Comprehensive Study Guide: One-Dimensional Kinematics

Foundations of One-Dimensional Kinematics

• Definition of Kinematics: Kinematics is formally defined as the study of motion without considering its causes or the forces that produce or change that motion.

• Etymology: The term originates from the Greek word for motion. It shares roots with English words such as "cinema" (moving pictures) and "kinesiology" (the study of human movement).

• Ubiquity of Motion: Motion is present across all scales in the physical universe:

  • Macroscopic examples include sports (a tennis game, an American football passing through the air), animal flight, and planetary exploration (a space-probe flyby of Neptune or Mars).

  • Physiological examples include the heart pumping blood through veins during rest.

  • Microscopic examples include continuous thermal vibrations of atoms and molecules in inanimate matter.

• Scope of One-Dimensional Kinematics: Focuses strictly on rectilinear motion—motion taking place along a straight line. Analysis of curved, two-dimensional, or three-dimensional paths (such as a car rounding a curve) is deferred to multi-dimensional kinematics.

Position, Displacement, and Distance

• Position:

  • Definition: The location of an object at any particular time specified relative to a convenient frame of reference.

  • Reference Frame: A coordinate system used to measure position and motion. Earth is the most common reference frame, where positions are measured relative to stationary terrestrial objects.

  • Moving Reference Frames: Reference frames can also be in motion relative to Earth. For example, describing a passenger's position inside an airplane uses the airplane as the reference frame rather than the Earth.

• Displacement:

  • Definition: Displacement is the net change in position of an object relative to its reference frame.

  • Mathematical Equation:

$\Delta x = x_{\text{f}} - x_0$

*   Symbol Meanings:
    *   $\Delta$ (capital Greek letter delta) denotes "change in" whatever physical quantity follows it.
    *   $\Delta x$ represents displacement (change in position).
    *   $x_0$ represents the initial position at the start of the interval.
    *   $x_{\text{f}}$ represents the final position at the end of the interval.
*   Units: The International System of Units (SI) unit for displacement is the meter ($\text{m}$). Other common length units include kilometers ($\text{km}$), miles ($\text{mi}$), and feet ($\text{ft}$).
*   Directionality: Displacement is a vector quantity possessing both magnitude and direction. In one-dimensional motion, direction is indicated by a positive ($+$) or negative ($-$) sign.
*   Sign Conventions: By standard convention, rightward or upward motion is designated as positive, whereas leftward or downward motion is designated as negative.

*   Worked Example (Professor Pacing):
    *   Initial position $x_0 = 1.5\,\text{m}$
    *   Final position $x_{\text{f}} = 3.5\,\text{m}$
    *   Displacement: $\Delta x = 3.5\,\text{m} - 1.5\,\text{m} = +2.0\,\text{m}$ (pointing $2.0\,\text{m}$ to the right).

*   Worked Example (Airplane Passenger):
    *   Initial position $x_0 = 6.0\,\text{m}$
    *   Final position $x_{\text{f}} = 2.0\,\text{m}$
    *   Displacement: $\Delta x = 2.0\,\text{m} - 6.0\,\text{m} = -4.0\,\text{m}$ (pointing $4.0\,\text{m}$ toward the rear).

• Distance vs. Distance Traveled:

  • Distance: The magnitude or size of displacement between two positions. It is a scalar quantity, expressed purely as a positive number with units, without directional signs.

  • Distance Traveled: The total length of the physical path traversed between two positions.

  • Key Distinction: Distance traveled is path-dependent and can be substantially larger than the magnitude of displacement.

  • Path-Independence of Displacement: Displacement depends solely on the initial and final position marks, independent of the intervening path taken.

• Worked Example (Cyclist Ride):

  • Scenario: A cyclist rides $3\,\text{km}$ west, turns around, and rides $2\,\text{km}$ east. Let East be positive ($+$) and West be negative ($-$).

  • (a) Displacement Calculation:

$\Delta x = -3\,\text{km} + 2\,\text{km} = -1\,\text{km}$

        (The displacement is $1\,\text{km}$ to the west). * (b) Distance Traveled Calculation:

$\text{Distance Traveled} = 3\,\text{km} + 2\,\text{km} = 5\,\text{km}$

*   (c) Magnitude of Displacement Calculation:

$|\Delta x| = |-1\,\text{km}| = 1\,\text{km}$

Vectors, Scalars, and Coordinate Systems

• Vector Quantities:

  • Definition: Any physical quantity that requires both magnitude (size) and direction for its complete description.

  • Examples: Displacement ($\Delta x = +2.0\,\text{m}$), Velocity ($90\,\text{km/h}$ due east), Force ($500\,\text{N}$ straight down).

  • Graphical Representation: Represented as arrows. The length of the arrow is directly proportional to the vector's magnitude, and the arrowhead indicates direction.

  • 1D Vector Direction: Denoted simply by a plus ($+$) or minus ($-$) sign.

• Scalar Quantities:

  • Definition: Any physical quantity specified completely by magnitude alone, possessing no direction.

  • Examples: Distance ($2.0\,\text{m}$), Speed limit ($90\,\text{km/h}$), Mass, Height ($1.8\,\text{m}$), Temperature ($-20^\circ\text{C}$), Energy ($250\,\text{kcal}$ / $250\,\text{Calories}$).

  • Negative Scalars: A negative sign on a scalar (such as $-20^\circ\text{C}$) denotes a specific point on a scale, not a direction in space.

  • Graphical Representation: Scalars are never represented by arrows.

• Coordinate Systems for One-Dimensional Motion:

  • A coordinate line must be assigned within the reference frame to give algebraic signs to directions.

  • Standard Conventions:

    • Horizontal axis ($x$): Right is positive ($+$), Left is negative ($-$).

    • Vertical axis ($y$): Upward is positive ($+$), Downward is negative ($-$).

*   Custom Coordinate Conventions: The positive direction can be defined arbitrarily to match the physical scenario (e.g., setting leftward motion as positive for a race moving left, or downward as positive for falling objects), provided the choice remains fixed throughout all calculations.

• Conceptual Check (Speed vs. Direction): Speed is a scalar quantity because it maintains its value regardless of changes in direction (such as a car rounding a corner at a constant speedometer reading). If it were a vector, any directional change would alter the vector even if the magnitude remained constant.

Time, Velocity, and Speed

• Physical Nature of Time:

  • Definition: Time is defined operationalistically by its measurement; it represents change or the interval over which change occurs.

  • SI Unit: The second ($\text{s}$).

  • Elapsed Time Formula:

$\Delta t = t_{\text{f}} - t_0$

*   Simplification Convention: Unless explicitly specified otherwise, the initial time $t_0$ is set to zero ($t_0 = 0\,\text{s}$), reducing the elapsed time equation to:

$\Delta t = t$

• Average Velocity:

  • Definition: Displacement divided by the time of travel.

  • Mathematical Equations:

$\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_{\text{f}} - x_0}{t_{\text{f}} - t_0}$

*   When $t_0 = 0$:

$\bar{v} = \frac{\Delta x}{t}$

*   Notation: The horizontal bar over $v$ ($\bar{v}$) denotes an average value.
*   Vector Property: Average velocity is a vector quantity with the same direction as the displacement $\Delta x$.
*   SI Units: Meters per second ($\text{m/s}$). Other units include $\text{km/h}$, $\text{mi/h}$ (mph), and $\text{cm/s}$.
*   Worked Calculation (Passenger moving backward in an airplane):

$\bar{v} = \frac{\Delta x}{t} = \frac{-4.0\,\text{m}}{5.0\,\text{s}} = -0.8\,\text{m/s}$

• Instantaneous Velocity vs. Instantaneous Speed:

  • Instantaneous Velocity ($v$): The velocity at a precise instant in time, or over an infinitesimally small time interval ($\Delta t \rightarrow 0$).

  • Instantaneous Speed: The magnitude of instantaneous velocity at a given moment. For instance, if $v = -3.0\,\text{m/s}$, the instantaneous speed is $3.0\,\text{m/s}$.

  • Speedometer Reading: A vehicle's speedometer measures instantaneous speed, presenting magnitude without directional information.

• Average Speed:

  • Definition: Total distance traveled divided by elapsed time.

$\text{Average Speed} = \frac{\text{Distance Traveled}}{\text{Elapsed Time}}$

*   Scalar Nature: Average speed is a scalar. It is **not** simply the magnitude of average velocity.

*   Worked Example (Round Trip to Store):
    *   A driver travels $3\,\text{km}$ to a store and returns home ($3\,\text{km}$) in $0.5\,\text{h}$ ($30\,\text{min}$).
    *   Total distance traveled = $6\,\text{km}$.
    *   Net displacement $\Delta x = 0\,\text{km}$.
    *   Average speed calculation:

$\text{Average Speed} = \frac{6\,\text{km}}{0.5\,\text{h}} = 12\,\text{km/h}$

    *   Average velocity calculation:

$\bar{v} = \frac{\Delta x}{t} = \frac{0\,\text{km}}{0.5\,\text{h}} = 0\,\text{km/h}$

• Worked Example (Commuter Train Baltimore to Washington, DC):

  • Scenario: A commuter train travels $40\,\text{miles}$ from Baltimore to Washington, DC, and back in $1\,\text{h}\,45\,\text{min}$ ($105\,\text{minutes}$).

  • (a) Average Velocity:

$\bar{v} = 0\,\text{m/s}$

        (Because initial and final positions are identical, $\Delta x = 0$). * (b) Average Speed in $\text{m/s}$: * Total distance = $40\,\text{miles} + 40\,\text{miles} = 80\,\text{miles}$. * Elapsed time = $105\,\text{minutes}$. * Dimensional Analysis Conversion:

$\frac{80\,\text{miles}}{105\,\text{minutes}} \times \frac{5280\,\text{feet}}{1\,\text{mile}} \times \frac{1\,\text{meter}}{3.28\,\text{feet}} \times \frac{1\,\text{minute}}{60\,\text{seconds}} = 20\,\text{m/s}$

Acceleration and Deceleration

• Average Acceleration:

  • Definition: The rate at which velocity changes over time.

  • Mathematical Equations:

$\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_{\text{f}} - v_0}{t_{\text{f}} - t_0}$

*   SI Units: Meters per second squared ($\text{m/s}^2$ or $\text{m}\cdot\text{s}^{-2}$), representing how many meters per second the velocity changes each second.
*   Vector Nature: Acceleration is a vector sharing the direction of the *change in velocity* ($\Delta v$). It can result from a change in speed, a change in direction, or both.

• Deceleration vs. Negative Acceleration:

  • Deceleration: Refers strictly to acceleration directed opposite to the direction of velocity. Deceleration always reduces speed.

  • Negative Acceleration: Refers strictly to acceleration pointing in the negative direction of the coordinate system. It may or may not reduce speed.

  • Four Conceptual Cases:

    1. Moving Right ($+v$), Speeding Up: Acceleration is Right ($+a$) $\rightarrow$ Positive acceleration, Non-decelerating.

    2. Moving Right ($+v$), Slowing Down: Acceleration is Left ($-a$) $\rightarrow$ Negative acceleration, Decelerating.

    3. Moving Left ($-v$), Slowing Down: Acceleration is Right ($+a$) $\rightarrow$ Positive acceleration, Decelerating.

    4. Moving Left ($-v$), Speeding Up: Acceleration is Left ($-a$) $\rightarrow$ Negative acceleration, Non-decelerating.

  • General Rule: If velocity $v$ and acceleration $a$ have the same sign, the object speeds up. If $v$ and $a$ have opposite signs, the object slows down.

• Worked Example 2.1 (Racehorse Leaving Gate):

  • Scenario: A racehorse accelerates from rest ($v_0 = 0$) to a velocity of $15.0\,\text{m/s}$ due west in $1.80\,\text{s}$. Let East be positive ($+$) and West be negative ($-$).

*   Knowns: $v_0 = 0\,\text{m/s}$, $v_{\text{f}} = -15.0\,\text{m/s}$, $\Delta t = 1.80\,\text{s}$.
*   Calculation:

$\Delta v = -15.0\,\text{m/s} - 0\,\text{m/s} = -15.0\,\text{m/s}$

$\bar{a} = \frac{\Delta v}{\Delta t} = \frac{-15.0\,\text{m/s}}{1.80\,\text{s}} = -8.33\,\text{m/s}^2$

*   Conclusion: The average acceleration is $8.33\,\text{m/s}^2$ due west.

• Instantaneous Acceleration:

  • Definition: Acceleration at a specific instant in time ($\Delta t \rightarrow 0$).

  • Approximation Method: Non-constant acceleration curves are divided into small discrete time intervals over which acceleration is approximately constant.

• Comprehensive Subway Train Case Studies:

*   **Example 2.2 (Displacement):**
    *   Part (a): $x_0 = 4.70\,\text{km}$, $x_{\text{f}} = 6.70\,\text{km} \rightarrow \Delta x = 6.70 - 4.70 = +2.00\,\text{km}$.
    *   Part (b): $x_0' = 5.25\,\text{km}$, $x_{\text{f}}' = 3.75\,\text{km} \rightarrow \Delta x' = 3.75 - 5.25 = -1.50\,\text{km}$.
*   **Example 2.3 (Distance Traveled):**
    *   Part (a): Distance traveled = $2.00\,\text{km}$.
    *   Part (b): Distance traveled = $1.50\,\text{km}$.
*   **Example 2.4 (Subway Train Speeding Up):**
    *   Scenario: Train accelerates from rest ($v_0 = 0$) to $30.0\,\text{km/h}$ in $20.0\,\text{s}$.

    *   $\Delta v = +30.0\,\text{km/h}$.
    *   Unit Conversion & Calculation:

$\bar{a} = \left(\frac{+30.0\,\text{km/h}}{20.0\,\text{s}}\right) \times \left(\frac{10^3\,\text{m}}{1\,\text{km}}\right) \times \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) = +0.417\,\text{m/s}^2$

*   **Example 2.5 (Subway Train Slowing Down):**
    *   Scenario: Train slows from $30.0\,\text{km/h}$ to a stop ($v_{\text{f}} = 0$) in $8.00\,\text{s}$.

    *   $\Delta v = 0 - 30.0\,\text{km/h} = -30.0\,\text{km/h}$.
    *   Unit Conversion & Calculation:

$\bar{a} = \left(\frac{-30.0\,\text{km/h}}{8.00\,\text{s}}\right) \times \left(\frac{10^3\,\text{m}}{1\,\text{km}}\right) \times \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) = -1.04\,\text{m/s}^2$

*   **Example 2.6 (Average Velocity of Leftward Trip):**
    *   Scenario: Train moves $\Delta x' = -1.50\,\text{km}$ in $t = 5.00\,\text{min}$ ($300\,\text{s}$).
    *   Calculation:

$\bar{v} = \frac{-1.50\,\text{km}}{5.00\,\text{min}} = -0.300\,\text{km/min} = -5.00\,\text{m/s}$

*   **Example 2.7 (Deceleration of Leftward Moving Train):**
    *   Scenario: Train moving left ($v_0 = -20.0\,\text{km/h}$) stops ($v_{\text{f}} = 0$) in $10.0\,\text{s}$.

    *   $\Delta v = 0 - (-20.0\,\text{km/h}) = +20.0\,\text{km/h}$.
    *   Unit Conversion & Calculation:

$\bar{a} = \left(\frac{+20.0\,\text{km/h}}{10.0\,\text{s}}\right) \times \left(\frac{10^3\,\text{m}}{1\,\text{km}}\right) \times \left(\frac{1\,\text{h}}{3600\,\text{s}}\right) = +0.556\,\text{m/s}^2$

Kinematic Equations for Constant Acceleration

• Notation Standards ($t_0 = 0$):

  • $t$ = elapsed time

  • $x_0$ = initial position

  • $x$ = final position

  • $\Delta x = x - x_0$ = displacement

  • $v_0$ = initial velocity

  • $v$ = final velocity

  • $\Delta v = v - v_0$ = change in velocity

  • $a$ = constant acceleration ($a = \bar{a}$)

• Derivations of Kinematic Equations:

  • Equation 1 (Average Velocity):

$\bar{v} = \frac{v_0 + v}{2}$

*   **Equation 2 (Displacement from Average Velocity):**

$x = x_0 + \bar{v}t = x_0 + \left(\frac{v_0 + v}{2}\right)t$

*   **Equation 3 (Final Velocity from Acceleration):**

$a = \frac{v - v_0}{t} \implies v = v_0 + at$

*   **Equation 4 (Final Position with Constant Acceleration):**
    *   Substitute Equation 3 into Equation 2:

$x = x_0 + \left(\frac{v_0 + v_0 + at}{2}\right)t = x_0 + v_0 t + \frac{1}{2}at^2$

*   **Equation 5 (Final Velocity Independent of Time):**
    *   Solve Equation 3 for time ($t = \frac{v - v_0}{a}$) and substitute into Equation 4:

$v^2 = v_0^2 + 2a(x - x_0)$

• Summary Table of Kinematic Equations:

$x = x_0 + \bar{v}t$

$\bar{v} = \frac{v_0 + v}{2}$

$v = v_0 + at$

$x = x_0 + v_0 t + \frac{1}{2}at^2$

$v^2 = v_0^2 + 2a(x - x_0)$

• Worked Example 2.8 (Jogger Run):

  • Scenario: A jogger runs with $\bar{v} = 4.00\,\text{m/s}$ for $t = 2.00\,\text{min}$ ($120\,\text{s}$), with $x_0 = 0$.

  • Calculation:

$x = 0 + (4.00\,\text{m/s})(120\,\text{s}) = 480\,\text{m}$

• Worked Example 2.9 (Airplane Deceleration):

  • Scenario: An airplane lands at $v_0 = 70.0\,\text{m/s}$ and decelerates at $a = -1.50\,\text{m/s}^2$ for $t = 40.0\,\text{s}$.

*   Calculation:

$v = v_0 + at = 70.0\,\text{m/s} + (-1.50\,\text{m/s}^2)(40.0\,\text{s}) = 10.0\,\text{m/s}$

• Real-World Application (ICBM vs. Space Shuttle Endeavor): An Intercontinental Ballistic Missile (ICBM) has higher initial acceleration than the Space Shuttle, but the Space Shuttle accelerates over a much longer time duration, achieving the higher final velocity required for Earth orbit.

• Worked Example 2.10 (Dragster Displacement):

  • Scenario: A dragster accelerates from rest ($v_0 = 0$, $x_0 = 0$) at $a = 26.0\,\text{m/s}^2$ for $t = 5.56\,\text{s}$.

*   Calculation:

$x = x_0 + v_0 t + \frac{1}{2}at^2 = 0 + 0 + \frac{1}{2}(26.0\,\text{m/s}^2)(5.56\,\text{s})^2 = 402\,\text{m}$

        ($402\,\text{m}$ is equivalent to one-quarter mile).

• Worked Example 2.11 (Dragster Final Velocity Without Time):

  • Scenario: Same dragster ($v_0 = 0$, $a = 26.0\,\text{m/s}^2$, $\Delta x = 402\,\text{m}$).

  • Calculation:

$v^2 = v_0^2 + 2a(x - x_0) = 0 + 2(26.0\,\text{m/s}^2)(402\,\text{m}) = 20904\,\text{m}^2/\text{s}^2$

$v = +\sqrt{20904} = +145\,\text{m/s}$

        ($145\,\text{m/s} \approx 522\,\text{km/h} \approx 324\,\text{mi/h}$).

• Worked Example 2.12 (Car Stopping Distances & Reaction Time):

  • Initial parameters: $v_0 = 30.0\,\text{m/s}$ ($\approx 110\,\text{km/h}$), final stopping velocity $v = 0\,\text{m/s}$.

  • (a) Dry Concrete ($a = -7.00\,\text{m/s}^2$):

$x - x_0 = \frac{v^2 - v_0^2}{2a} = \frac{0 - (30.0\,\text{m/s})^2}{2(-7.00\,\text{m/s}^2)} = 64.3\,\text{m}$

*   (b) Wet Concrete ($a = -5.00\,\text{m/s}^2$):

$x - x_0 = \frac{0 - (30.0\,\text{m/s})^2}{2(-5.00\,\text{m/s}^2)} = 90.0\,\text{m}$

*   (c) Including Driver Reaction Time ($t_{\text{reaction}} = 0.500\,\text{s}$):
    *   Distance traveled during reaction time (constant speed):

$\Delta x_{\text{reaction}} = v_0 t = (30.0\,\text{m/s})(0.500\,\text{s}) = 15.0\,\text{m}$

    *   Total Stopping Distance (Dry) = $64.3\,\text{m} + 15.0\,\text{m} = 79.3\,\text{m}$.
    *   Total Stopping Distance (Wet) = $90.0\,\text{m} + 15.0\,\text{m} = 105\,\text{m}$.

• Worked Example 2.13 (Freeway Ramp Acceleration Time):

  • Scenario: $x_0 = 0$, $x = 200\,\text{m}$, $v_0 = 10.0\,\text{m/s}$, $a = 2.00\,\text{m/s}^2$.

  • Equation setup:

$200 = 10.0 t + \frac{1}{2}(2.00)t^2 \implies t^2 + 10.0 t - 200 = 0$

*   Applying Quadratic Formula ($a=1$, $b=10.0$, $c=-200$):

$t = \frac{-10.0 \pm \sqrt{(10.0)^2 - 4(1)(-200)}}{2(1)} = \frac{-10.0 \pm \sqrt{900}}{2} = \frac{-10.0 \pm 30.0}{2}$

*   Solutions: $t = 10.0\,\text{s}$ or $t = -20.0\,\text{s}$. Discarding the unphysical negative root yields $t = 10.0\,\text{s}$.

Problem-Solving Strategies for Kinematics

• Systematic 6-Step Problem Solving Framework:

  1. Examine the Physical Situation: Identify the involved physical principles. Draw a clear diagram, establish a coordinate frame, and specify the positive direction.

  2. List Knowns: Formally list all explicit and implicit given quantities (e.g., "starts from rest" means $v_0 = 0$, "comes to a stop" means $v = 0$).

  3. List Unknowns: Identify the exact physical variable(s) to be computed.

  4. Select Equations: Identify equations linking knowns to unknowns. Select equations containing only one unknown variable whenever possible.

  5. Execute Substitution: Substitute numerical values along with their associated units. Solve algebraically and check unit consistency.

  6. Sanity Check: Evaluate the final result. Verify that the magnitude, sign, and units are physically reasonable.

• Evaluating Unreasonable Results:

  • Unreasonable outputs usually indicate flawed premises or unphysical assumptions.

  • Example: A runner accelerating at $a = 0.4\,\text{m/s}^2$ for $t = 100\,\text{s}$ yields $v = 40\,\text{m/s}$ ($\approx 150\,\text{km/h}$). The mathematical operation is correct, but the premise that a human can maintain constant acceleration for 100 seconds is physically impossible.

Free-Falling Objects and Gravitational Motion

• Gravitational Acceleration Fundamentals:

  • Universality: In the absence of air resistance and friction, all objects near Earth's surface accelerate downward toward Earth's center at the exact same constant acceleration, regardless of mass.

  • Lunar Verification: Apollo 15 astronaut David R. Scott dropped a hammer and a feather simultaneously on the Moon (where air resistance is zero and $g_{\text{Moon}} = 1.67\,\text{m/s}^2$); both objects struck the surface at the same instant.

• Free-Fall Definition: Motion occurring purely under the influence of gravitational force, free from atmospheric drag or friction.

• Standard Value of Acceleration Due to Gravity ($g$):

  • Symbol: $g$

  • Average Earth Surface Value: $g = 9.80\,\text{m/s}^2$

  • Terrestrial Variation: Ranges from $9.78\,\text{m/s}^2$ to $9.83\,\text{m/s}^2$ depending on latitude, altitude, topography, and sub-surface geology.

• Free-Fall Coordinate Conventions:

  • If Upward is defined as Positive ($+$): Acceleration $a = -g = -9.80\,\text{m/s}^2$

  • If Downward is defined as Positive ($+$): Acceleration $a = +g = +9.80\,\text{m/s}^2$

• Kinematic Equations for Free-Fall (Upward = Positive, $a = -g$):

$v = v_0 - gt$

$y = y_0 + v_0 t - \frac{1}{2}gt^2$

$v^2 = v_0^2 - 2g(y - y_0)$

• Worked Example 2.14 (Rock Thrown Upward from Cliff):

  • Scenario: A rock is thrown upward from a cliff edge with $v_0 = +13.0\,\text{m/s}$, $y_0 = 0$, a = -9.80\,\text{m/s}^2$.\n\n\n\n * At t = 1.00\,\text{s}:\n * y_1 = (13.0)(1.00) - \frac{1}{2}(9.80)(1.00)^2 = +8.10\,\text{m}\n * v_1 = 13.0 - (9.80)(1.00) = +3.20\,\text{m/s} (Moving upward).\n * At t = 2.00\,\text{s}:\n * y_2 = (13.0)(2.00) - \frac{1}{2}(9.80)(2.00)^2 = +6.40\,\text{m}\n * v_2 = 13.0 - (9.80)(2.00) = -6.60\,\text{m/s} (Moving downward).\n * At t = 3.00\,\text{s}:\n * y_3 = (13.0)(3.00) - \frac{1}{2}(9.80)(3.00)^2 = -5.10\,\text{m}\n * v_3 = 13.0 - (9.80)(3.00) = -16.4\,\text{m/s} (Below starting level, moving downward).\n * Peak Height: Occurs when v = 0 \implies t = \frac{13.0}{9.80} = 1.33\,\text{s}$, corresponding to peak displacement$y = 8.62\,\text{m}.\n\n\n\n* **Worked Example 2.15 (Rock Thrown Downward):**\n * Scenario: Rock thrown downward from cliff with v_0 = -13.0\,\text{m/s}$, evaluated at position$y = -5.10\,\text{m}.\n * Calculation:\n\nv^2 = v_0^2 - 2g(y - y_0) = (-13.0\,\text{m/s})^2 - 2(9.80\,\text{m/s}^2)(-5.10\,\text{m} - 0) = 169 + 99.96 = 268.96\,\text{m}^2/\text{s}^2\n\nv = -\sqrt{268.96} = -16.4\,\text{m/s}\n\n * Physical Principle: The speed of a falling object at a given height depends only on its initial speed and vertical position, regardless of whether it was initially thrown straight up or straight down.\n\n\n\n* **Worked Example 2.16 (Precision Measurement of Local g):**\n * Scenario: A metal sphere falls 1.0000\,\text{m}$from rest ($v_0 = 0$) in$0.45173\,\text{s}.\n * Calculation:\n\ny - y_0 = -1.0000\,\text{m}\n\n-1.0000\,\text{m} = 0 + \frac{1}{2}a(0.45173\,\text{s})^2\n\na = \frac{2(-1.0000\,\text{m})}{(0.45173\,\text{s})^2} = -9.8010\,\text{m/s}^2\n\n * Result: Local gravitational magnitude g = 9.8010\,\text{m/s}^2.\n\n\n\n# Graphical Analysis of One-Dimensional Motion\n\n* **Linear Functional Forms:**\n * Standard Equation for a Line:\n\ny = mx + b\n\n * Slope (m):\n\nm = \frac{\text{Rise}}{\text{Run}} = \frac{\Delta y}{\Delta x}\n\n * Intercept (b$): The point where the graphed line crosses the vertical axis ($y$at$x = 0).\n\n\n\n* **Position vs. Time Graph (a = 0, Constant Velocity):**\n * Vertical axis = Position (x$), Horizontal axis = Time ($t).\n * **Slope of Position vs. Time Graph:** Represents Velocity (v).\n * **Vertical Intercept:** Represents Initial Position (x_0).\n * Graph Equation: x = x_0 + vt\n\n\n\n * **Worked Example 2.17 (Jet Car Velocity from Graph):**\n * Data points selected from straight line graph: (0.50\,\text{s}, 525\,\text{m})$and$(6.40\,\text{s}, 2000\,\text{m}).\n * Slope Calculation:\n\n\bar{v} = \frac{\Delta x}{\Delta t} = \frac{2000\,\text{m} - 525\,\text{m}}{6.40\,\text{s} - 0.50\,\text{s}} = \frac{1475\,\text{m}}{5.90\,\text{s}} = 250\,\text{m/s}\n\n        (250\,\text{m/s} = 900\,\text{km/h} = 560\,\text{mi/h}).\n\n* **Graphs with Constant Acceleration (a eq 0):**\n * **Position vs. Time (x$vs.$t):** A curved line (parabola). The slope of a tangent line drawn at any point equals the **instantaneous velocity** at that specific instant.\n * **Velocity vs. Time (v$vs.$t$):** A straight line. The **slope** of a velocity vs. time graph equals **acceleration** ($a = \frac{\Delta v}{\Delta t}$). The vertical intercept equals initial velocity ($v_0).\n * **Acceleration vs. Time (a$vs.$t):** A horizontal straight line representing constant acceleration value.\n\n\n\n * **Worked Example 2.18 (Instantaneous Velocity via Tangent Line):**\n * Finding instantaneous velocity at t = 25\,\text{s}$on a curved$x$vs.$t graph.\n * Endpoints of tangent line drawn at t = 25\,\text{s}$:$(19\,\text{s}, 1300\,\text{m})$and$(32\,\text{s}, 3120\,\text{m}).\n * Slope Calculation:\n\nv = \frac{3120\,\text{m} - 1300\,\text{m}}{32\,\text{s} - 19\,\text{s}} = \frac{1820\,\text{m}}{13\,\text{s}} = 140\,\text{m/s}\n\n\n\n* **Graphs with Changing Acceleration:**\n * As acceleration decreases toward zero, the velocity vs. time curve levels off horizontally, and the corresponding position vs. time curve approaches a linear line.\n * **Worked Example 2.19 (Calculating Acceleration from Velocity Graph):**\n * Tangent line drawn at t = 25\,\text{s}$on non-linear$v$vs.$t curve.\n * Tangent endpoints: (10\,\text{s}, 210\,\text{m/s})$and$(50\,\text{s}, 260\,\text{m/s}).\n * Slope Calculation:\n\na = \frac{260\,\text{m/s} - 210\,\text{m/s}}{50\,\text{s} - 10\,\text{s}} = \frac{50\,\text{m/s}}{40\,\text{s}} = 1.25\,\text{m/s}^2\n\n\n\n* **Conceptual Graphing (Ship Entering Harbor):**\n * Motion Description: Ship maintains constant speed, then decelerates at a constant heavy rate, then transitions to a smaller constant deceleration rate until fully stopped.\n * Velocity Graph: Horizontal segment, followed by a steep negative slope segment, followed by a shallow negative slope segment ending at v = 0.\n * Acceleration Graph: Zero value segment, followed by a large negative horizontal step, followed by a smaller negative horizontal step.\n\n\n\n\n\n# Comprehensive Kinematics Glossary\n\n* **Acceleration:** The rate of change in velocity over time.\n* **Acceleration Due to Gravity:** The constant acceleration of an object in free-fall caused by Earth's gravitational pull (g = 9.80\,\text{m/s}^2).\n* **Average Acceleration:** The change in velocity divided by the time interval over which that change occurs.\n* **Average Speed:** Total distance traveled divided by total elapsed time.\n* **Average Velocity:** Displacement divided by the elapsed time over which displacement occurs.\n* **Deceleration:** Acceleration in the direction opposite to velocity; results in a decrease in speed.\n* **Dependent Variable:** The variable being measured (typically plotted along the vertical y axis).\n* **Displacement:** The change in position of an object (\Delta x = x_{\text{f}} - x_0).\n* **Distance:** The magnitude or size of displacement between two positions.\n* **Distance Traveled:** The total length of the path traveled between two positions.\n* **Elapsed Time:** The difference between ending time and beginning time (\Delta t = t_{\text{f}} - t_0).\n* **Free-Fall:** The state of movement that results exclusively from gravitational force, without air resistance or friction.\n* **Independent Variable:** The variable with respect to which the dependent variable is measured (typically plotted along the horizontal x axis).\n* **Instantaneous Acceleration:** Acceleration evaluated at a specific point in time.\n* **Instantaneous Speed:** The absolute magnitude of the instantaneous velocity.\n* **Instantaneous Velocity:** Velocity evaluated at a specific instant in time, or over an infinitesimally small time interval.\n* **Kinematics:** The study of motion without considering its underlying causes or forces.\n* **Model:** A simplified description containing only those elements necessary to analyze a physical situation.\n* **Position:** The specific location of an object relative to a reference frame at a particular time.\n* **Scalar:** A physical quantity fully described by magnitude, possessing no direction.\n* **Slope:** The ratio of the difference in vertical values (rise) to the difference in horizontal values (run) between two points on a line.\n* **Time:** Change, or the continuum/interval over which change occurs.\n* **Vector:** A physical quantity described by both magnitude and direction.\n* **y-Intercept:** The value of y$where a graph line intersects the vertical axis ($x = 0$$).