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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: Relating position, velocity, and acceleration via definite and indefinite integration, finding velocity from acceleration and position from velocity, solving for constants of integration with initial conditions, and calculating net displacement and total distance traveled.

You should already know: The derivative relationship between position, velocity, and acceleration; Basic integration rules for polynomial, trigonometric, and exponential functions; How to apply the Fundamental Theorem of Calculus to evaluate definite integrals.

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 AB 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?

This topic connects the derivative relationships between motion variables you learned in differential calculus to integration, leveraging the fact that integration reverses differentiation. According to the AP Calculus AB Course and Exam Description (CED), this topic falls under Unit 8: Applications of Integration, which accounts for 6–12% of the total exam weight. It appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as part of a multi-concept FRQ that also tests other integration applications.

The core idea is that velocity is the rate of change of position, and acceleration is the rate of change of velocity. Since integration gives the net change of a function over an interval given its rate of change, we use integration to recover position from velocity and velocity from acceleration, usually with given initial conditions to solve for unknown constants of integration. A key exam focus is distinguishing between net displacement (net change in position) and total distance traveled over an interval, which tests conceptual understanding of what integration actually measures.

2. Recovering Velocity and Position from Acceleration with Initial Conditions

We start from the fundamental derivative relationships you already know: $v(t)=\frac{ds}{dt}$ and $a(t)=\frac{dv}{dt}$, where $s(t)$ = position at time $t$, $v(t)$ = velocity, and $a(t)$ = acceleration. To reverse these derivatives, we integrate step-by-step: if we know $a(t)$, we integrate to get the general form of $v(t)$: $$v(t)=\int a(t)dt+C$$ An initial condition (almost always a given value of velocity at some time $t_0$, $v(t_0)=v_0$) lets us solve for the constant $C$. Once we have a fully solved $v(t)$, we integrate again to get position: $$s(t)=\int v(t)dt+D$$ We then use a given initial position $s(t_0)=s_0$ to solve for the second constant $D$. This works for any starting time $t_0$, not just $t=0$: substitute $t=t_0$ into the general function to solve for the constant regardless of when the motion starts.

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 $C$: Substitute $t=1,v(1)=4$: $3(1{)}^2-2(1)+C=4⟹1+C=4⟹C=3$, so $v(t)=3t^2-2t+3$
3. 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 $D$: $1^3-1^2+3(1)+D=-2⟹3+D=-2⟹D=-5$
5. Final position function: $\boxed{s(t)=t^3-t^2+3t-5}$

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.

3. 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 and total distance traveled.

Net displacement is the net change in position from the start to the end of the interval, which is the signed integral of velocity, because velocity is positive when moving in the positive direction and negative when moving in the negative direction: $$\text{Net Displacement}=s(b)-s(a)={\int }_a^{b}v(t)dt$$ Negative velocity contributes negative displacement, so it cancels out positive displacement from motion in the opposite direction.

Total distance traveled, by contrast, is the total length of the path the particle took, regardless of direction. All motion counts as positive distance, so we integrate the absolute value of velocity: $$\text{Total Distance Traveled}={\int }_a^b|v(t)|dt$$ To compute this integral, first find all points in $[a,b]$ where $v(t)=0$ (where the particle changes direction), split the integral into subintervals where $v(t)$ is entirely positive or entirely negative, drop the absolute value by adding a negative sign to intervals where $v(t)$ is negative, then add the results.

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: $${\int }_0^5(t^2-4)dt={\left[\frac{t^3}{3}-4t\right]}_0^5=\left(\frac{125}{3}-20\right)=\frac{65}{3}\approx 21.67\text{ m}$$
2. For total distance, first find where $v(t)=0$ on $[0,5]$: $t^2-4=0⟹t=2$ (we discard $t=-2$ for non-negative time).
3. Check the sign of $v(t)$: $v(t)<0$ on $[0,2]$ and $v(t)>0$ on $[2,5]$, so $|v(t)|=-(t^2-4)$ on $[0,2]$.
4. Compute the split integral: $$\begin{align*} \int_0^5 |t^2 - 4| dt &= \int_0^2 -(t^2 - 4) dt + \int_2^5 (t^2 - 4) dt \ &= \left[-\frac{t^3}{3} + 4t\right]_0^2 + \left[\frac{t^3}{3} - 4t\right]_2^5 \ &= \frac{16}{3} + 27 = \frac{97}{3} \approx 32.33 \text{ m} \end{align*}$$
5. 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 total distance smaller than |net displacement|, you know you made a sign error when splitting the integral.

4. Finding Velocity/Position at a Point Using the Fundamental Theorem

A common AP exam problem asks for the velocity or position at a specific time $t=b$, given an initial value at $t=a$, but does not require the full velocity or position function. In these cases, you do not need to solve for the constant of integration at all—you can use the Fundamental Theorem of Calculus Part 2 directly to get the result faster and avoid common constant-of-integration errors.

The relationships are straightforward:

• Change in velocity over $[a,b]$: $v(b)=v(a)+{\int }_a^{b}a(t)dt$
• Change in position over $[a,b]$: $s(b)=s(a)+{\int }_a^{b}v(t)dt$

This method is especially useful for multiple-choice questions where you only need a final numerical value, not an expression for the full function.

Worked Example

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

1. We only need $v(10)$, not the full velocity function, so we use the Fundamental Theorem form directly: $v(10)=v(0)+{\int }_0^{10}a(t)dt$.
2. Substitute the known values: $v(10)=100+{\int }_0^{10}2e^{0.1t}dt$.
3. Compute the definite integral: The antiderivative of $2e^{0.1t}$ is $20e^{0.1t}$, so evaluating from 0 to 10 gives $20e^1-20e^0=20(e-1)\approx 34.36$.
4. Add the initial velocity: $v(10)\approx 100+34.36=134.36$ 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.

5. Common Pitfalls (and how to avoid them)

• 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: When finding where $v(t)=0$ to split the integral for total distance, including roots of $v(t)=0$ that are outside the interval of integration. Why: Students solve for all roots automatically and split at all roots, even ones less than $a$ or greater than $b$, 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 of integration 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. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

A particle moving along the x-axis has acceleration given by $a(t)=-\sin t$ for $t\ge 0$. If the initial velocity is $v(0)=2$ and the initial position is $s(0)=5$, what is the position at $t=\pi$? A) $3+\pi$ B) $5+\pi$ C) $5-\pi$ D) $1+3\pi$

Worked Solution: First integrate acceleration to get velocity: $v(t)=\int -\sin tdt=\cos t+C$. Use the initial condition $v(0)=2$: $\cos (0)+C=1+C=2$, so $C=1$, giving $v(t)=\cos t+1$. Next integrate velocity to get position: $s(t)=\int (\cos t+1)dt=\sin t+t+D$. Use $s(0)=5$ to get $D=5$. Evaluate at $t=\pi$: $s(\pi )=\sin (\pi )+\pi +5=0+\pi +5=5+\pi$. The correct answer is B.

Question 2 (Free Response)

A particle moves along the y-axis with velocity given by $v(t)=t^3-6t^2+8t$ for $0\le t\le 4$. (a) Find the net displacement of the particle over the interval $[0,4]$. (b) Find the total distance traveled by the particle over the interval $[0,4]$. (c) Given that the initial position of the particle is $s(0)=-3$, find the position of the particle at all times when the velocity is equal to zero.

Worked Solution: (a) Net displacement is the definite integral of $v(t)$ over $[0,4]$: $$\begin{align*} \int_0^4 (t^3 - 6t^2 + 8t) dt &= \left[\frac{t^4}{4} - 2t^3 + 4t^2\right]_0^4 \ &= 64 - 128 + 64 = 0 \end{align*}$$ Net displacement is $\boxed{0}$.

(b) Factor $v(t)=t(t-2)(t-4)$, so the only turning point inside $[0,4]$ is at $t=2$. $v(t)>0$ on $(0,2)$ and $v(t)<0$ on $(2,4)$, so: $$\begin{align*} \text{Total Distance} &= \int_0^2 v(t) dt + \int_2^4 -v(t) dt \ &= 4 - (-4) = 8 \end{align*}$$ Total distance traveled is $\boxed{8}$ units.

(c) Velocity is zero at $t=0$, $t=2$, $t=4$. The general position function is $s(t)=\frac{t^4}{4}-2t^3+4t^2+C$. Using $s(0)=-3$, $C=-3$. Evaluating:

• $s(0)=-3$ (given)
• $s(2)=4-16+16-3=1$
• $s(4)=64-128+64-3=-3$

Positions when $v(t)=0$ are $\boxed{-3}$ at $t=0$, $\boxed{1}$ at $t=2$, and $\boxed{-3}$ at $t=4$.

Question 3 (Application / Real-World Style)

A hiker is walking along a straight trail leading away from their campsite. Their velocity in miles per hour is given by $v(t)=\frac{1}{t+1}+2\sin \left(\frac{\pi t}{4}\right)$ for $0\le t\le 4$ hours, where $t=0$ is the starting time at the campsite. How far is the hiker from the campsite after 4 hours of walking? Round your answer to the nearest tenth of a mile.

Worked Solution: The hiker starts at the campsite, so $s(0)=0$. The position at $t=4$ is: $$s(4)=s(0)+{\int }_0^4\left(\frac{1}{t+1}+2\sin \left(\frac{\pi t}{4}\right)\right)dt$$ The antiderivative is $\ln (t+1)-\frac{8}{\pi }\cos \left(\frac{\pi t}{4}\right)$. Evaluating from 0 to 4: $$\begin{align*} \left(\ln 5 - \frac{8}{\pi}\cos(\pi)\right) - \left(\ln 1 - \frac{8}{\pi}\cos(0)\right) &= \ln 5 + \frac{8}{\pi} + \frac{8}{\pi} \ &\approx 1.609 + 5.093 = 6.702 \end{align*}$$ After 4 hours of walking, the hiker is approximately 6.7 miles from their starting campsite.

7. Quick Reference Cheatsheet

8. 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 the relationship between position, velocity, and acceleration via integration, you will struggle to generalize this reasoning to other rate-of-change problems, which are common in both MCQ and FRQ sections of the AP exam. This topic also reinforces the core inverse relationship between differentiation and integration that unites the entire course. Follow-up topics to study next:

• Average value of a function
• Net change as the integral of a rate of change
• Area between two curves