Introduction to Kinematics - College Physics 2E Chapter 2

Kinematics Overview: This field involves the description of motion without necessarily considering the forces that cause that motion. It is the focus of Chapter 2 in College Physics 2E.
Scalars vs. Vectors: * Scalar Quantities: These are quantities defined by magnitude only. * Examples: distance, temperature, mass, and speed. * Vector Quantities: These are quantities defined by both magnitude and direction. * Examples: position, displacement, velocity, and acceleration. * Notation: Vectors are typically denoted with an arrow over the variable, such as $\vec{d}$ or $\vec{v}$ . * It is critical to distinguish between the two, as a value can be given as a scalar or a vector depending on whether direction is included.

Position: This refers to the specific location of an object in space. It is always defined relative to a known reference point (often the origin in a coordinate system).
Displacement: This is defined as the change in position. It is a vector quantity that represents the final position relative to the initial position. * Formula: $\Delta d = d_f - d_i$ * Characteristics: Displacement only accounts for the starting and ending points. The path taken between those points ("what happens in the middle") does not affect the displacement.
Directional Conventions: * In general, moving away from the reference point is considered Positive (+). * Moving towards the reference point is considered Negative (-). * Example: An airplane passenger walking from the front to the back of a plane may have a negative displacement relative to the cockpit if the cockpit is the reference point.

Distance: The total length of the path traveled between two positions. This is a scalar quantity and does not include direction.
Displacement: The net change in position (magnitude and direction). This is a vector.
Comparison: Distance is always greater than or equal to displacement magnitude ( $\text{Distance} \ge \text{Displacement}$ ). These words are not synonymous; distance includes every step of the travel while displacement only cares about the net result.
Illustrative Example: A cyclist rides $4\,km$ West and then turns around to ride $6\,km$ East. * Displacement: $(-4\,km) + 6\,km = 2\,km$ East. * Distance: $4\,km + 6\,km = 10\,km$ . * Magnitude of Displacement: $2\,km$ .

Time: Represents the duration or interval of time, calculated as $\Delta t = t_f - t_i$ . * It is simpler to assume the motion starts at $t_i = 0$ , in which case $t = \Delta t$ . * Time is never a negative value.
Velocity: The rate at which position changes. * Average Velocity Formula: $\bar{v} = \frac{\Delta d}{\Delta t} = \frac{d_f - d_i}{t_f - t_i}$ . * Velocity is a vector quantity and can be negative if the change in position is toward the reference point.
Average Speed: A scalar quantity defined as the total distance traveled over time. * Commuter Train Example: A train goes from Baltimore to Washington, DC, and back in $2\,\text{hours}$ and $15\,\text{minutes}$ . The distance between stations is $60\,km$ . * Displacement: $0\,km$ (because it returned to the start). * Average Velocity: $0\,m/s$ . * Total Distance: $120\,km = 120,000\,m$ . * Total Time: $135\,\text{minutes} = 8,100\,s$ . * Average Speed: $\frac{120,000\,m}{8,100\,s} \approx 14.81\,m/s$ .
Instantaneous Velocity: The velocity at a specific point in time. It is visualized as the slope of a position-time graph at a specific point. If velocity is constant, the average velocity equals the instantaneous velocity.

Definition: The rate at which velocity changes over time. * Formula: $\bar{a} = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{t}$ . * Acceleration is a vector quantity.
Acceleration vs. Deceleration: * When velocity and acceleration are in the same direction, the object speeds up. * When velocity and acceleration are in opposite directions, it is called deceleration (the object slows down). * Note: Acceleration in the negative direction results in speeding up if the object was already moving in the negative direction.
Racehorse Example: A horse accelerates from rest ( $0\,m/s$ ) to a velocity of $15.0\,m/s$ due West in $1.80\,s$ . * Average acceleration: $\frac{15.0\,m/s - 0\,m/s}{1.80\,s} = 8.33\,m/s^2$ West.
Instantaneous Acceleration: Represented by the slope of a velocity-time graph. A horizontal line on a velocity-time graph indicates constant acceleration.

These equations apply only when acceleration ( $a$ ) is constant (e.g., gravity $g = -9.81\,m/s^2$ ).

Primary Equations: 1. $d_f = d_i + v_i t + \frac{1}{2} a t^2$ 2. $v_f = v_i + a t$ 3. $v_f^2 = v_i^2 + 2 a (d_f - d_i)$ 4. $d_f = d_i + \bar{v}t = d_i + \frac{1}{2}(v_i + v_f)t$

Sketch the problem and list all given/inferred variables ( $d_i, d_f, v_i, v_f, a, t$ ).
Determine requirements: Identify the variable needed for the solution.
Select equation: Choose an equation with only one unknown.
Solve: Substitute values with units and evaluate for the unknown.

Example: Jogger: * Variables: $v = 4.00\,m/s$ , $t = 2.00\,\text{min} (120\,s)$ , $d_i = 0$ . * Solution: $d_f = 0 + 4(120) = 480\,m$ .
Example: Airplane Landing: * Variables: $v_i = 70.0\,m/s$ , $a = -1.50\,m/s^2$ , $t = 40.0\,s$ . * Solution: $v_f = 70 + (-1.5)(40) = 70 - 60 = 10\,m/s$ .
Example: Dragster: * Variables: $a = 26\,m/s^2$ , $v_i = 0$ , $t = 5.56\,s$ . * Solution: $d_f = 0 + 0(5.56) + \frac{1}{2}(26)(5.56)^2 \approx 401.88\,m$ .
Example: Stopping Distances: * Scenario: $v_i = 30.0\,m/s$ , $v_f = 0$ . Dry concrete ( $a = -7\,m/s^2$ ) vs. Wet concrete ( $a = -5\,m/s^2$ ). * Calculation: Use $v_f^2 = v_i^2 + 2 a d$ to find distance $d$ . * Account for reaction time: Multiply $v_i$ by reaction time ( $0.500\,s$ ) and add this displacement to the braking distance.
Quadratic Situations: When solving for time $t$ in $d_f = d_i + v_i t + \frac{1}{2} a t^2$ , move constants to one side to form $\frac{1}{2} a t^2 + v_i t + \Delta d = 0$ and apply the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Discard negative time results.

Gravitational Acceleration ( $g$ ): Falling objects experience a constant vertical acceleration of $9.8\,m/s^2$ (downward).
Assumptions: Air resistance and friction are negligible.
Direction: Downward is typically assigned a negative value.
Example Case: A person on a high cliff throws a rock straight up at $13.0\,m/s$ . * The rock misses the cliff and falls. * To find position/velocity at $t = 3.00\,s$ : * $y = y_i + v_i t - \frac{1}{2} g t^2$ * $v = v_i - g t$