1D and 2D Kinematics: Speed, Velocity, Displacement, and Vectors

Kinematics in One Dimension: Speed and Average Velocity

• Definitions of Speed and Distance

  • Speed is defined as distance traveled divided by the elapsed time: $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$.
  • An everyday example of measuring average speed is taking the total odometer reading on a car (distance) and dividing it by the elapsed clock time.
  • More precisely, this quotient represents the average speed over a given time interval.

• Distance versus Displacement

  • In motion restricted to a straight line without reversing direction, distance traveled is equal to the magnitude of displacement.
  • If an object oscillates or moves back and forth (such as an insect moving back and forth along a path), the total distance traveled becomes greater than the magnitude of displacement.
  • Displacement is a vector that measures the straight-line distance and direction from an initial position to a final position, whereas distance measures the full cumulative length of the path taken.

• Motion Diagram Example: Jane's Constant Walk

  • Scenario setup: Jane walks to the right along the positive x-axis ($+x$, measured in meters, $\text{m}$) at a constant rate, covering $3\,\text{m}$ in $3\,\text{s}$.
  • Initial condition: At time $t = 0\,\text{s}$, Jane passes the $x = 1\,\text{m}$ mark.
  • Calculating speed: $\text{Speed} = \frac{3\,\text{m}}{3\,\text{s}} = 1\,\text{m/s}$. The ratio indicates that Jane moves $1\,\text{m}$ for every $1\,\text{s}$ that elapses.
  • Concept of negative time ($t = -1\,\text{s}$): Negative time values represent tracking motion before the stopwatch or clock was started at $t = 0\,\text{s}$. Moving backward in time by $1\,\text{s}$ corresponds to moving backward in position to the left by $1\,\text{m}$.
  • Mapping timestamps to positions from $t = -1\,\text{s}$ to $t = 4\,\text{s}$:
  • At $t = -1\,\text{s}$, position $x = 0\,\text{m}$.
  • At $t = 0\,\text{s}$, position $x = 1\,\text{m}$.
  • At $t = 1\,\text{s}$, position $x = 2\,\text{m}$.
  • At $t = 2\,\text{s}$, position $x = 3\,\text{m}$.
  • At $t = 3\,\text{s}$, position $x = 4\,\text{m}$.
  • At $t = 4\,\text{s}$, position $x = 5\,\text{m}$.

Average Velocity and One-Dimensional Problem Solving

• Average Velocity Mathematical Formulation

  • Displacement arrows in a motion diagram point in the direction of motion. For constant velocity, segment arrows are evenly spaced.
  • While problem solving often focuses on the initial important point and the final important point, the overall displacement $\Delta x$ is the straight-line vector from start to finish.
  • Average velocity is defined as displacement divided by the elapsed time interval: $v_{\text{avg}} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$.
  • Standard physics notation often omits the word "average," but average velocity is the rigorous physical term.

• Full Trip Calculation for Jane's Walk

  • Initial position $x_i = 0\,\text{m}$ at initial time $t_i = -1\,\text{s}$.
  • Final position $x_f = 5\,\text{m}$ at final time $t_f = 4\,\text{s}$.
  • Overall displacement: $\Delta x = x_f - x_i = 5\,\text{m} - 0\,\text{m} = +5\,\text{m}$.
  • Total time interval: $\Delta t = t_f - t_i = 4\,\text{s} - (-1\,\text{s}) = 5\,\text{s}$ (accounting for 5 distinct 1-second intervals).
  • Average velocity calculation: $v_{\text{avg}} = \frac{+5\,\text{m}}{5\,\text{s}} = +1\,\text{m/s}$.
  • Significance of signs: Positive ($+$) and negative ($-$) signs specify physical direction. Moving to the right in the positive x-direction yields a positive displacement and positive velocity.

• Example Problem: Frank's Motion

  • Scenario setup:
  • At clock time $t = 12\,\text{s}$, Frank is at position $x_i = +25\,\text{m}$.
  • Five seconds later ($\Delta t = 5\,\text{s}$, clock time $t = 17\,\text{s}$), Frank is at position $x_f = +20\,\text{m}$.
  • Displacement calculation: $\Delta x = x_f - x_i = 20\,\text{m} - 25\,\text{m} = -5\,\text{m}$.
  • Velocity calculation: $v = \frac{\Delta x}{\Delta t} = \frac{-5\,\text{m}}{5\,\text{s}} = -1\,\text{m/s}$.
  • Physical interpretation: The negative sign ($-1\,\text{m/s}$) indicates that Frank is moving to the left (negative x-direction).
  • Irrelevant vs. relevant temporal data: The explicit clock reading at the start ($t = 12\,\text{s}$) or end ($t = 17\,\text{s}$) is unnecessary. Only the time interval elapsed during motion ($\Delta t = 5\,\text{s}$) is required for calculating velocity.
  • Penalty warning: Omitting physical units (such as $\text{m}$, $\text{s}$, or $\text{m/s}$) in calculations or final answers leads to lost points on assessments.

Vectors and Two-Dimensional Kinematics

• Coordinate Conventions for Two-Dimensional Maps

  • Map representations set the positive x-axis ($+x$) as East and the positive y-axis ($+y$) as North, with distances measured in miles ($\text{mi}$) or meters ($\text{m}$).

• Geometric Analysis of Non-Straight Motion: Jenny's Path

  • Scenario setup: Jenny starts at the coordinate origin $(0,0)$, runs $1\,\text{mi}$ Northeast, and then runs $1\,\text{mi}$ South.
  • Angle interpretation: Unspecified "Northeast" motion splits Quadrant 1 evenly at a $45^\circ$ angle relative to the positive x-axis.
  • Right triangle geometry analysis:
  • The first displacement leg forms a right triangle with the x-axis, where the $1\,\text{mi}$ path is the hypotenuse.
  • The hypotenuse is strictly the longest side of a right triangle. Thus, the vertical leg from the top point down to the x-axis is less than $1\,\text{mi}$ ($1\,\text{mi} \times \sin(45^\circ) \approx 0.707\,\text{mi}$).
  • Running $1\,\text{mi}$ directly South extends past the x-axis, placing Jenny in Quadrant 4 (Southeast).
  • Visual nature of displacement:
  • Displacement cannot be computed by simple scalar subtraction ($1\,\text{mi} - 1\,\text{mi} \neq 0\,\text{mi}$).
  • Displacement is a graphical arrow drawn directly from the start point (origin) to the final endpoint.
  • Drawing displacement reveals the necessary mathematical operations (such as trigonometry or the Pythagorean theorem) rather than plain algebraic subtraction.

Scalar versus Vector Quantities

• Scalar Quantities

  • A scalar is a standard numerical value accompanied by units that possesses magnitude but no direction.
  • Example - Temperature: A reading of $75^\circ\text{F}$ or $-10^\circ\text{F}$ (e.g., winter conditions in Minnesota). The positive or negative signs on temperature do not indicate spatial directions like East or West.
  • Example - Speed: Distance per time derived purely from scalar values (odometer distance and clock time).

• Vector Quantities

  • A vector is a quantity characterized by both a magnitude (size) and a direction.
  • Example - Displacement: Distance moved in a specified direction.
  • Example - Velocity: Speed moved in a specified direction.
  • Example - Force: Push or pull strength measured in Newtons ($\text{N}$) in a specified direction.

• Proper Application of the Term "Magnitude"

  • Magnitude describes the size or length of a specific vector and must always be linked to that vector type.
  • Magnitude of displacement is measured in units of distance ($\text{m}$).
  • Magnitude of velocity is measured in units of speed ($\text{m/s}$).
  • Magnitude of force is measured in units of force ($\text{N}$).
  • Writing "magnitude = number" without specifying the associated vector quantity is improper.

• Path Distance versus Displacement along Sidewalks

  • If Jane walks East on a sidewalk along a street and then turns North on another sidewalk, her distance is the cumulative length along the sidewalks.
  • Her displacement is the straight-line vector (hypotenuse) connecting her starting point directly to her ending point across the corner.

Trigonometric Principles and Unit Circle Fundamentals

• Wave Behavior and Quadrant Values of Trigonometric Functions
  • Trigonometric functions repeat across quarters of the unit circle ($360^\circ$ total) and alternate between maximum and minimum values of $1$ and $-1$.
  • Sine function key quadrant values:
  • $\sin(0^\circ) = 0$
  • $\sin(90^\circ) = 1$ (peak value at a quarter circle, $90^\circ$)
  • $\sin(180^\circ) = 0$ (halfway around circle, $180^\circ$)
  • $\sin(270^\circ) = -1$ (trough value at three-quarter circle, $270^\circ$)
  • $\sin(360^\circ) = 0$ (full circle completion, $360^\circ$)
  • Cosine function key quadrant values:
  • $\cos(0^\circ) = 1$
  • $\cos(90^\circ) = 0$
  • $\cos(180^\circ) = -1$
  • $\cos(270^\circ) = 0$
  • $\cos(360^\circ) = 1$
  • Standard quadrant angles ($0^\circ$, $90^\circ$, $180^\circ$, $270^\circ$, $360^\circ$) should be recognized conceptually without relying on a calculator.

Two-Dimensional Vector Problem Solving Example

• Example Problem: Cyclist Net Displacement

  • Scenario setup: A cyclist travels $1080\,\text{m}$ East, then turns and travels $1430\,\text{m}$ North.
  • Visual diagram construction:
  • Starting point $S$ located at the origin.
  • Eastward leg: Horizontal arrow pointing East of length $1080\,\text{m}$.
  • Northward leg: Vertical arrow pointing North from the end of the Eastward leg of length $1430\,\text{m}$.
  • Endpoint $E$ at the terminus of the Northward leg.
  • Net displacement vector drawn directly from $S$ to $E$, with the arrowhead pointing at E$.\n * Magnitude calculation via the Pythagorean theorem:\n * \text{Magnitude} = \sqrt{a^2 + b^2}\n * \text{Magnitude} = \sqrt{(1080\,\text{m})^2 + (1430\,\text{m})^2}\n * \text{Magnitude} = \sqrt{1166400\,\text{m}^2 + 2044900\,\text{m}^2} = \sqrt{3211300\,\text{m}^2} \approx 1792\,\text{m}\n * Direction angle (\theta) calculation:\n * Vector angle placement rule: The direction angle \theta$must always be placed at the **tail** of the displacement vector (where the observer/object starts at point$S).\n * Trigonometric ratio: \tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\n * Opposite side = Northward leg = 1430\,\text{m}\n * Adjacent side = Eastward leg = 1080\,\text{m}\n * Angle formula: \theta = \arctan\left(\frac{1430\,\text{m}}{1080\,\text{m}}\right) = \arctan(1.32407) \approx 52.9^\circ\n * Complete directional description: 52.9^\circ$$ North of East.

• Calculator Operations and Settings

  • Mode verification: Always verify whether the calculator is set to Degrees or Radians mode prior to performing trigonometric calculations.
  • Computing inverse trigonometric functions in Radians mode when reporting an answer in degrees produces incorrect numerical values.