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

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: position functions, instantaneous velocity as the first derivative of position, instantaneous acceleration as the second derivative of position, speed, interpreting direction of motion, identifying when a particle is at rest, and sign conventions for straight-line motion.

You should already know: Basic derivative rules including the chain rule. Interpreting the sign of a derivative on an interval. Solving polynomial and trigonometric equations for 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 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 Straight-line motion: position, velocity, acceleration?

Straight-line motion describes the movement of a particle or object along a single straight axis (almost always the x-axis for AP problems), where signs of all quantities encode direction (positive = right/up, negative = left/down per standard AP convention). This topic is part of Unit 4: Contextual Applications of Differentiation, which contributes 10-15% of the total AP Calculus AB exam score per the official College Board CED. It appears regularly on both multiple-choice (MCQ) and free-response (FRQ) sections, typically as 1-2 MCQ questions and one or more parts of a longer FRQ, testing both computational and conceptual understanding. The core idea is that differentiation describes how position changes over time: velocity is the instantaneous rate of change of position, and acceleration is the instantaneous rate of change of velocity. This topic connects calculus to real-world kinematics and builds the foundation for later applications of integration to motion and related rates problems.

2. Position and Instantaneous Velocity

Position of a particle at time $t$ is given by a function $s(t)$, which returns the coordinate of the particle on the straight axis, with $t\ge 0$ (time is always non-negative in AP motion problems). Displacement between $t=a$ and $t=b$ is $s(b)-s(a)$, which measures net change in position, distinct from total distance traveled.

Instantaneous velocity $v(t)$ at time $t$ is defined as the first derivative of the position function with respect to time, because it measures the instantaneous rate of change of position: $$v(t)=s^{\prime }(t)=\frac{ds}{dt}$$

The sign of velocity directly tells us direction of motion: if $v(t)>0$, the particle is moving in the positive direction (right or up); if $v(t)<0$, it is moving in the negative direction (left or down); if $v(t)=0$, the particle is instantaneously at rest. Common exam questions ask for velocity at a specific time, or to find all times when the particle is at rest.

Worked Example

The position of a particle moving along the x-axis is given by $s(t)=t^3-6t^2+9t+2$ for $t\ge 0$, where $s$ is measured in centimeters and $t$ in seconds. (a) Find the velocity of the particle at $t=2$ seconds. (b) At what times $t>0$ is the particle at rest?

1. Velocity is the first derivative of position, so we differentiate using the power rule: $v(t)=s^{\prime }(t)=3t^2-12t+9$.
2. For part (a), substitute $t=2$ into $v(t)$: $v(2)=3(2{)}^2-12(2)+9=12-24+9=-3$ cm/s.
3. For part (b), the particle is at rest when $v(t)=0$, so we set the derivative equal to zero: $3t^2-12t+9=0$.
4. Factor the quadratic: $3(t^2-4t+3)=3(t-1)(t-3)=0$, giving solutions $t=1$ and $t=3$.
5. Both solutions satisfy $t>0$, so the particle is at rest at $t=1$ second and $t=3$ seconds.

Exam tip: Always include units in your final answer for FRQ motion problems. AP graders dock 1 point for missing or incorrect units on contextual questions, so add units to every final answer by default.

3. Acceleration: Second Derivative of Position

Acceleration $a(t)$ is the instantaneous rate of change of velocity with respect to time, so it is the derivative of velocity, which makes it the second derivative of the original position function: $$a(t)=v^{\prime }(t)=\frac{dv}{dt}=s^{\prime \prime }(t)=\frac{d^{2}s}{d{t}^2}$$

The sign of acceleration tells us whether velocity is increasing or decreasing at a given time: if $a(t)>0$, velocity is increasing; if $a(t)<0$, velocity is decreasing. A key conceptual skill tested heavily on the AP exam is determining if a particle is speeding up or slowing down at a given time. This rule relies on both velocity and acceleration: if $v(t)$ and $a(t)$ have the same sign (both positive or both negative), the particle is speeding up; if they have opposite signs, the particle is slowing down.

Worked Example

For the particle with position $s(t)=t^3-6t^2+9t+2$ ($t\ge 0$, units cm, s), find the acceleration at $t=1.5$ seconds, and determine if the particle is speeding up or slowing down at that time.

1. We already found velocity from the previous example: $v(t)=3t^2-12t+9$. Calculate $v(1.5)=3(2.25)-12(1.5)+9=-2.25$ cm/s, so velocity is negative at $t=1.5$.
2. Acceleration is the derivative of velocity, so differentiate $v(t)$ to get $a(t)=6t-12$ cm/s².
3. Substitute $t=1.5$ to find $a(1.5)=6(1.5)-12=-3$ cm/s², so acceleration is also negative.
4. Since $v(t)$ and $a(t)$ have the same sign (both negative), the particle is speeding up at $t=1.5$ seconds.

Exam tip: Never assume that negative acceleration means the particle is slowing down. The sign of acceleration alone only describes how velocity is changing; you always need to compare the sign of acceleration to the sign of velocity to check if speed is increasing or decreasing.

4. Speed

Many students confuse velocity and speed, but they are distinct quantities for AP Calculus. Velocity is a 1D vector, meaning it has a sign that encodes direction. Speed is a scalar, meaning it is always non-negative and carries no information about direction. Speed is defined as the absolute value of velocity: $$\text{Speed}(t)=|v(t)|$$

While the definition is simple, AP regularly tests the derivative of speed (the rate of change of speed) and uses speed for conceptual questions about speeding up and slowing down. For $v(t)\ne 0$, the derivative of speed can be written as: $$\frac{d}{dt}|v(t)|=\frac{v(t)\cdot v^{\prime }(t)}{|v(t)|}=\frac{v(t)a(t)}{|v(t)|}$$

The sign of this derivative confirms the speeding up/slowing down rule: the derivative is positive (speed increasing) when $v(t)$ and $a(t)$ have the same sign, and negative (speed decreasing) when they have opposite signs.

Worked Example

A particle moving along the x-axis has velocity $v(t)=4\cos (2t)$ at time $t\ge 0$, where velocity is measured in m/s. (a) What is the speed of the particle at $t=\frac{\pi }{3}$? (b) What is the rate of change of speed at that time?

1. For part (a), speed is the absolute value of velocity. First calculate $v\left(\frac{\pi }{3}\right)=4\cos \left(\frac{2\pi }{3}\right)=4\left(-\frac{1}{2}\right)=-2$ m/s.
2. Take the absolute value to get speed: $\left|v\left(\frac{\pi }{3}\right)\right|=|-2|=2$ m/s.
3. For part (b), first find acceleration: $a(t)=v^{\prime }(t)=-8\sin (2t)$. At $t=\frac{\pi }{3}$, $a\left(\frac{\pi }{3}\right)=-8\sin \left(\frac{2\pi }{3}\right)=-4\sqrt{3}$ m/s².
4. Use the derivative of speed formula: $\frac{d}{dt}|v(t)|=\frac{v(t)a(t)}{|v(t)|}=\frac{(-2)(-4\sqrt{3})}{|-2|}=4\sqrt{3}$ m/s². So the rate of change of speed is $4\sqrt{3}$ m/s².

Exam tip: Circle the word "speed" every time it appears in a question, to remind yourself to take the absolute value of velocity. It is one of the most common avoidable mistakes on AP motion questions.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Concluding that a particle changes direction at $t=a$ just because $v(a)=0$. Why: Students memorize "particle changes direction when velocity is zero" but forget that velocity must change sign at that point; a particle can stop momentarily without reversing direction. Correct move: Always test the sign of $v(t)$ on intervals on either side of any $v(t)=0$ solution before concluding direction changes.
• Wrong move: Stating that a particle is slowing down because acceleration is negative. Why: Students associate "negative acceleration" with "deceleration" and assume it means slowing down regardless of velocity's sign. Correct move: Always write down the sign of both $v(t)$ and $a(t)$ for the given time, then check for same (speeding up) or opposite (slowing down) signs.
• Wrong move: Writing velocity as the answer when asked for acceleration. Why: When working quickly, students forget that acceleration is the derivative of velocity, so they stop at the first derivative of position. Correct move: Write $s(t)\to v(t)=s^{\prime }\to a(t)=v^{\prime }=s^{\prime \prime }$ at the top of every motion problem to label derivatives clearly.
• Wrong move: Keeping negative time solutions when solving for when the particle is at rest. Why: Students solve the equation $v(t)=0$ but forget the context that $t\ge 0$ for all motion problems. Correct move: Cross out any negative solutions for $t$ immediately after solving, since time starts at $t=0$.
• Wrong move: Forgetting to take the absolute value of velocity when calculating speed. Why: Students are used to working with velocity, so they just copy the velocity value (including the negative sign) as the answer for speed. Correct move: Add a step to take absolute value immediately after calculating velocity whenever speed is requested.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

A particle moves along the x-axis with position function $s(t)=e^t-3t$ for $t\ge 0$. What is the speed of the particle at $t=1$? A) $e-3$ B) $3-e$ C) $e$ D) $3$

Worked Solution: First, recall that speed is the absolute value of velocity, and velocity is the first derivative of position. Differentiate $s(t)$ to get $v(t)=s^{\prime }(t)=e^t-3$. Substitute $t=1$ to find $v(1)=e-3\approx -0.282$, which is negative. Take the absolute value to get speed: $|e-3|=3-e$, since $e<3$. The correct answer is B.

Question 2 (Free Response)

A marble moves along a straight horizontal track with position given by $s(t)=t^4-10t^3+24t^2+5$ for $t\ge 0$, where $s$ is in meters and $t$ in seconds. (a) Find the velocity function $v(t)$, and calculate $v(2)$. (b) Find all non-negative values of $t$ for which the particle is at rest. (c) Determine if the particle is speeding up or slowing down at $t=1.5$. Justify your answer.

Worked Solution: (a) Velocity is the first derivative of position, so we use the power rule: $$v(t)=s^{\prime }(t)=4t^3-30t^2+48t$$ Substitute $t=2$: $$v(2)=4(8)-30(4)+48(2)=32-120+96=8\text{ m/s}$$

(b) The particle is at rest when $v(t)=0$. Factor $v(t)$: $$4t^3-30t^2+48t=2t(2t^2-15t+24)=0$$ Solutions are $t=0$, and roots from the quadratic formula: $$t=\frac{15\pm \sqrt{225-192}}{4}=\frac{15\pm \sqrt{33}}{4}$$ All solutions are non-negative, so the particle is at rest at $t=0$, $t\approx 2.31$ s, and $t\approx 5.19$ s.

(c) First find acceleration, the derivative of velocity: $a(t)=12t^2-60t+48$. At $t=1.5$: $v(1.5)=4(3.375)-30(2.25)+48(1.5)=18$ m/s (positive), and $a(1.5)=12(2.25)-60(1.5)+48=-15$ m/s² (negative). Since $v(t)$ and $a(t)$ have opposite signs, the particle is slowing down at $t=1.5$.

Question 3 (Application / Real-World Style)

An elevator in a tall building moves vertically along a straight line. Its position (height above the ground floor, in meters) at time $t$ seconds after starting its ascent is given by $s(t)=-\frac{1}{15}{t}^3+t^2$ for $0\le t\le 15$. What is the acceleration of the elevator 5 seconds after it starts moving? Interpret your answer in context.

Worked Solution: First, find velocity as the first derivative of position: $v(t)=s^{\prime }(t)=-\frac{1}{5}{t}^2+2t$. Acceleration is the derivative of velocity: $a(t)=v^{\prime }(t)=-\frac{2}{5}t+2$. Substitute $t=5$: $a(5)=-\frac{2}{5}(5)+2=-2+2=0$ m/s². Interpretation: 5 seconds after starting, the velocity of the elevator is not changing, meaning the elevator has finished speeding up from rest and is now moving at a constant speed before it begins slowing down for its destination floor.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundation for all motion problems that you will encounter for the rest of the AP Calculus AB course. Immediately next, you will learn how to use integration to find position from velocity, and calculate total distance traveled from velocity, which is a common FRQ topic that relies entirely on your ability to connect derivatives of position, velocity, and acceleration. Without mastering the relationships between these three functions from this chapter, solving integral motion problems will be extremely difficult, as you will not be able to correctly interpret the quantities you calculate. This topic also builds conceptual understanding of derivatives as rates of change, which is core to all contextual applications of differentiation, including related rates.

Follow-up topics for further study: Rates of change in applied contexts Integral applications to motion Related rates