Dynamics: Worked Example Walkthroughs

Abstract

This article demonstrates how to connect fundamental kinematic concepts—position, velocity, and acceleration—through concrete worked examples. By systematically deriving and interpreting functions that describe particle motion, we illustrate the calculus-based reasoning that underpins classical dynamics. The approach emphasizes conceptual understanding alongside computational technique.

Background

Dynamics is the study of motion and the forces that produce it. At its foundation lies kinematics: the description of motion without reference to causes. Three quantities form the core of kinematic analysis: position (where an object is), velocity (how fast it is moving), and acceleration (how its velocity is changing) ^{[acceleration]}.

These quantities are related through differentiation. Position as a function of time, $s(t)$, yields velocity when differentiated with respect to time. Velocity, in turn, yields acceleration when differentiated. This hierarchical relationship—position → velocity → acceleration—is the backbone of kinematic problem-solving ^{[velocity-function]}, ^{[acceleration]}.

In practice, we often begin with either a position function or an acceleration function and work outward to find the others. Understanding how to move fluidly between these representations is essential for analyzing real-world motion.

Key Results

Differentiation chain in kinematics:

Given a position function $s(t)$, we obtain:

• Velocity: $v(t) = \frac{ds}{dt}$ ^{[velocity-function]}
• Acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$ ^{[acceleration]}

Average velocity over an interval:

For any time interval $[t_1, t_2]$, 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 provides a single summary value for motion over a period, even when instantaneous velocity varies ^{[average-velocity]}.

Critical points in motion:

Velocity zeros (where $v(t) = 0$) mark turning points where the particle changes direction. These are essential for computing total distance traveled, as distance accumulates the absolute value of displacement across direction changes ^{[velocity-function]}.

Worked Examples

Example 1: Deriving Velocity and Acceleration from Position

Given: A particle's position is described by ^{[position-function]}: $s(t) = 10t^2 + 20 \text{ mm}$

Find: The velocity and acceleration functions.

Solution:

Differentiate the position function to obtain velocity ^{[velocity-function]}: $v(t) = \frac{ds}{dt} = 20t \text{ mm/s}$

Differentiate the velocity function to obtain acceleration ^{[acceleration-function]}: $a(t) = \frac{dv}{dt} = 20 \text{ mm/s}^2$

Interpretation: The acceleration is constant at 20 mm/s². The velocity increases linearly with time, starting at zero when $t = 0$. Since $v(t) = 20t$ is always non-negative for $t \geq 0$, the particle never reverses direction—it continuously accelerates in the positive direction ^{[velocity-function]}.

---

Example 2: Finding Turning Points and Distance Traveled

Given: A particle's velocity is ^{[velocity-function]}: $v(t) = 2t - 6 \text{ m/s}$

Find: When does the particle stop? Over the interval $[0, 5]$ s, what is the total distance traveled?

Solution:

Set $v(t) = 0$ to find turning points: $2t - 6 = 0 \implies t = 3 \text{ s}$

The particle stops at $t = 3$ s. For $0 \leq t < 3$, velocity is negative (motion in the negative direction). For $t > 3$, velocity is positive (motion in the positive direction) ^{[velocity-function]}.

To find total distance, we must integrate the absolute value of velocity. First, find the position function by integrating velocity: $s(t) = \int (2t - 6) \, dt = t^2 - 6t + C$

Assuming $s(0) = 0$, we have $C = 0$, so $s(t) = t^2 - 6t$.

Evaluate position at the critical points:

• $s(0) = 0$
• $s(3) = 9 - 18 = -9$ m
• $s(5) = 25 - 30 = -5$ m

Total distance is the sum of absolute displacements: $\text{Distance} = |s(3) - s(0)| + |s(5) - s(3)| = |-9 - 0| + |-5 - (-9)| = 9 + 4 = 13 \text{ m}$

Interpretation: Although the particle ends at position $-5$ m (a net displacement of $-5$ m), it travels a total distance of 13 m because it reverses direction at $t = 3$ s ^{[velocity-function]}.

---

Example 3: Computing Average Velocity

Given: The position function from Example 1: $s(t) = 10t^2 + 20$ mm. Consider the interval $[1, 3]$ s.

Find: The average velocity over this interval.

Solution:

Evaluate position at the endpoints:

• $s(1) = 10(1)^2 + 20 = 30$ mm
• $s(3) = 10(3)^2 + 20 = 110$ mm

Apply the average velocity formula ^{[average-velocity]}: $v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{110 - 30}{3 - 1} = \frac{80}{2} = 40 \text{ mm/s}$

Interpretation: Over the 2-second interval from $t = 1$ to $t = 3$, the particle's position changes by 80 mm. The average rate of change is 40 mm/s. Note that the instantaneous velocity at $t = 1$ is $v(1) = 20$ mm/s and at $t = 3$ is $v(3) = 60$ mm/s; the average lies between these values ^{[average-velocity]}.

---

References

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

AI Disclosure

This article was drafted with the assistance of an AI language model. All mathematical claims and conceptual statements are grounded in cited class notes derived from Engineering Mechanics: Dynamics (Hibbeler, 14th ed.). The worked examples were constructed and verified by the AI to illustrate the cited concepts; they do not appear verbatim in the source material. The author reviewed the final text for technical accuracy and clarity.