# Straight-line motion: position, velocity, acceleration

> AP Calculus BC · Unit 4: Contextual Applications of Differentiation
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u4-straight-line-motion-position-velocity/

This subtopic covers derivative relationships between position, velocity, and acceleration for one-dimensional straight-line motion. You will learn to interpret direction, speed, acceleration, and identify direction changes for moving particles.

**Prerequisites:** First and second derivative computation for common functions; Sign testing of functions on intervals; Factoring polynomials to find roots

## Learning objectives

- Relate position, velocity, and acceleration via differentiation
- Interpret the sign of velocity to determine direction of motion
- Determine when a particle is speeding up or slowing down
- Identify all times a particle changes direction on a straight line
- Distinguish between velocity and speed

## Position and Instantaneous Velocity

Straight-line motion models a particle constrained to move along a single axis, with all motion quantities expressed as single-valued functions of time $t \geq 0$. This is a foundational application of derivatives to real-world rates, appearing on both multiple-choice and free-response sections of the AP Calculus BC exam.

**Position** — Displacement of the particle from a fixed origin at time $t$. Positive values mean the particle is to the right/forward of the origin, negative values mean left/backward.

*Notation:* $s(t)$

**Instantaneous Velocity** — The instantaneous rate of change of position with respect to time, calculated as the first derivative of position. The sign of velocity indicates direction of motion.

*Notation:* $v(t)$

$$v(t) = s'(t) = \lim_{\Delta t \to 0} \frac{s(t+\Delta t) - s(t)}{\Delta t}$$

$v(t) > 0$ means the particle moves right/forward, $v(t) < 0$ means it moves left/backward, and $v(t) = 0$ 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 $t \geq 0$, 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 position, so differentiate using the power rule:

   $$v(t) = s'(t) = 3t^2 - 12t + 9$$
2. Evaluate $v(t)$ at $t=2$:

   $$v(2) = 3(2)^2 - 12(2) + 9 = 12 - 24 + 9 = -3$$
3. Interpret the result: the velocity at $t=2$ is $-3$ cm/s. The negative sign indicates the particle is moving to the left.

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

## Acceleration and Speed Change

Acceleration is the instantaneous rate of change of velocity with respect to time, so it connects directly to both position and velocity via differentiation.

**Instantaneous Acceleration** — The rate of change of velocity with respect to time, equal to the derivative of velocity and the second derivative of position.

*Notation:* $a(t)$

$$a(t) = v'(t) = s''(t)$$

A common misconception is that positive acceleration means speeding up and negative acceleration means slowing down. This is incorrect. Whether a particle speeds up or slows down depends on the relative signs of $v(t)$ and $a(t)$:

- If $v(t)$ and $a(t)$ have the **same sign**: the particle is speeding up
- If $v(t)$ and $a(t)$ have **opposite signs**: the particle is slowing down

**Worked example:** For the same particle from the previous example, $s(t) = t^3 - 6t^2 + 9t + 2$ for $t \geq 0$, 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 know $v(t) = 3t^2 - 12t + 9$. Differentiate to get acceleration:

   $$a(t) = 6t - 12 \text{ cm/s}^2$$
2. $a(0.5) = -9$ cm/s², $v(0.5) = 3.75$ cm/s
3. $a(4) = 12$ cm/s², $v(4) = 9$ 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.

> **tip**
>
> Never assume positive acceleration means speeding up. Always compare the signs of $v(t)$ and $a(t)$ directly.

## Speed and Direction Change

**Speed** — The non-negative magnitude of velocity, with no direction associated with it.

*Notation:* $\text{speed}(t)$

$$\text{speed}(t) = |v(t)|$$

A common AP exam question asks for all times when a particle changes direction. For continuous velocity (always the case in AP problems), a particle can only change direction if velocity changes sign, which can only happen at points where $v(t) = 0$. Follow this process to confirm direction change:

1. Find all $t>0$ where $v(t) = 0$ (candidate times)
2. Test the sign of $v(t)$ on intervals on either side of each candidate
3. Confirm a sign change occurs: only then does the particle change direction

**Worked example:** For the position function $s(t) = t^3 - 6t^2 + 9t + 2$, $t \geq 0$, 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 is:

   $$\text{speed}(2) = |v(2)| = 3 \text{ cm/s}$$
2. Find candidate times by setting $v(t) = 0$:

   $$3t^2 - 12t + 9 = 3(t-1)(t-3) = 0 \implies t=1, t=3$$
3. Test the sign of $v(t)$ on intervals: $0 < t < 1$, $v>0$; $1 < t < 3$, $v<0$; $t>3$, $v>0$.
4. Velocity changes sign at both $t=1$ and $t=3$, so the particle changes direction at both times.

> **tip**
>
> Never automatically conclude direction change just because $v(t) = 0$. Always check for a sign change.

## Concept Check

**Check your understanding**

Test your understanding with these AP-style questions:

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

   - $-e^{-\pi}$
   - $-e^{-\pi} + 1$
   - $-e^{-\pi} - 1$
   - $e^{-\pi}$

   *Answer:* $-e^{-\pi}$

   *Why:* Correct. Use the product rule to differentiate: $v(t) = e^{-t}(-\sin t + \cos t)$, so substituting $t=\pi$ gives $v(\pi) = -e^{-\pi}$.

2. A particle moves along the $x$-axis for $t \geq 0$ 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. (b) Find all times $t \geq 0$ where the particle is at rest. (c) At $t=2$, is the particle speeding up or slowing down? Justify your answer.

   *Why:* A particle is at rest if and only if $v(t) = 0$, regardless of position or acceleration. Always compare signs of $v$ and $a$ to determine speed change.

## Common pitfalls

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

## Cheatsheet

| Category | Formula / Rule | Notes |
| --- | --- | --- |
| Position | $s(t)$ | Displacement from origin; positive = right/forward, negative = left/backward |
| Instantaneous Velocity | $v(t) = s'(t)$ | Sign indicates direction; zero = particle is at rest |
| Instantaneous Acceleration | $a(t) = v'(t) = s''(t)$ | Rate of change of velocity; sign indicates if velocity is increasing/decreasing |
| Speed | $\text{speed}(t) = \|v(t)\|$ | Non-negative magnitude of velocity; no direction |
| Speed Change Rule | Same sign $v(t), a(t)$ = speeding up; opposite = slowing down | Never use only acceleration sign to determine this |
| Direction Change Condition | Change direction iff $v(t)=0$ and $v(t)$ changes sign | $a(t)=0$ does not imply direction change |
| Particle at Rest | $v(t) = 0$ | Independent of position or acceleration values |

## What's next

This topic is the foundational prerequisite for all motion-related content later in the AP Calculus BC course. Immediately after mastering 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 here, 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 contributes significantly to your total exam score. The same relationships you learned here extend directly to higher-dimensional motion, so mastering the one-dimensional case makes advanced motion topics much easier to learn.

- [Rates of change in applied contexts other than motion](https://www.owlsprep.com/study/ap-calculus-bc-u4-rates-of-change-in-applied/)
- [Introduction to Related Rates](https://www.owlsprep.com/study/ap-calculus-bc-u4-introduction-to-related-rates/)
- [Solving Related Rates Problems](https://www.owlsprep.com/study/ap-calculus-bc-u4-solving-related-rates-problems/)

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