Physics Slides #1: Comprehensive Notes on Kinematics and ics and One Dimensional Motion

Representations of Motion

Motion Diagram: This is a multiple exposure picture that shows the positions of an object at different points in time.
Motion Graph: A graphical plot typically showing position versus time ( $x$ vs. $t$ ).
Kinematic Graphs: Velocity or acceleration graphs which plot velocity ( $v$ ) or acceleration ( $a$ ) against time ( $t$ ).
Equations: Mathematical functions that define the dynamics of the object, such as $x(t)$ and $v(t)$ .

Position, Displacement, and Distance

Position Vectors (1D): * $\vec{x}$ is the position vector relative to the origin ( $\vec{x} = 0$ ). The specific name of the variable (e.g., "x") is arbitrary. * The unit vector is denoted as $\hat{i}$ , which represents the direction. * The formula for a position vector is $\vec{x} = x\hat{i}$ , where $x$ is the "x-component" and can be positive or negative. * The origin ( $x=0$ ) is a special point used as a reference to allow for consistent problem solving.
Displacement (Change in Position): * Displacement ( $\Delta \vec{x}$ ) is always defined as "Final - Initial." * Vector representation for points 3 and 4: $\Delta \vec{x}_{34} = \vec{x}_4 - \vec{x}_3$ . * This can be rearranged to show the final position as the sum of initial position and change: $\vec{x}_4 = \vec{x}_3 + \Delta \vec{x}_{34}$ .
Distance vs. Displacement: * Displacement ( $\Delta \vec{x}$ ): The net change in position. If an object returns to its starting point ( $\vec{x}_{10} - \vec{x}_1 = 0$ ), the displacement is zero. * Distance ( $d$ ): The total path length traveled. If an object moves from point 1 to point 6 (distance $d$ ) and back to point 1 (distance $d$ ), the total distance is $2d$ .

Velocity and Motion Graphs

Average Velocity ( $v_{ave}$ ): * Defined as the displacement divided by the time interval: $\vec{v}_{ave} = \frac{\Delta \vec{x}}{\Delta t}$ . * Specific formula for a segment between $t_3$ and $t_4$ : $\vec{v}_{ave, 34} = \frac{\vec{x}_4 - \vec{x}_3}{t_4 - t_3} = \frac{\Delta \vec{x}_{34}}{\Delta t}$ . * Notation: $\vec{x}(t_1)$ is the position at a specific time, while $\vec{x}(t)$ is position as a function of any time.
Predicting the Future: Knowing the average velocity allows for the calculation of future positions: $\vec{x}_4 = \vec{v}_{ave} \cdot (t_4 - t_3) + \vec{x}_3$ .
Average Speed: Defined as total distance per unit time: $v_{ave} = \frac{\text{distance}}{\Delta t} = \frac{2d}{\Delta t}$ .
Instantaneous Velocity: * Fundamental Definition: $\vec{v}(t) = \frac{d}{dt} \vec{x}(t)$ . * On a position vs. time graph, velocity is the Slope of the line tangent to the curve at any point ( $\text{Slope} = \frac{\Delta x}{\Delta t}$ ).

The Constant Velocity Model (1D)

In this model, the velocity is constant, meaning the instantaneous velocity is equal to the average velocity for any time interval ( $v = v_{average}$ ).
The derivation for the position function is: * $\frac{d x(t)}{dt} = v = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$ * $x(t_2) = x(t_1) + v \cdot (t_2 - t_1)$
If setting $t_1 = 0$ and initial position $x(t_1) = x(0) = x_i$ , the function becomes: $x(t) = x_i + v \cdot t$ .
Graph Interpretations: * If x_i < 0 and v > 0, the object starts behind the origin and moves forward. * If x_i > 0 and v < 0, the object starts in front of the origin and moves backward.

Acceleration

Definition: Acceleration is the rate of change of velocity over change in time.
Average Acceleration ( $\vec{a}_{ave}$ ): $\vec{a}_{ave} = \frac{\vec{v}(t_f) - \vec{v}(t_i)}{t_f - t_i} = \frac{\Delta \vec{v}}{\Delta t}$ .
Instantaneous Acceleration ( $\vec{a}(t)$ ): * The limit as $\Delta t$ approaches zero: $\vec{a} = \frac{d}{dt} \vec{v}(t)$ . * It is the second derivative of position with respect to time: $\vec{a}(t) = \frac{d^2}{dt^2} \vec{x}(t)$ . * On a velocity vs. time graph, acceleration is the slope of the tangent line.

Constant Acceleration Kinematic Equations

For motion with constant acceleration ( $a$ ):

Velocity-Time Relation: $v = v_0 + at$
Position-Time Relation: $x = x_0 + v_0t + \frac{1}{2}at^2$
Velocity-Displacement Relation: $v^2 = v_0^2 + 2a(x - x_0)$
Average Velocity Displacement: $x - x_0 = \frac{v_f + v_i}{2}t$
Average Velocity Definition: $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x - x_0}{t}$
Arithmetic Average Velocity: $\bar{v} = \frac{v_0 + v}{2}$

General Forms using initial time ( $t_i$ ):

$x(t) = \frac{1}{2}a \cdot (t - t_i)^2 + v_i \cdot (t - t_i) + x_i$
$v^2(t) - v^2(t_i) = 2a \cdot (x(t) - x(t_i))$
$x(t) - x(t_i) = \frac{v(t) + v(t_i)}{2}(t - t_i)$

Problem-Solving Methodology

Visualize: Use pictures and motion graphs to understand the problem and define variables.
Identify Variables: List all known variables, unknown variables, and what is specifically being asked for.
Identify Physics: Determine the physical principles (e.g., constant velocity vs. constant acceleration).
Consider Relations: Identify equations containing the knowns and unknowns. For complex motions, break them into constant acceleration pieces.
Solvability Check: Determine if the number of equations matches the number of unknowns.
Execution and Units: Solve algebraically before plugging in numbers. Always check units and perform a reasonableness check.
Warning: Do NOT start by simply hunting for any equation that fits the numbers.

Example Exercises and Case Studies

Example: Ralph Driving to the Mont

Scenario: Ralph drives $700\,\text{m}$ from University at a speed limit of $30\,\text{mi/h}$ .
Conversion: * $v = 30\,\text{mi/h} \times \frac{1000\,\text{m}}{0.621\,\text{mi}} \times \frac{1\,\text{hr}}{3600\,\text{s}} = 13.4\,\text{m/s}$ .
Time Calculation: * Given: $t_i = 0, x_i = 0, x_f = 700\,\text{m}$ . * $t_f = \frac{x_f}{v} = \frac{700\,\text{m}}{13.4\,\text{m/s}} = 52.2\,\text{s}$ .

Example: Ralph Catching George

Scenario: George drives at $10\,\text{m/s}$ starting $100\,\text{m}$ ahead of Ralph ( $x_0 = 100\,\text{m}$ ). Ralph starts at the origin ( $x_R = 0$ ). They both travel to "The Mont" ( $x_f = 700\,\text{m}$ ).
Interaction: They must meet at the same position ( $x_f$ ) at the same time ( $t_f$ ).
Equations: * Ralph: $x_f = v_R \cdot t_f$ * George: $x_f = v_G \cdot t_f + x_0$
Solution: * Solve for $t_f$ : $t_f = \frac{x_f - x_0}{v_G} = \frac{700 - 100}{10} = 60\,\text{s}$ . * Solve for $v_R$ : $v_R = \frac{x_f}{t_f} = \frac{x_f}{\frac{x_f - x_0}{v_G}} = v_G \frac{x_f}{x_f - x_0}$ . * $v_R = 10 \frac{700}{700 - 100} = 11.7\,\text{m/s}$ .

Example: Airplane Landing

Scenario: Airplane lands with $v_0 = 70.0\,\text{m/s}$ , accelerates opposite to motion at $a = -1.50\,\text{m/s}^2$ for $40.0\,\text{s}$ .
Calculation: * $v_f = v_0 + at$ * $v_f = 70\,\text{m/s} + (-1.5\,\text{m/s}^2 \times 40\,\text{s}) = 10\,\text{m/s}$ .

Example: Cheetah and Gazelle

Scenario: Gazelle passes at constant $v_G = 10\,\text{m/s}$ . Cheetah accelerates from rest at $a_C = 4\,\text{m/s}^2$ .
Calculation (a): How long to catch the gazelle? * $x_G = v_G t$ * $x_C = \frac{1}{2} a_C t^2$ * $v_G t = \frac{1}{2} a_C t^2 \Rightarrow t = \frac{2v_G}{a_C} = \frac{2 \times 10}{4} = 5\,\text{s}$ .
Calculation (b): Displacement? * $x = \frac{1}{2} (4\,\text{m/s}^2)(5\,\text{s})^2 = 50\,\text{m}$ .

Questions & Discussion

Interactive Question 1: In a motion diagram, what are the red dots? * Answer: A) Position.
Interactive Question 2: In a coordinate system, what are the green arrows between positions? * Answer: B) Displacement.
Classroom Exercise (Motion Graphs): Describe motion on an $x$ vs $t$ graph without using "Speed" or "Velocity." * If displacement increases at an increasing rate, the object is accelerating. * If displacement increases at a constant rate, the object moves at a steady pace. * If displacement increases at a decreasing rate, the object is decelerating (velocity is decreasing).
Graph Signs Problem: Describe the signs of $v$ and $a$ in various states: 1. $v = 0, a = 0$ : At rest. 2. v > 0, a > 0: Moving forward, speeding up. 3. v > 0, a = 0: Moving forward at constant speed. 4. v > 0, a < 0: Moving forward, slowing down. 5. v = 0, a < 0: Momentarily at rest, accelerating backward. 6. v < 0, a < 0: Moving backward, speeding up (in negative direction).
Drag Race Exercises: * A) Given $t = 10.2\,\text{s}$ , $x = 400\,\text{m}$ , find $a$ : Use $x_f = \frac{1}{2}a \cdot t_f^2$ . * B) Given $t = 12.5\,\text{s}$ , $x = 400\,\text{m}$ , find $v_f$ : Use $x_f - x_i = \frac{v_f + v_i}{2}(t_f - t_i)$ . * C) Given $x = 75\,\text{m}$ , $v_0 = 0$ , $v_f = 30\,\text{m/s}$ , find $a$ : Use $v_f^2 - v_i^2 = 2a(x_f - x_i)$ . * D) Given $a = 7.5\,\text{m/s}^2$ , $x = 400\,\text{m}$ , find final speed: Use $v_f^2 - v_i^2 = 2a(x_f - x_i)$ . * E) Given $t = 6.5\,\text{s}$ , $v_i = 0$ , $v_f = 30\,\text{m/s}$ , find distance: Use $x_f - x_i = \frac{v_f + v_i}{2}(t_f - t_i)$ .