4.2 Straight-Line Motion: Connecting Position, Velocity, and Acceleration

What is straight-line motion in AP Calculus?

Straight-line (rectilinear) motion connects three functions through derivatives: position $x(t)$, velocity $v(t) = x'(t)$, and acceleration $a(t) = v'(t) = x''(t)$. Once you can read signs and take derivatives, you can tell which direction an object moves, when it speeds up or slows down, and how far it is from the start.

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Why This Matters for the AP Calculus Exam

Motion problems are one of the most common ways the AP Calculus exam tests whether you understand what a derivative actually means. They show up in both multiple-choice and free-response settings, and they reward you for choosing the right procedure (differentiate to go from position to velocity to acceleration), reading signs correctly, and explaining your answer with correct units. Treating "rate of change" as a signal to differentiate is the core skill here, and the same structure carries into related rates and other applied problems later in this unit.

Key Takeaways

• Velocity is the derivative of position: $v(t) = x'(t)$. Acceleration is the derivative of velocity and the second derivative of position: $a(t) = v'(t) = x''(t)$.
• Velocity is signed. Positive velocity means moving in the positive direction; negative velocity means moving in the negative direction.
• Speed is the absolute value of velocity: speed $= |v(t)|$. Speed is never negative.
• An object speeds up when $v(t)$ and $a(t)$ have the same sign, and slows down when they have opposite signs.
• Units matter: if position is in meters and time is in seconds, velocity is in m/s and acceleration is in m/s$^2$.
• A particle changes direction where velocity changes sign (often where $v(t) = 0$), not just where velocity is large or small.

Derivatives and Motion

The derivative gives the instantaneous rate of change of a function at a specific point. That single idea is what lets you move between position, velocity, and acceleration.

Velocity and Speed

If $x(t)$ gives an object's position as a function of time, then its derivative $x'(t)$ gives the velocity at any instant. You will often see this written as $v(t)$, since velocity is the rate of change of position with respect to time.

Instantaneous velocity is always signed:

• When velocity is negative, the object is moving in the negative direction (often drawn as left).
• When velocity is positive, the object is moving in the positive direction (often drawn as right).

If you only want the speed, take the absolute value of the velocity. Speed has no direction, so it is never negative.

Acceleration

The derivative of velocity gives acceleration, since acceleration is the rate at which velocity changes. Because velocity is already the first derivative of position, acceleration is the second derivative of position:

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

• Positive acceleration means velocity is increasing over time.
• Negative acceleration means velocity is decreasing over time.
• Zero acceleration means velocity is constant.

A reliable way to decide whether an object is speeding up or slowing down is to compare the signs of velocity and acceleration:

• When $v(t)$ and $a(t)$ have the same sign, the object is speeding up.
• When $v(t)$ and $a(t)$ have opposite signs, the object is slowing down.

Rectilinear Motion: Practice

Problems

Question 1:

A particle moves along the x-axis. The function $x(t)$ gives the particle's position at any time $t \geq 0$.

$x(t) = 13t - 9$

What is the particle's acceleration $a(t)$ at $t = 3$?

Question 2:

A particle moves along the x-axis. The function $x(t)$ gives the particle's position at any time $t \geq 0$.

$x(t) = 4t^2 - 3t + 16$

What is the particle's velocity $v(t)$ at $t = 4$?

Solutions

Question 1:

Acceleration is the rate of change of velocity, which is the rate of change of position. So find the second derivative of $x(t) = 13t - 9$:

$a(t) = x''(t) = \frac{d}{dt}\left[\frac{d}{dt}[13t - 9]\right] = 0$

The acceleration is constant at $0$ for all $t \geq 0$, so the acceleration at $t = 3$ is $0$.

Question 2:

Velocity is the rate of change of position, so find the first derivative of $x(t) = 4t^2 - 3t + 16$:

$v(t) = x'(t) = \frac{d}{dt}[4t^2 - 3t + 16] = 8t - 3$

The velocity is $v(t) = 8t - 3$ for all $t \geq 0$, so its velocity at $t = 4$ is $8(4) - 3 = 29$.

How to Use This on the AP Calculus Exam

MCQ

• Translate the wording first. "Velocity" means take one derivative of position; "acceleration" means take two.
• Watch for sign questions. "Moving left" or "moving in the negative direction" means $v(t) < 0$, not that position is negative.
• For "speeding up" or "slowing down," check the signs of both $v(t)$ and $a(t)$ together. Same signs means speeding up; opposite signs means slowing down.
• To find when a particle changes direction, look for where $v(t) = 0$ and check that velocity actually switches sign there.

Free Response

• Show the derivative steps clearly so the reader can follow how you got velocity and acceleration. Clear work makes your reasoning easy to follow.
• Include units in your final answers (for example, m/s for velocity, m/s$^2$ for acceleration). Correct units are important for clear exam work.
• When a question asks you to interpret an answer, write a sentence in context, not just a number. For example, "at $t = 4$ the particle is moving in the positive direction at 29 units per second."

Common Trap

• Confusing speed and velocity. Velocity keeps its sign; speed is the absolute value. A particle with velocity $-29$ has a speed of $29$.

Common Misconceptions

• Negative acceleration always means slowing down. Not true. If velocity is also negative, the object is actually speeding up. Compare the signs of $v(t)$ and $a(t)$ instead of looking at acceleration alone.
• Speed and velocity are the same thing. Velocity is signed and shows direction; speed is the absolute value of velocity and is never negative.
• A particle stops moving wherever acceleration is zero. Zero acceleration only means velocity is constant, not zero. The particle is momentarily at rest where $v(t) = 0$, not where $a(t) = 0$.
• Position being negative means the object moves backward. Direction of motion depends on the sign of velocity, not the sign of position. An object at a negative position can still be moving in the positive direction.
• You always take only one derivative. Acceleration requires the second derivative of position. Read the question carefully to see how many derivatives you need.

Related AP Calculus Guides

• Unit 4 Overview: Contextual Applications of Differentiation
• 4.1 Interpreting the Meaning of the Derivative in Context
• 4.4 Intro to Related Rates
• 4.6 Approximating Values of a Function Using Local Linearity and Linearization
• 4.5 Solving Related Rates Problems
• 4.3 Rates of Change in Applied Contexts other than Motion