Quantitative One-Dimensional Kinematics Study Guide

Quantitative Kinematics in One Dimension

• The primary goal of one-dimensional (1D) kinematics is to establish a precise quantitative model of motion along a straight line.
• Motion can be described using multiple complementary representations:
  • Verbal descriptions of physical motion.
  • Data tables listing time and position coordinates.
  • Motion diagrams using dots and vector arrows.
  • Position-versus-time graphs ($x$ vs. $t$).
  • Algebraic equations expressing position as a function of time ($x(t)$).

Motion Diagrams and Velocity Vector Analysis

• A motion diagram represents the position of an object at equal, discrete time intervals ($\Delta t$).
• Dots: Indicate the spatial location of the object at each time step.
• Arrows (Vectors): Drawn attached to the dots to represent the object's velocity (magnitude and direction) at each position.
• Interpreting Motion Diagrams:
  • Constant Speed: Dots are spaced equally apart, and velocity vectors maintain a constant length.
  • Speeding Up: Spacing between consecutive dots increases, and velocity vectors become longer over time.
  • Slowing Down: Spacing between consecutive dots decreases, and velocity vectors become shorter over time.
  • Direction of Motion: Indicated by the direction in which the velocity vector arrows point.

• Analysis of Partial Motion Diagram (Leftward Motion):
  • Vector arrows point toward the left, establishing that the direction of motion is to the left.
  • The length of the vector arrows decreases from right to left, indicating that the object's speed is decreasing.
  • Physical Description: An object moving to the left and slowing down.

• Determining Missing Velocity Vectors:
  • When dots are spaced uniformly along a horizontal axis, the displacement in each equal time interval is identical.
  • Uniform displacement per unit time corresponds to motion at constant velocity.
  • Missing arrows for rightward uniform motion must be pointing to the right and remain a constant size.

Quantitative Motion Modeling: Google Maps Application

• Real-World Route Parameters:
  • Starting Location: Smith Hall, Rutgers University (Newark campus).
  • Destination: New Jersey Performing Arts Center (NJPAC).
  • Walking Distance ($\Delta x$): $0.6\,\text{miles}$.
  • Estimated Walking Time ($\Delta t$): $13\,\text{min}$ (primary route) or $15\,\text{min}$ (alternate route).
• Algorithmic Kinematics Model:
  • Navigation tools compute estimated travel durations by assuming a constant average human walking speed ($v_{\text{walk}}$).
  • Kinematic Relationship:     $\Delta t = \frac{\Delta x}{v_{\text{walk}}}$
  • Assuming an average walking velocity $v_{\text{walk}} \approx 1.2\,\text{m/s}$ ($\approx 2.7\,\text{mph}$), the duration is computed directly from path distance.

Empirical Motion Analysis: Dual Buggy Experiment

• Experimental Protocol:
  • Two battery-powered buggies (Buggy A and Buggy B) move across a flat surface.
  • Beanbags are placed along the path every $\Delta t = 2\,\text{s}$ to mark instantaneous buggy positions.
• Reference Frame and Coordinate System:
  • Origin ($x = 0\,\text{m}$): Defined at a fixed spatial reference point along the linear path.
  • Coordinate Axis: Positive directional axis established along the path of forward travel.
  • Data Collection Table:

• Key Conceptual Distinctions in Kinematics:
  • Instantaneous Position vs. Displacement:
  • Question 1: "Where is the buggy at $10\,\text{seconds}$?"
    • Refers to instantaneous position $x(10\,\text{s})$, a single coordinate point on the spatial axis at time reading $t = 10\,\text{s}$.
  • Question 2: "How far does the buggy travel in the first $10\,\text{seconds}$?"
    • Refers to displacement $\Delta x$ or total path distance $d$ over the interval $\Delta t = 10\,\text{s}$.
    • Mathematical definition:         $\Delta x = x(10\,\text{s}) - x(0\,\text{s})$
  • Time Instant vs. Time Interval:
  • A time instant ($t = 10\,\text{s}$) represents a single point on a clock.
  • A time interval ($\Delta t = 10\,\text{s}$) represents an elapsed duration ($\Delta t = t_f - t_i$).
  • Why Position and Distance differ:
  • Position $x(t)$ depends entirely on the choice of spatial origin.
  • Distance traveled equals final position $x(t)$ only if initial position $x(0\,\text{s}) = 0\,\text{m}$. If $x(0\,\text{s}) \neq 0\,\text{m}$, then x(10\,\text{s}) \neq \Delta x$.\n- **Calculating Speed:**\n - Speed is determined by calculating the change in position per unit time:\n    v = \frac{\Delta x}{\Delta t}\n\n# Graphical and Mathematical Kinematic Equations\n\n- **Data Visualization:**\n - Plotting position x(t)$on the vertical y-axis against time$t on the horizontal x-axis yields a position-versus-time graph.\n - A straight line on a position-time graph signifies motion at **constant velocity** (zero acceleration).\n- **Linear Equation Parameters:**\n  x(t) = m \cdot t + b\n - **Slope (m$):** Represents physical velocity ($v). \n    \text{Slope} = \frac{\Delta x}{\Delta t} = v\n - Unit: meters per second (\text{m/s}).\n - **Vertical Intercept (b$):** Represents initial position ($x_0$) at clock reading$t = 0\,\text{s}.\n - Unit: meters (\text{m}).\n- **General Kinematic Model for Constant Velocity:**\n  x(t) = x_0 + v \cdot t\n  \Delta x = v \cdot \Delta t\n\n# Matching Motion Diagrams, Position-Time Graphs, and Equations\n\n\n\n\n\n- **Quantitative Analysis of Graph Lines:**\n - **Line (A):**\n - Initial position: x_0 = +2.0\,\text{m}\n - Position at t = 5.0\,\text{s}$:$x = +12.0\,\text{m}\n - Slope / Velocity: \n      v_A = \frac{12.0\,\text{m} - 2.0\,\text{m}}{5.0\,\text{s} - 0.0\,\text{s}} = +2.0\,\text{m/s}\n - Equation: \n      x_A(t) = (2.0\,\text{m/s})t + 2.0\,\text{m}\n - **Line (B):**\n - Initial position: x_0 = -2.0\,\text{m}\n - Position at t = 5.0\,\text{s}$:$x = +8.0\,\text{m}\n - Slope / Velocity: \n      v_B = \frac{8.0\,\text{m} - (-2.0\,\text{m})}{5.0\,\text{s} - 0.0\,\text{s}} = +2.0\,\text{m/s}\n - Equation: \n      x_B(t) = (2.0\,\text{m/s})t - 2.0\,\text{m}\n - **Line (C):**\n - Initial position: x_0 = +2.0\,\text{m}\n - Position at t = 5.0\,\text{s}$:$x = -4.0\,\text{m}\n - Slope / Velocity: \n      v_C = \frac{-4.0\,\text{m} - 2.0\,\text{m}}{5.0\,\text{s} - 0.0\,\text{s}} = -1.2\,\text{m/s}\n - Equation: \n      x_C(t) = (-1.2\,\text{m/s})t + 2.0\,\text{m}\n - **Line (D):**\n - Initial position: x_0 = +10.0\,\text{m}\n - Position at t = 5.0\,\text{s}$:$x = -6.0\,\text{m}\n - Slope / Velocity: \n      v_D = \frac{-6.0\,\text{m} - 10.0\,\text{m}}{5.0\,\text{s} - 0.0\,\text{s}} = -3.2\,\text{m/s}\n - Equation: \n      x_D(t) = (-3.2\,\text{m/s})t + 10.0\,\text{m}\n- **Motion Diagram Correlations:**\n - **Diagram 1:** Closely spaced leftward arrows \implies Slow negative velocity, matching **Line (C)**.\n - **Diagram 2:** Rightward arrows starting at negative position \implies Positive velocity with negative intercept, matching **Line (B)**.\n - **Diagram 3:** Widely spaced leftward arrows starting from far right \implies Fast negative velocity, matching **Line (D)**.\n - **Diagram 4:** Rightward arrows starting at positive position \implies Positive velocity with positive intercept, matching **Line (A)**.\n\n# Predictive Kinematics: Passing Buggies Problem\n\n- **Problem Setup:**\n - Two buggies are separated by initial distance d = 3.0\,\text{m}.\n - Buggies are released simultaneously, moving directly toward each other at constant speeds v_1$and$v_2.\n- **System of Equations Derivation:**\n - Origin x = 0\,\text{m} set at Buggy 1's starting point.\n - Equation for Buggy 1 (moving right):\n    x_1(t) = v_1 \cdot t\n - Equation for Buggy 2 (starts at x_0 = 3.0\,\text{m}, moving left):\n    x_2(t) = 3.0\,\text{m} - v_2 \cdot t\n- **Solving for Intersection / Passing Point:**\n - Buggies pass each other when their positions match: x_1(t_{\text{pass}}) = x_2(t_{\text{pass}}).\n    v_1 \cdot t_{\text{pass}} = 3.0\,\text{m} - v_2 \cdot t_{\text{pass}}\n    (v_1 + v_2)t_{\text{pass}} = 3.0\,\text{m}\n    t_{\text{pass}} = \frac{3.0\,\text{m}}{v_1 + v_2}\n - Passing Position Formula:\n    x_{\text{pass}} = v_1 \cdot \left(\frac{3.0\,\text{m}}{v_1 + v_2}\right)\n\n# Critical Conceptual Definitions & Exit Ticket Breakdown\n\n- **Flaw in "Speed is Distance over Time":**\n - Stating "distance over time" incorrectly implies dividing by a single instant in time (t$) rather than an elapsed duration or time interval ($\Delta t).\n- **Formal Physics Definitions:**\n - **Average Speed:** A scalar magnitude defined as total path distance traveled divided by total elapsed time interval:\n    \text{Average Speed} = \frac{\text{Total Distance}}{\Delta t}\n - **Average Velocity:** A vector quantity defined as displacement (change in position) divided by time interval:\n    \vec{v}{\text{avg}} = \frac{\Delta \vec{x}}{\Delta t} = \frac{\vec{x}_f - \vec{x}_i}{t_f - t_i}\n - **Instantaneous Speed:** The absolute magnitude of the instantaneous velocity vector at a specific moment:\n    v(t) = \left|\frac{dx}{dt}\right| = \lim{\Delta t \rightarrow 0} \left|\frac{\Delta x}{\Delta t}\right|\n\n# After Class Practice Problems and Solutions\n\n## Practice Problem 1: Comparative Motion on Position-Time Graph\n\n\n\n- **Scenario Description:**\n - Graph displays position versus time for Sarah and Michael walking on a straight track.\n - Both traces are parallel lines with positive slopes.\n - Sarah's line has a higher vertical intercept (x_{0,\text{Sarah}} > x_{0,\text{Michael}}).\n- **Analysis:**\n - Speed is determined by line slope (v = \frac{\Delta x}{\Delta t}).\n - Parallel lines possess equal slopes, meaning v_{\text{Sarah}} = v_{\text{Michael}}.\n- **Correct Choice:**\n - **C. Sarah and Michael are walking at the same speed.**\n\n## Practice Problem 2: Verbal and Diagrammatic Motion Description\n\n\n\n- **Graph 1:**\n - Line features a constant negative slope extending from positive position across zero to negative position.\n - Verbal Description: Object starts at a positive position, moves left (negative direction) at constant speed, crosses origin, and continues leftward.\n - Motion Diagram: Dots spaced uniformly with constant-length arrows pointing left.\n- **Graph 2:**\n - Line slopes downward to zero position, then slopes upward.\n - Verbal Description: Object starts at positive position, moves left at constant speed toward origin, instantly turns around at origin, and moves right at constant speed.\n - Motion Diagram: Uniform leftward arrows heading to origin, followed by uniform rightward arrows heading away.\n- **Graph 3:**\n - Line exhibits steep negative slope followed by a shallower negative slope.\n - Verbal Description: Object moves left at high constant speed, then instantly reduces speed to a lower constant speed while still moving left.\n - Motion Diagram: Widely spaced dots with long leftward vectors initially, changing to closely spaced dots with shorter leftward vectors.\n- **Graph 4:**\n - Line starts at negative position with positive slope, then becomes flat (zero slope) at positive position.\n - Verbal Description: Object starts at negative position, moves right at constant speed across origin, stops completely, and stays stationary at positive position.\n - Motion Diagram: Evenly spaced rightward vectors transitioning to a stationary point with zero velocity.\n\n## Practice Problem 3: Analytical Two-Object Meeting Problem\n\n- **Problem Statement:**\n - Object A position: \n    x_A(t) = 10.0\,\text{m} - (4.0\,\text{m/s})t\n - Object B position (benchmark model starting at origin): \n    x_B(t) = (2.0\,\text{m/s})t\n- **Meeting Condition:**\n - Set positions equal: x_A(t) = x_B(t).\n    10.0\,\text{m} - (4.0\,\text{m/s})t = (2.0\,\text{m/s})t\n    10.0\,\text{m} = (6.0\,\text{m/s})t\n    t = \frac{10.0}{6.0}\,\text{s} = \frac{5}{3}\,\text{s} \approx 1.67\,\text{s}\n- **Meeting Position:**\n - Substitute meeting time back into x_B(t):\n    x = (2.0\,\text{m/s}) \cdot \left(\frac{5}{3}\,\text{s}\right) = \frac{10.0}{3}\,\text{m} \approx 3.33\,\text{m}\n\n## Practice Problem 4: Motion Diagram to Kinematic Models\n\n\n\n- **Diagram Parameter Extraction (\Delta t = 1.0\,\text{s}):**\n - Toy Car 1 (top): Moves right 2.0\,\text{m}$per second starting from origin$x_0 = 0.0\,\text{m}.\n - Toy Car 2 (bottom): Moves right 1.0\,\text{m}$per second starting from position$x_0 = 3.0\,\text{m}.\n- **Mathematical Equations:**\n - Car 1: \n    x_1(t) = (2.0\,\text{m/s})t\n - Car 2: \n    x_2(t) = 3.0\,\text{m} + (1.0\,\text{m/s})t\n- **Graphical Representation:**\n - Plot of two lines intersecting at t = 3.0\,\text{s}$, at position$x = 6.0\,\text{m}$$.

Practice Problem 5: Textbook Practice Assignments

• Chapter 2 Practice Questions:
  • Questions: 2, 3, 4, 18, 20, 29, 30
• Chapter 2 Numerical Problems:
  • Problems: 23, 24, 33