Dynamics: Reference Tables and Quick Lookups

Abstract

This article consolidates core kinematic relationships from introductory dynamics into a unified reference. We present the definitions and formulas for position, velocity, and acceleration functions, along with worked examples demonstrating how these quantities relate through differentiation and integration. The material is organized for quick lookup and practical problem-solving.

Background

Kinematics—the study of motion without regard to forces—forms the foundation of dynamics. Three quantities dominate kinematic analysis: position (where an object is), velocity (how fast it is moving), and acceleration (how quickly its velocity changes). Each is defined as a rate of change with respect to time, creating a natural hierarchy of derivatives ^{[position-function]}, ^{[velocity-function]}, ^{[acceleration-function]}.

In practice, a dynamics problem often provides one of these quantities as a function of time, and the task is to find the others. Understanding the relationships between them—and having quick access to the formulas—accelerates problem-solving and reduces errors.

Key Results

Definition: Acceleration

Acceleration quantifies the rate at which velocity changes over time ^{[acceleration]}:

$a = \frac{dv}{dt}$

where $a$ is acceleration, $v$ is velocity, and $t$ is time. Acceleration can be constant or vary with time. When acceleration is expressed as a function of time, integrating it yields velocity; differentiating velocity yields acceleration.

Definition: Velocity

Velocity is the rate of change of position with respect to time ^{[velocity-function]}:

$v(t) = \frac{ds}{dt}$

where $s$ is position. A positive velocity indicates motion in the positive direction; negative velocity indicates motion in the opposite direction. The points where $v(t) = 0$ mark instantaneous stops and are critical for determining total distance traveled and direction changes.

Definition: Position

Position describes the location of a particle along a line as a function of time ^{[position-function]}:

$s(t) = \text{(function of } t \text{)}$

Position is the fundamental quantity; velocity and acceleration are its first and second derivatives, respectively.

Average Velocity

When only the net displacement over a time interval is known, average velocity provides a summary measure ^{[average-velocity]}:

$v_{\text{avg}} = \frac{\Delta s}{\Delta t}$

where $\Delta s$ is the change in position and $\Delta t$ is the change in time. Average velocity differs from instantaneous velocity when motion is non-uniform.

Worked Examples

Example 1: Deriving Velocity from Position

Given: Position function $s(t) = 10t^2 + 20$ (in millimeters) ^{[position-function]}

Find: Velocity as a function of time.

Solution:

Differentiate the position function with respect to time:

$v(t) = \frac{ds}{dt} = \frac{d}{dt}(10t^2 + 20) = 20t \text{ mm/s}$

At $t = 0$ s, $v = 0$ mm/s (particle at rest).
At $t = 2$ s, $v = 40$ mm/s (particle moving in positive direction).

Example 2: Deriving Acceleration from Velocity

Given: Velocity function $v(t) = 2t - 6$ (in m/s) ^{[velocity-function]}

Find: Acceleration as a function of time.

Solution:

Differentiate the velocity function:

$a(t) = \frac{dv}{dt} = \frac{d}{dt}(2t - 6) = 2 \text{ m/s}^2$

The acceleration is constant at $2 \text{ m/s}^2$. The particle is always speeding up in the positive direction (when $v > 0$) or slowing down in the negative direction (when $v < 0$).

Finding the stopping point:

Set $v(t) = 0$:

$2t - 6 = 0 \implies t = 3 \text{ s}$

At $t = 3$ s, the particle momentarily stops before reversing direction.

Example 3: Integrating Acceleration to Find Velocity

Given: Acceleration function $a(t) = 2t - 1$ (in m/s²) and initial condition $v(0) = 5$ m/s ^{[acceleration-function]}

Find: Velocity as a function of time.

Solution:

Integrate the acceleration function:

$v(t) = \int a(t) \, dt = \int (2t - 1) \, dt = t^2 - t + C$

Apply the initial condition $v(0) = 5$:

$5 = 0 - 0 + C \implies C = 5$

Therefore:

$v(t) = t^2 - t + 5 \text{ m/s}$

At $t = 1$ s, $v(1) = 1 - 1 + 5 = 5$ m/s.
At $t = 2$ s, $v(2) = 4 - 2 + 5 = 7$ m/s.

Quick Reference Table

Quantity	Definition	Formula
Position	Location along a line	$s(t)$ (given or derived)
Velocity	Rate of change of position	$v(t) = \frac{ds}{dt}$
Acceleration	Rate of change of velocity	$a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
Average Velocity	Net displacement per unit time	$v_{\text{avg}} = \frac{\Delta s}{\Delta t}$

References

• ^{[acceleration]}
• ^{[velocity-function]}
• ^{[position-function]}
• ^{[acceleration-function]}
• ^{[average-velocity]}

AI Disclosure

This article was drafted with AI assistance from class notes (Zettelkasten). All mathematical claims and definitions are cited to source notes. The worked examples and table are original compositions based on the cited material. The author reviewed all content for technical accuracy before publication.