Position, velocity, acceleration via integration — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: This chapter covers relating one-dimensional position, velocity, and acceleration via definite and indefinite integration, solving for position from velocity, velocity from acceleration, calculating displacement, and total distance traveled for rectilinear motion.

You should already know: Derivative relationships between position, velocity, and acceleration. Basic indefinite and definite integration rules. The Fundamental Theorem of Calculus.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Position, velocity, acceleration via integration?

Position, velocity, acceleration via integration is the application of integration to reverse the derivative relationships between these three motion quantities learned in differential calculus. For rectilinear (straight-line) motion, the primary context tested on the AP exam, we use standard notation: $s(t)$ (or $x(t)$) gives the position of a particle relative to a fixed origin at time $t\ge 0$, velocity $v(t)=s^{\prime }(t)$, and acceleration $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$.

When you are given acceleration or velocity and need to find velocity or position, integration is the required tool. This topic also requires 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. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often paired with other topics like differential equations or parametric motion.

2. Finding Velocity from Acceleration and Position from Velocity with Initial Conditions

The core relationship between these quantities starts with the derivative definitions: acceleration is the rate of change of velocity, and velocity is the rate of change of position. To reverse this, we integrate: $$\frac{dv}{dt}=a(t)⟹v(t)=\int a(t)dt+C$$ $$\frac{ds}{dt}=v(t)⟹s(t)=\int v(t)dt+C$$ The constant of integration $C$ is solved using an initial condition: a known value of velocity (usually initial velocity $v(0)=v_0$) or position (usually initial position $s(0)=s_0$) at a given starting time. We can also write this in definite integral form to avoid carrying an unknown constant: $v(t)=v(0)+{\int }_0^{t}a(\tau )d\tau$ and $s(t)=s(0)+{\int }_0^{t}v(\tau )d\tau$, where $\tau$ is a dummy variable of integration.

Intuition: Integration accumulates all the small changes in acceleration over time to get the total change in velocity, then adding the starting velocity gives the current velocity. The same logic applies to getting position from velocity.

Worked Example

A particle moving along the x-axis has acceleration $a(t)=6t-\sin t$ for $t\ge 0$. 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 $v(t)=3t^2+\cos t+1$.
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

Net displacement is defined as the net change in a particle’s position between two times $t=a$ and $t=b$, or $\Delta s=s(b)-s(a)$. 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. Unlike finding the full position function, 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 $v(t)=t^2-4t+3$ for $0\le t\le 4$. Find the net displacement of the particle over the interval.

1. Recall displacement is the definite integral of velocity over the interval, so $\Delta s={\int }_0^4(t^2-4t+3)dt$.
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

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: $$D={\int }_a^b|v(t)|dt$$ To compute this integral:

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

Worked Example

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

1. Find turning points by solving $v(t)=0$: $t^2-4t+3=(t-1)(t-3)=0⟹t=1,t=3$, 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 $F(t)=\frac{t^3}{3}-2t^2+3t$:
   • ${\int }_0^{1}v(t)dt=F(1)-F(0)=\frac{4}{3}$
   • ${\int }_1^3-v(t)dt=-(F(3)-F(1))=\frac{4}{3}$
   • ${\int }_3^{4}v(t)dt=F(4)-F(3)=\frac{4}{3}$
5. Add the results: $D=\frac{4}{3}+\frac{4}{3}+\frac{4}{3}=4$ units.

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 (and how to avoid them)

• 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 $a(t)\to v(t)\to s(t)/\text{displacement}/\text{distance}$ 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. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

A particle moving along a straight line has acceleration $a(t)=2e^{2t}$ for $t\ge 0$. If the initial velocity is $v(0)=3$, what is $v(1)$? A) $e^2$ B) $e^2+2$ C) $2e^2+1$ D) $3e^2$

Worked Solution: To find velocity from acceleration, we first integrate $a(t)$: $v(t)=\int 2e^{2t}dt=e^{2t}+C$. Next, apply the initial condition $v(0)=3$ to solve for $C$: $v(0)=e^0+C=1+C=3$, so $C=2$. Substitute $t=1$ to get $v(1)=e^{2(1)}+2=e^2+2$. The correct answer is B.

Question 2 (Free Response)

A particle moves along the x-axis with velocity given by $v(t)=2\cos (\pi t)+2t$ for $0\le t\le 2$. At time $t=0$, the particle is at position $s(0)=1$. (a) Find the position function $s(t)$ for all $0\le t\le 2$. (b) Find the net displacement of the particle over the interval $0\le t\le 2$. (c) Find the total distance traveled by the particle over the interval $0\le t\le 2$.

Worked Solution: (a) We use the definite integral form for position: $s(t)=s(0)+{\int }_0^{t}v(\tau )d\tau$. Integrate term-by-term: $${\int }_0^t(2\cos (\pi \tau )+2\tau )d\tau ={\left[\frac{2}{\pi }\sin (\pi \tau )+{\tau }^2\right]}_0^t=\frac{2}{\pi }\sin (\pi t)+t^2$$ Add the initial position: $s(t)=1+\frac{2}{\pi }\sin (\pi t)+t^2$.

(b) Net displacement is ${\int }_0^{2}v(t)dt={\left[\frac{2}{\pi }\sin (\pi t)+t^2\right]}_0^2=(\frac{2}{\pi }\sin (2\pi )+4)-0=4$. Net displacement is 4 units.

(c) First find zeros of $v(t)$ on $[0,2]$: $2\cos (\pi t)+2t=0⟹\cos (\pi t)=-t$. For $t>0$, $-t<-1$ when $t>1$, and $\cos (\pi t)\ge -1$ for all $t$. For $0<t<1$, $\cos (\pi t)>-1>-t$, so $v(t)$ is always positive on $[0,2]$. Thus $|v(t)|=v(t)$, so total distance equals displacement, which is 4 units.

Question 3 (Application / Real-World Style)

A rocket is launched vertically from the ground, starting from rest (initial velocity = 0 at $t=0$). The acceleration of the rocket for the first 10 seconds of flight is $a(t)=12+0.3t^2$ m/s², where $t$ is time in seconds. What is the height of the rocket 10 seconds after launch? Include units and interpret your result.

Worked Solution: First, find velocity by integrating acceleration: $v(t)=\int (12+0.3t^2)dt=12t+0.1t^3+C$. Initial velocity $v(0)=0$, so $C=0$, giving $v(t)=12t+0.1t^3$. Next, find height (position) by integrating velocity: $s(t)=\int (12t+0.1t^3)dt=6t^2+0.025t^4+D$. Initial position $s(0)=0$, so $D=0$. Evaluate at $t=10$: $s(10)=6(10^2)+0.025(10^4)=600+250=850$. 10 seconds after launch, the rocket is 850 meters above the ground launch pad.

7. Quick Reference Cheatsheet

8. 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. Next topics to study after this chapter: Parametric motion modeling Vector-valued function integration First-order differential equations