle what the question asks for and write the corresponding formula next to the question immediately after reading it to confirm.

Wrong move:

Forgetting to flip the sign of $v (t)$ on intervals where $v (t) < 0$ when calculating total distance.

Why:

You remember to split the interval but rush and leave $v (t)$ as-is, leading to a total distance that is too low.

Correct move:

After testing the sign on each subinterval, explicitly write $∣ v (t) ∣ = + v (t)$ or $∣ v (t) ∣ = - v (t)$ under each integral before integrating.

Wrong move:

Solving for the constant of integration when calculating definite displacement or total distance.

Why:

You are used to finding position functions, so you add an unnecessary step that introduces arithmetic errors.

Correct move:

For any definite quantity over an interval, use the definite integral directly; only solve for $C$ if the question explicitly asks for the full position or velocity function.

Wrong move:

Integrating acceleration directly to find displacement or distance, instead of first integrating acceleration to get velocity.

Why:

The question gives you acceleration and you forget the chain of relationships, skipping a step.

Correct move:

Write the relationship chain $a (t) \to v (t) \to s (t) / displacement / distance$ before starting any calculations.

Wrong move:

Using the initial position constant to solve for $C$ in the velocity function after integrating acceleration.

Why:

Rushing through multiple integration steps mixes up which initial condition goes with which integration.

Correct move:

Solve for $C$ immediately after integrating acceleration to get velocity before moving on to integrate velocity for position.

6. Quick Reference Cheatsheet

Category	Formula	Notes
Velocity from Acceleration	$v (t) = v_{0} + \int_{0}^{t} a (τ) d τ$	$v_{0} = v (0)$ is initial velocity; works for all $t \geq 0$
Position from Velocity	$s (t) = s_{0} + \int_{0}^{t} v (τ) d τ$	$s_{0} = s (0)$ is initial position
Indefinite Velocity	$v (t) = \int a (t) d t + C$	Solve for $C$ using a known initial velocity
Indefinite Position	$s (t) = \int v (t) d t + C$	Solve for $C$ using a known initial position
Net Displacement $[a, b]$	$Δ s = \int_{a}^{b} v (t) d t$	Can be positive/negative/zero; no initial position needed
Total Distance $[a, b]$	$D = \int_{a}^{b} ∣ v (t) ∣ d t$	Always non-negative; split at all turning points in $(a, b)$
Core Derivative Relationship	$a (t) = v^{'} (t) = s^{''} (t)$	Reversed by integration to get velocity and position
Sign Convention	Positive $v$ = positive direction movement	Standard for all AP rectilinear motion problems

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 · BC MCQ
Total distance from velocity function
2022 · BC FRQ
Position from acceleration with initial conditions

What's Next

This topic is the foundation for understanding motion in higher dimensions, a core tested topic on AP Calculus BC. Mastering the relationship between position, velocity, and acceleration via integration is required to solve problems involving parametric motion and vector-valued functions, where you analyze two-dimensional motion by applying the same rules to each directional component separately. It also connects directly to differential equations, where solving for position given acceleration and initial conditions is a common multi-part FRQ prompt. Without solidifying the relationships and key difference between displacement and distance here, you will struggle to extend these ideas to more complex motion contexts.