Straight-line motion: position, velocity, acceleration — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The derivative relationships between position, velocity, and acceleration for one-dimensional straight-line motion; definitions of speed, direction of motion, rest, and direction change for moving particles on a line.

You should already know: How to compute first and second derivatives of common function types. How to test the sign of a function on an interval. How to factor polynomials to find roots.

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 Straight-line motion: position, velocity, acceleration?

Straight-line motion is the foundational application of derivatives to modeling real-world rate of change, and it appears in both multiple-choice (MCQ) and free-response (FRQ) sections of the AP Calculus BC exam. Per the AP Course and Exam Description (CED), this topic is part of Unit 4: Contextual Applications of Differentiation, which contributes 10–15% of the total AP exam score. In straight-line motion, an object or particle is constrained to move along a single axis (almost always the $x$-axis for problems), so all motion quantities (position, velocity, acceleration) are single-valued functions of time $t\ge 0$. The core insight uniting this topic is that velocity and acceleration are just instantaneous rates of change of position and velocity, respectively, so they are calculated directly via differentiation. AP exam questions on this topic range from routine derivative calculations to conceptual questions about interpreting direction and speed, and this content is a prerequisite for all later motion-related topics in the course, including parametric motion and integration of motion functions.

2. Position and Instantaneous Velocity

Position is defined as $s(t)$, the displacement of the particle from a fixed origin at time $t$. By convention, positive values of $s(t)$ mean the particle is to the right (or forward, for linear motion problems) of the origin, while negative values mean it is to the left (or backward). Velocity is the instantaneous rate of change of position with respect to time, so by the definition of the derivative, we get: $$v(t)=s^{\prime }(t)=\underset{\Delta t\to 0}{\lim }\frac{s(t+\Delta t)-s(t)}{\Delta t}$$ Unlike average velocity (which measures change over an interval of time), instantaneous velocity gives the speed and direction of the particle at a single moment in time, which is what the AP exam almost always asks for. The sign of $v(t)$ directly tells us direction: $v(t)>0$ means position is increasing, so the particle moves right/forward; $v(t)<0$ means position is decreasing, so the particle moves left/backward; $v(t)=0$ means the particle is momentarily at rest.

Worked Example

A particle moves along the $x$-axis with position function $s(t)=t^3-6t^2+9t+2$ for $t\ge 0$, where $s$ is measured in centimeters and $t$ in seconds. Find the velocity of the particle at $t=2$ seconds, and state the direction of motion at that time.

1. Velocity is defined as the first derivative of the position function, so we differentiate $s(t)$ using the power rule: $v(t)=s^{\prime }(t)=3t^2-12t+9$.
2. Evaluate $v(t)$ at $t=2$: $v(2)=3(2{)}^2-12(2)+9=12-24+9=-3$.
3. The velocity at $t=2$ is $-3$ cm/s.
4. The negative sign indicates the particle is moving to the left toward the origin at 2 seconds.

Exam tip: Always include units in your answer for FRQ questions, and explicitly connect the sign of velocity to direction—AP readers require the interpretation, not just a numerical value.

3. Acceleration and Speed Change

Acceleration is the instantaneous rate of change of velocity with respect to time, so it is the derivative of velocity and the second derivative of position: $$a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$$ Acceleration is measured in units of distance per time squared (e.g., cm/s²). A common misconception is that positive acceleration means speeding up and negative acceleration means slowing down, but this is incorrect. Acceleration only tells us how velocity is changing: the sign of acceleration relative to the sign of velocity determines if speed is increasing or decreasing. If $v(t)$ and $a(t)$ have the same sign, the particle is speeding up; if they have opposite signs, the particle is slowing down. This rule holds for all straight-line motion problems on the AP exam.

Worked Example

For the same particle from the previous example, $s(t)=t^3-6t^2+9t+2$ for $t\ge 0$, find the acceleration at $t=0.5$ and $t=4$, and determine if the particle is speeding up or slowing down at each time.

1. We already found $v(t)=3t^2-12t+9$, so acceleration is the derivative of velocity: $a(t)=6t-12$ cm/s².
2. Evaluate at $t=0.5$: $a(0.5)=6(0.5)-12=-9$ cm/s², and $v(0.5)=3(0.25)-12(0.5)+9=3.75$ cm/s.
3. Evaluate at $t=4$: $a(4)=6(4)-12=12$ cm/s², and $v(4)=3(16)-12(4)+9=9$ cm/s.
4. Compare signs: At $t=0.5$, $v$ is positive and $a$ is negative (opposite signs), so the particle is slowing down. At $t=4$, $v$ and $a$ are both positive (same sign), so the particle is speeding up.

Exam tip: Never assume that positive acceleration means speeding up and negative acceleration means slowing down. Always compare the signs of $v(t)$ and $a(t)$ directly to answer speeding up/slowing down questions.

4. Speed and Direction Change

Speed is defined as the magnitude of velocity, so it is always non-negative (it has no direction): $$\text{speed}(t)=|v(t)|$$ A common AP exam question asks for all times when a particle changes direction. A particle can only change direction if its velocity changes sign, and (for continuous velocity, which is always the case on AP problems) velocity can only change sign at points where $v(t)=0$. The process to find direction change is: (1) find all $t>0$ where $v(t)=0$, (2) test the sign of $v(t)$ on intervals on either side of each candidate, (3) confirm if a sign change occurs. If the sign changes, the particle changes direction; if not, it just stops momentarily and continues moving in the same direction.

Worked Example

For the position function $s(t)=t^3-6t^2+9t+2$, $t\ge 0$, find the speed at $t=2$, and determine all times where the particle changes direction.

1. Speed is the absolute value of velocity: we already found $v(2)=-3$ cm/s, so speed at $t=2$ is $|v(2)|=3$ cm/s.
2. Find all candidate times for direction change by setting $v(t)=0$: $3t^2-12t+9=3(t-1)(t-3)=0$, so candidates are $t=1$ and $t=3$.
3. Test the sign of $v(t)$ on each interval: $0<t<1$ gives $v>0$, $1<t<3$ gives $v<0$, and $t>3$ gives $v>0$.
4. Velocity changes sign at both $t=1$ and $t=3$, so the particle changes direction at $t=1$ second and $t=3$ seconds.

Exam tip: Never automatically conclude that a particle changes direction just because $v(t)=0$. Always check the sign of $v(t)$ on both sides—for example, $s(t)=(t-2{)}^4$ has $v(2)=0$ but $v$ never changes sign, so no direction change occurs.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Concluding a particle is speeding up because acceleration is positive, or slowing down because acceleration is negative. Why: Students confuse the sign of acceleration with the change in speed, relying on an intuitive rule that only holds when velocity is already positive. Correct move: Always compare the sign of $v(t)$ and $a(t)$: same sign = speeding up, opposite signs = slowing down, every time.
• Wrong move: Concluding a particle changes direction at any time where $a(t)=0$. Why: Students mix up the conditions for direction change, confusing when velocity is zero with when acceleration is zero. Correct move: Direction change requires a sign change in velocity, which can only occur at times where $v(t)=0$ (for continuous velocity). Always check zeros of velocity, not acceleration, for direction change.
• Wrong move: Reporting velocity instead of speed, or leaving a negative sign when asked for speed. Why: Students forget that speed is a magnitude, so they just write down $v(t)$ instead of $|v(t)|$. Correct move: When a question asks for speed, immediately take the absolute value of your calculated velocity before reporting the answer.
• Wrong move: Automatically marking all times $v(t)=0$ as direction changes, without checking for a sign change. Why: Many problems have $v(t)=0$ at a point where velocity touches zero but does not cross it, so no direction change occurs. Correct move: After finding all $t$ where $v(t)=0$, always test the sign of $v(t)$ on intervals on either side of the critical time to confirm a sign change.
• Wrong move: Interpreting $s(t)<0$ as the particle being at rest. Why: Students confuse position and velocity, thinking negative position means no movement. Correct move: A particle is at rest if and only if $v(t)=0$, regardless of the value of position $s(t)$. Always check velocity, not position, to confirm if the particle is at rest.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

A particle moves along the $x$-axis with position $s(t)=e^{-t}\sin t$ for $t\ge 0$. What is the velocity of the particle at $t=\pi$? A. $-e^{-\pi }$ B. $-e^{-\pi }+1$ C. $-e^{-\pi }-1$ D. $e^{-\pi }$

Worked Solution: Velocity is the first derivative of position, so we use the product rule to differentiate $s(t)$: $v(t)=-e^{-t}\sin t+e^{-t}\cos t=e^{-t}(-\sin t+\cos t)$. Substitute $t=\pi$: $\sin \pi =0$ and $\cos \pi =-1$, so $v(\pi )=e^{-\pi }(-0+(-1))=-e^{-\pi }$. The correct answer is A.

Question 2 (Free Response)

A particle moves along the $x$-axis for $t\ge 0$ with position given by $s(t)=2t^3-15t^2+24t+10$, where $s$ is measured in meters and $t$ in minutes. (a) Find the velocity and acceleration functions for the particle. (b) Find all times $t\ge 0$ where the particle is at rest. (c) At $t=2$, determine if the particle is speeding up or slowing down. Justify your answer.

Worked Solution: (a) Velocity is the first derivative of position: $$v(t)=s^{\prime }(t)=6t^2-30t+24\text{ m/min}$$ Acceleration is the derivative of velocity: $$a(t)=v^{\prime }(t)=12t-30{\text{ m/min}}^2$$ (b) The particle is at rest when $v(t)=0$. Factor $v(t)$: $6(t^2-5t+4)=6(t-1)(t-4)=0$. The non-negative solutions are $t=1$ and $t=4$, so the particle is at rest at $t=1$ minute and $t=4$ minutes. (c) Evaluate $v(2)=6(4)-30(2)+24=-12$ m/min, and $a(2)=12(2)-30=-6$ m/min². Both $v(2)$ and $a(2)$ are negative, so they have the same sign. Therefore, the particle is speeding up at $t=2$.

Question 3 (Application / Real-World Style)

A small remote-controlled car moves in a straight line along a test track. Its position relative to the starting line (measured in meters) at time $t$ (measured in seconds after starting) is given by $s(t)=-t^3+9t^2-18t+10$ for $0\le t\le 5$. An engineer wants to know when the car changes direction during the test run. Find all times in $0<t<5$ when the car changes direction, and interpret your result in context.

Worked Solution:

1. Velocity is the first derivative of position: $v(t)=-3t^2+18t-18=-3(t^2-6t+6)$.
2. Set $v(t)=0$ to find candidate times: solve $t^2-6t+6=0$ with the quadratic formula: $t=\frac{6\pm \sqrt{12}}{2}=3\pm \sqrt{3}$. Both values are between $0$ and $5$, so candidates are $t\approx 1.27$ s and $t\approx 4.73$ s.
3. Test the sign of $v(t)$: for $0<t<3-\sqrt{3}$, $v(t)<0$; for $3-\sqrt{3}<t<3+\sqrt{3}$, $v(t)>0$; for $3+\sqrt{3}<t<5$, $v(t)<0$. Velocity changes sign at both candidates.
4. Interpretation: The car changes direction twice during the test: once around 1.27 seconds after starting (from backward to forward motion) and again around 4.73 seconds after starting (from forward to backward motion).

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for all motion-related content later in the AP Calculus BC course. Immediately after this, you will study how to use integration to solve motion problems, connecting position, velocity, and acceleration via the Fundamental Theorem of Calculus. Without a solid grasp of the derivative relationships covered in this chapter, you will struggle to interpret and solve integrated motion problems, which appear frequently on the FRQ section of the BC exam. This topic also lays the groundwork for two-dimensional parametric and vector-valued motion, a key BC-only topic that makes up a significant portion of the exam score. The same derivative relationships you learned here extend directly to higher-dimensional motion, so mastering the one-dimensional case makes more advanced motion topics much easier to learn.

• Related rates
• Integration of motion functions
• Parametric and vector-valued motion