Dynamics: Key Theorems and Proofs

Abstract

This article develops the foundational relationships in particle kinematics through calculus. We establish how position, velocity, and acceleration relate via differentiation, derive average velocity from displacement, and demonstrate these concepts through a concrete worked example. The treatment is rigorous but accessible, suitable for undergraduate engineering mechanics.

Background

Dynamics concerns the motion of bodies under the influence of forces. Before analyzing forces, we must establish the kinematic framework—the purely geometric description of motion. This framework rests on three linked quantities: position, velocity, and acceleration ^{[position-function]}, ^{[velocity-function]}, ^{[acceleration]}.

In one-dimensional motion along a line, we describe a particle's location by a position function $s(t)$, measured in some unit (e.g., millimeters or meters) at time $t$. From this single function, calculus yields both velocity and acceleration through successive differentiation.

Key Results

Theorem 1: Velocity as the Derivative of Position

Statement: If a particle's position is given by $s(t)$, then its velocity is ^{[velocity-function]}: $v(t) = \frac{ds}{dt}$

This is the instantaneous rate of change of position. Velocity is a signed quantity: positive values indicate motion in the positive direction, negative values indicate motion in the negative direction.

Significance: The velocity function reveals not only how fast the particle moves, but also when it changes direction. Points where $v(t) = 0$ are turning points, critical for determining total distance traveled versus net displacement.

Theorem 2: Acceleration as the Derivative of Velocity

Statement: If a particle's velocity is given by $v(t)$, then its acceleration is ^{[acceleration]}: $a(t) = \frac{dv}{dt}$

Acceleration quantifies how quickly velocity changes. It answers the question: is the particle speeding up or slowing down?

Significance: Acceleration is the bridge between kinematics and dynamics. In dynamics proper, Newton's second law relates acceleration to applied forces. Here, we treat acceleration as a kinematic quantity derived from motion data.

Theorem 3: Average Velocity over an Interval

Statement: Over a time interval from $t_1$ to $t_2$, the average velocity is ^{[average-velocity]}: $v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

This is the total displacement divided by elapsed time.

Significance: Average velocity summarizes motion over an interval without regard to intermediate variations. It is useful for quick estimates and for understanding overall behavior when instantaneous velocity is unknown or difficult to measure.

Corollary: Chain Rule in Kinematics

Combining Theorems 1 and 2 yields a second-order relationship: $a(t) = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{ds}{dt}\right) = \frac{d^2s}{dt^2}$

Acceleration is the second derivative of position. This relationship is fundamental to solving differential equations in dynamics.

Worked Example

Consider a particle whose position is given by ^{[position-function]}: $s(t) = 10t^2 + 20 \text{ (mm)}$

Step 1: Find the velocity function.

Differentiate with respect to time: $v(t) = \frac{ds}{dt} = 20t \text{ (mm/s)}$

Step 2: Find the acceleration function.

Differentiate velocity: $a(t) = \frac{dv}{dt} = 20 \text{ (mm/s}^2\text{)}$

Note that acceleration is constant in this example.

Step 3: Calculate average velocity from $t = 0$ to $t = 2$ seconds.

Position at $t = 0$: $s(0) = 10(0)^2 + 20 = 20$ mm

Position at $t = 2$: $s(2) = 10(2)^2 + 20 = 40 + 20 = 60$ mm

Displacement: $\Delta s = 60 - 20 = 40$ mm

Time interval: $\Delta t = 2 - 0 = 2$ s

Average velocity: $v_{\text{avg}} = \frac{40}{2} = 20 \text{ mm/s}$

Step 4: Verify using instantaneous velocity.

At $t = 0$: $v(0) = 20(0) = 0$ mm/s

At $t = 2$: $v(2) = 20(2) = 40$ mm/s

Since acceleration is constant, the average velocity equals the arithmetic mean of initial and final velocities: $v_{\text{avg}} = \frac{v(0) + v(2)}{2} = \frac{0 + 40}{2} = 20 \text{ mm/s}$

This confirms our result.

References

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

AI Disclosure

This article was drafted with AI assistance from class notes (Zettelkasten). All mathematical statements and theorems are grounded in cited source material. The worked example and explanatory text were generated and structured by an AI language model under human direction. The author retains responsibility for technical accuracy and has verified all claims against the source notes.