Extending One-Dimensional Kinematics to Accelerating Objects

Objectives and Core Model Extension

• The primary goal of this unit is to extend the current model of one-dimensional motion to encompass accelerating objects.
• Key kinematic quantities—position ($x$), velocity ($v$), and acceleration ($a$)—are analyzed across verbal, visual, graphical, and mathematical representations.

Position-Time Graphs and Kinematic Warm-Up Exercises

• Interpretation of Constant Velocity Position-Time Graph:

*   **Question:** Which description matches the position-time graph?
    *   A. An object moving first in the negative direction, then in the positive direction.
    *   B. An object slowing down, then speeding up.
    *   C. An object constantly moving in the negative direction.
    *   D. An object constantly moving in the positive direction.
*   **Analysis:** The slope of a position-versus-time graph represents the velocity ($v$). Because the line is straight and has a positive slope throughout its entire path, the object is constantly moving in the positive direction. Crossing the horizontal time axis indicates moving from a negative position coordinate to a positive position coordinate, not a change in direction.

• Mathematical Representation of Position:

*   **Question:** Which equation correctly represents the position of car 2?
    *   $x(t) = (2\text{ m/s})t + 2\text{ m}$
    *   $x(t) = (1.5\text{ m/s})t + 9\text{ m}$
    *   $x(t) = (-3\text{ m/s})t + 9\text{ m}$
    *   $x(t) = (-1.5\text{ m/s})t + 9\text{ m}$
    *   $x(t) = (-2\text{ m/s})t + 9\text{ m}$

• Physical Significance of Position Intercepts:

  • Scenario: A student collects data from a moving buggy and derives the mathematical equation:         $x(t) = (-0.45\text{ m/s})t + 0.90\text{ m}$
  • Question: What happens at $t = 2.0\text{ s}$?
    • The buggy stops
    • The student stopped collecting data
    • The buggy reaches the origin
    • The buggy changes directions
  • Derivation: Substituting $t = 2.0\text{ s}$ into the equation yields:         $x(2.0\text{ s}) = (-0.45\text{ m/s})(2.0\text{ s}) + 0.90\text{ m}$$x(2.0\text{ s}) = -0.90\text{ m} + 0.90\text{ m}$$x(2.0\text{ s}) = 0\text{ m}$
  • Conclusion: At $t = 2.0\text{ s}$, the position of the buggy is $0\text{ m}$, meaning the buggy reaches the origin.

• Graphing and Diagramming from Kinematic Functions:

  • Mathematical Model:$x(t) = 4.0\text{ m} - \frac{2.0\text{ m}}{\text{s}}t$
  • Initial Conditions: The initial position is $x_0 = 4.0\text{ m}$ and the constant velocity is $v = -2.0\text{ m/s}$.
  • Motion Diagram: Represented by equally spaced dots along the $x$-axis moving in the negative direction, with constant-length velocity vectors pointing leftward.
  • Position-Time Graph: A straight line starting at an intercept of $4.0\text{ m}$ at $t = 0\text{ s}$ with a uniform negative slope of $-2.0\text{ m/s}$, crossing the time axis at $t = 2.0\text{ s}$.

Free-Fall Laboratory Observations

• Characteristics of Free-Fall Motion:
  • When an object is dropped from rest in a laboratory setting (neglecting air resistance), it undergoes uniform downward acceleration under the influence of gravity.
  • The position changes non-linearly over time (quadratic dependence on time), while the magnitude of its velocity increases uniformly in the downward direction at a rate of approximately $g = 9.8\text{ m/s}^2$

Velocity-Time Graphs and Displacement Analysis

• Calculating Displacement from Velocity-Time Graphs:

  • For an object moving at constant velocity, displacement (\text{\Delta} x) over a given time interval (\text{\Delta} t = t_2 - t_1) is calculated directly by:         \text{\Delta} x = v \times \text{\Delta} t
  • Sample Calculation: For an object traveling at a constant velocity of $v = 3\text{ m/s}$ between $t = 1\text{ s}$ and $t = 3\text{ s}$:         \text{\Delta} t = 3\text{ s} - 1\text{ s} = 2\text{ s}         \text{\Delta} x = (3\text{ m/s}) \times (2\text{ s}) = 6\text{ m}
  • Options Evaluated:
    • $3\text{ m}$
    • $6\text{ m}$ (Correct)
    • $9\text{ m}$
    • $12\text{ m}$

• Geometric Meaning of Displacement:

  • The displacement of an object on a velocity-versus-time graph is visually and mathematically represented by the bounded area under the curve (between the velocity line and the horizontal time axis) over the specified time interval.
  • Experimental Verification: Comparing displacement derived from a position-versus-time graph (\text{\Delta} x = x_f - x_i) against the calculated geometric area under the corresponding velocity-versus-time graph (\text{Area} = v \text{\Delta} t) confirms that the two metrics produce identical values within experimental margin. The hypothesis is supported by experiment.

Motion with Constant Acceleration: The Fan Cart Experiment

• Observations of a Fan Cart:

  • Verbal Representation: A cart attached to a running fan experiences a constant net horizontal force, causing it to speed up continuously as it travels down the track.
  • Motion Diagram: The distance between consecutive positions increases at successive equal time intervals. Velocity vectors lengthen progressively in the direction of motion.
  • Position-versus-Time Graph: Displays a parabolic curve opening upwards, reflecting a continuously increasing slope (increasing instantaneous velocity).
  • Velocity-versus-Time Graph: Displays a linear function with a non-zero positive slope.

• Mathematical Expression of Velocity:

  • Linear model for velocity under uniform acceleration:         $v(t) = a t + v_0$
  • Slope ($a$): Represents the constant acceleration of the cart in units of $\text{m/s}^2$.
  • Vertical Intercept ($v_0$): Represents the initial velocity of the cart at $t = 0\text{ s}$ in units of $\text{m/s}$.

• Displacement Derivation for Accelerating Objects:

*   **Hypothesis:** The displacement of an accelerating object remains equal to the geometric area under its velocity-versus-time graph.
*   **Experimental Result:** Tested and **supported by experiment**.
*   **Geometric Derivation:**
    *   The region under a linear velocity graph between initial time $t_i$ and final time $t_f$ (over duration \text{\Delta} t) forms a trapezoid bounded by initial velocity $v_i$ and final velocity $v_f$.
    *   Decomposing the trapezoid into a rectangle and a right triangle:

            \text{Area}_{\text{rectangle}} = v_i \text{\Delta} t             \text{Area}_{\text{triangle}} = \frac{1}{2} (v_f - v_i) \text{\Delta} t * Substituting acceleration a = \frac{v_f - v_i}{\text{\Delta} t}, which yields (v_f - v_i) = a \text{\Delta} t:             \text{Area}_{\text{triangle}} = \frac{1}{2} (a \text{\Delta} t) \text{\Delta} t = \frac{1}{2} a (\text{\Delta} t)^2 * Summing both geometric areas yields the total displacement formula:             \text{\Delta} x = v_i \text{\Delta} t + \frac{1}{2} a (\text{\Delta} t)^2 * Alternatively, using average velocity $\bar{v} = \frac{v_i + v_f}{2}$:             \text{\Delta} x = \frac{v_i + v_f}{2} \text{\Delta} t

Kinematics on an Inclined Plane

• Motion Analysis of a Cart Pushed Up an Incline:

  • Setup: A cart is given an initial push up a smooth inclined track. It moves upward, slows down to a momentary stop at its apex, and then rolls back down the incline.
  • Coordinate System: $+ ext{x}$ direction defined up the incline.

• Sign Conventions and Patterns Across Motion Phases:

    | Phase of Motion | Sign of Position ($x$) | Sign of Velocity ($v$) | Sign of Acceleration ($a$) | Speeding Up or Slowing Down |     | :--- | :--- | :--- | :--- | :--- |     | On the way up | Positive ($x > 0$) | Positive ($v > 0$) | Negative ($a < 0$) | Slowing down |     | Highest Point | Positive ($x > 0$) | Zero ($v = 0$) | Negative ($a < 0$) | Momentarily at rest |     | On the way down | Positive ($x > 0$) | Negative ($v < 0$) | Negative ($a < 0$) | Speeding up |

• Clarification on Acceleration and Motion Direction:

  • Question: Does positive acceleration always mean an object is speeding up?
  • Answer: No. The sign of acceleration alone does not determine whether an object is speeding up or slowing down; it only indicates the direction of the acceleration vector relative to the coordinate system.
  • General Rule 1 (Speeding Up): Objects speed up when velocity and acceleration share the same sign (both positive, or both negative).
  • General Rule 2 (Slowing Down): Objects slow down when velocity and acceleration have opposite signs (one positive and one negative).

• State of the Object at the Highest Point:

  • Question: Which statement was true about the cart at its highest point?
    • Its velocity was zero and its acceleration was non-zero
    • Its acceleration was zero and its velocity was non-zero
    • Both its velocity and acceleration were zero
    • Both its velocity and acceleration were non-zero
  • Explanation: At the peak, velocity instantaneously passes through zero ($v = 0\text{ m/s}$) as it changes direction. However, gravity continuously pulls the object down the ramp, maintaining a constant, non-zero downward acceleration ($a \neq 0$).

Summary Principles, Exit Ticket, and Assigned Practice

• Exit Ticket Analysis:
  • Claim: