Comprehensive Study Guide to Measurement, Kinematics, and One-Dimensional Motion

Course Administration and Technical Setup

Required Course Materials and Access:
- Access to WebAssign must be purchased for coursework.
- In-class participation marks are administered through Poll Everywhere.
- Submissions for in-class responses occur at https://pollev.com/adamz.
- When logging into Poll Everywhere, single sign-on (SSO) must be completed using University of Saskatchewan credentials (USask email address).
- If a prompt stating "Pre-registration required. Please log in to the Poll Everywhere account registered with Brian Zulkoskey to respond" appears, select "Log in with SSO" and authenticate with USask credentials.
- The Math-Readiness-Quiz must be completed prior to September 13th through the laboratory Canvas course.

Language, Dimensions, and Physical Standards

Precision of Language in Physics:
- Everyday conversational terms often conflate distinct physical quantities, requiring strict operational definitions in physics.
- Velocity and speed are not synonyms: speed is a scalar representing the magnitude of motion, whereas velocity is a vector incorporating both magnitude and spatial direction.
- Mass and weight represent fundamentally different physical quantities: mass is an intrinsic measure of inertia and quantity of matter ( $[M]$ ), whereas weight is the gravitational force acting on that mass ( $[M][L]/[T]^2$ ).
Fundamental Base Dimensions and SI Standards:
- Physical quantities are measured against standardized fundamental units.
- All physical quantities can be expressed in terms of seven base dimensions: length ( $[L]$ ), mass ( $[M]$ ), time ( $[T]$ ), electric charge (or electric current), temperature, amount of substance, and luminous intensity.
- In classical mechanics, the foundational base dimensions and their corresponding SI fundamental units are:
- Length ( $[L]$ ): meter ( $\text{m}$ )
- Mass ( $[M]$ ): kilogram ( $\text{kg}$ )
- Time ( $[T]$ ): second ( $\text{s}$ )
Components of Physical Quantities:
- Every physical quantity in an analysis must include three essential parts:
- The numerical value itself.
- The physical units associated with the measurement.
- The directional orientation in space whenever the quantity is a vector.

Significant Figures and Scientific Notation

Measurement Precision and Uncertainty:
- The numerical precision of any measured quantity is constrained by the limitations and resolution of the measuring instrument.
Operational Rules for Mathematical Operations:
- Multiplication and Division: The number of significant figures in the final product or quotient is governed by the factor possessing the lowest number of significant figures (the least accurate factor).
- Addition and Subtraction: The result must be rounded to match the smallest number of decimal places contained in any term of the operation.
- Multi-Step and Mixed Calculations:
- Intermediate values must never be rounded during intermediate calculation steps to prevent truncation error accumulation.
- When mixing mathematical operations across complex calculations, round the final result to match the same number of significant figures as the initial input value with the fewest significant figures.
Calculation Examples:
- Multiplication: $x = (2.50\,\text{m/s})(1.2\,\text{s}) = 3.0\,\text{m}$ (The factor $1.2\,\text{s}$ possesses 2 significant figures, constraining the result to 2 significant figures).
- Addition and Subtraction: $d = 2.50\,\text{m} + 4.25\,\text{m} - 2.0\,\text{m} = 4.75\,\text{m} = 4.8\,\text{m}$ (The term $2.0\,\text{m}$ is known to one decimal place, dictating that the final value must be rounded to one decimal place).
- Multi-Step Conversion Problem:
- Given $x = \frac{y}{z}$ , where $y = 3.25 \times 10^3\,\text{m}$ (3 significant figures) and $z = 4.7\,\text{h}$ (2 significant figures).
- Converting the time $z$ to seconds: $z = 4.7\,\text{h} \times \left(\frac{60\,\text{min}}{1\,\text{h}}\right) \times \left(\frac{60\,\text{s}}{1\,\text{min}}\right) = 4.7 \times 3600\,\text{s} = 16,920\,\text{s}$
- Determining $x$ in $\text{m/s}$ : $x = \frac{3.25 \times 10^3\,\text{m}}{16,920\,\text{s}} = 0.19208\,\text{m/s}$
- Standard conversion factors (such as $60\,\text{s/min}$ ) are exact definition numbers and do not limit the significant figures of the result.
- Rounding to 2 significant figures (constrained by $4.7\,\text{h}$ ): $x = 0.19\,\text{m/s}$
Scientific Notation Standards:
- Any real number can be uniquely expressed in scientific notation as the product of a coefficient between 1 and 10 and an integer power of 10.
- Explicit significant figures representation: if the value $12,500\,\text{m}$ is known to 4 significant figures, it is unambiguously written as $1.250 \times 10^4\,\text{m}$ .
- Calculators should be configured to display results directly in scientific notation to monitor precision accurately.

Metric Prefixes and Unit Conversions

Standard Metric Unit Prefixes:

| Prefix | Symbol | Power of 10 | | :--- | :--- | :--- | | tera | $\text{T}$ | $10^{12}$ | | giga | $\text{G}$ | $10^{9}$ | | mega | $\text{M}$ | $10^{6}$ | | kilo | $\text{k}$ | $10^{3}$ | | centi | $\text{c}$ | $10^{-2}$ | | milli | $\text{m}$ | $10^{-3}$ | | micro | $\mu$ | $10^{-6}$ | | nano | $\text{n}$ | $10^{-9}$ |

Unit Conversion Principles:
- Combining physical quantities requires establishing uniform units across all terms.
- Conversion factors are expressed as dimensional fractions equivalent to 1 (e.g., $1\,\text{m} = 100\,\text{c}\text{m} \implies \frac{100\,\text{cm}}{1\,\text{m}} = 1$ ), oriented so that undesired units cancel algebraically.
- When a physical unit carries an exponential power, the conversion factor must be raised to that identical power.
- Worked Conversion Example ( $1.92\,\text{m}^3/\text{s}$ to $\text{cm}^3/\text{min}$ ): $1.92\,\frac{\text{m}^3}{\text{s}} \times \left(\frac{100\,\text{cm}}{1\,\text{m}}\right)^3 \times \left(\frac{60\,\text{s}}{1\,\text{min}}\right)$ $= 1.92 \times (100)^3 \times 60\,\frac{\text{cm}^3}{\text{min}}$ $= 1.92 \times 10^6 \times 60\,\frac{\text{cm}^3}{\text{min}} = 1.152 \times 10^8\,\frac{\text{cm}^3}{\text{min}}$ $= 1.15 \times 10^8\,\frac{\text{cm}^3}{\text{min}}$

Dimensional Analysis

Algebraic Behavior of Dimensions:
- Dimensions are handled strictly as algebraic entities ( $[L]$ , $[M]$ , $[T]$ ).
- An equation is dimensionally valid if and only if both sides have identical net dimensions. Consequently, the resulting units on both sides of the equation must be identical.
- Dimensional analysis verifies the structural validity of physical equations, but cannot determine dimensionless numerical coefficients (such as $\frac{1}{2}$ , $2$ , or $\pi$ ).
Dimensional Verification Problem (PollEverywhere Q2):
- Task: Identify the dimensionally correct equation for distance $x$ ( $[x] = [L]$ ) given mass $[m] = [M]$ , speed $[v] = [L]/[T]$ , acceleration $[a] = [L]/[T]^2$ , and time $[t] = [T]$ .
- Evaluating potential equations:
- Testing $x = at^2$ : $[x] = [a][t]^2 \implies [L] = \left(\frac{[L]}{[T]^2}\right)([T])^2 = [L]$ (Dimensionally correct)
- Testing $x = at$ : $[x] = [a][t] \implies [L] = \left(\frac{[L]}{[T]^2}\right)([T]) = \frac{[L]}{[T]} \neq [L]$ (Incorrect)
- Testing $x = vt^2$ : $[x] = [v][t]^2 \implies [L] = \left(\frac{[L]}{[T]}\right)([T])^2 = [L][T] \neq [L]$ (Incorrect)
- Testing $x = v^2 t$ : $[x] = [v]^2[t] \implies [L] = \left(\frac{[L]}{[T]}\right)^2([T]) = \frac{[L]^2}{[T]} \neq [L]$ (Incorrect)
- Testing $x = mvt$ : $[x] = [m][v][t] \implies [L] = [M]\left(\frac{[L]}{[T]}\right)([T]) = [M][L] \neq [L]$ (Incorrect)

Coordinate Systems and Trigonometry

Coordinate System Definitions:
- Any spatial coordinate system requires:
- A designated fixed origin point ( $O$ ).
- A specified set of labeled axes or directions.
- Consistent mathematical rules for mapping coordinates relative to the origin and axes.
- Coordinate systems are required to define vector quantities in space.
Two-Dimensional Systems:
- Cartesian Coordinates: Points are identified by an ordered pair $(x, y)$ representing perpendicular displacements from the coordinate axes.
- Plane Polar Coordinates: Points are defined as $(r, \theta)$ , where $r$ represents the straight-line radial distance from the origin and $\theta$ represents the angle measured counterclockwise from a fixed reference line ( $\theta = 0^\circ$ , typically the positive $x$ -axis).

Trigonometric Coordinate Transformations:
- Using a right triangle where $r$ forms the hypotenuse, $x$ is the adjacent side along the horizontal axis, and $y$ is the opposite vertical side: $r^2 = x^2 + y^2 \implies r = \sqrt{x^2 + y^2}$ $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{r}$ $\cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r}$ $\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} = \frac{y}{x} = \frac{\sin(\theta)}{\cos(\theta)}$

Trigonometric relationships on a right triangle

Transformation Relations (PollEverywhere Q3 & Q4 / P115-Q3 & P115-Q4):
- Solving for Cartesian coordinates in terms of plane polar coordinates: $x = r\cos(\theta)$ $y = r\sin(\theta)$

Kinematics in One Dimension: Displacement, Speed, and Velocity

One-Dimensional Vectors and Displacement:
- Displacement (denoted $\Delta x$ ) is defined as the change in position of an object along a coordinate axis: $\Delta x = x_f - x_i$
- Unlike distance, displacement is a vector quantity that contains sign and directional information.

Displacement Examples Along a Number Line:
- Example 1 (Motion from position A to position B):
- Initial position: $x_i = x_A = +30\,\text{m}$
- Final position: $x_f = x_B = +50\,\text{m}$
- Displacement: $\Delta x = x_B - x_A = +50\,\text{m} - (+30\,\text{m}) = +20\,\text{m}$
- Example 2 (Motion from position B to position A):
- Initial position: $x_i = x_B = +50\,\text{m}$
- Final position: $x_f = x_A = +30\,\text{m}$
- Displacement: $\Delta x = x_A - x_B = +30\,\text{m} - (+50\,\text{m}) = -20\,\text{m}$
- Example 3 (Motion from position F to position D):
- Initial position: $x_i = x_F = -60\,\text{m}$
- Final position: $x_f = x_D = 0\,\text{m}$
- Displacement: $\Delta x = x_D - x_F = 0\,\text{m} - (-60\,\text{m}) = +60\,\text{m}$
Average Speed Versus Average Velocity:
- Average speed is a scalar defined as total path length divided by elapsed time: $\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}$
- Average velocity ( $\bar{v}$ ) is a vector defined as net displacement divided by elapsed time: $\bar{v} = \frac{\Delta x}{\Delta t} = \frac{x_f - x_i}{t_f - t_i}$
Comparative Example: Saskatoon-Regina Round Trip:
- A vehicle departs Saskatoon, travels $250\,\text{km}$ to the Regina airport, and returns $250\,\text{km}$ to Saskatoon in a total elapsed time of $5.00\,\text{h}$ .
- Average Speed Calculation (PollEverywhere Q6 / Q9): $\text{Average Speed} = \frac{250\,\text{km} + 250\,\text{km}}{5.00\,\text{h}} = \frac{500\,\text{km}}{5.00\,\text{h}} = 100\,\text{km/h}$
- Average Velocity Calculation (PollEverywhere Q7 / Q10):
- Because the vehicle starts and terminates at Saskatoon, $x_f = x_i$ . $\Delta x = x_f - x_i = 0\,\text{km}$ $\bar{v} = \frac{0\,\text{km}}{5.00\,\text{h}} = 0\,\text{km/h}$

Graphical Analysis of Motion

Slope of Position Versus Time ( $x$ vs. $t$ ):
- The slope of a line segment connecting two points on a position-time graph represents the average velocity ( $\bar{v} = \frac{\Delta x}{\Delta t}$ ).
- The slope of the tangent line to the curve at any specific time point represents the instantaneous velocity ( $v$ ).
Average Velocity Calculations on Linear Segments:
- Given points $A\,(0\,\text{s}, -20\,\text{m})$ , $B\,(10\,\text{s}, 0\,\text{m})$ , and $C\,(30\,\text{s}, 40\,\text{m})$ : $\bar{v}_{A \rightarrow B} = \frac{x_B - x_A}{t_B - t_A} = \frac{0\,\text{m} - (-20\,\text{m})}{10\,\text{s} - 0\,\text{s}} = \frac{+20\,\text{m}}{10\,\text{s}} = +2\,\text{m/s}$

Position versus time graph with multi-segment motion

Multi-Interval Analysis of Position-Time Graphs:
- Interval 1 ( $0\,\text{s}$ to $2.00\,\text{s}$ ): $\bar{v} = \frac{+10.0\,\text{m} - 0.00\,\text{m}}{2.00\,\text{s} - 0.00\,\text{s}} = +5.00\,\text{m/s}$
- Interval 2 ( $0\,\text{s}$ to $4.00\,\text{s}$ ): $\bar{v} = \frac{+5.00\,\text{m} - 0.00\,\text{m}}{4.00\,\text{s} - 0.00\,\text{s}} = +1.25\,\text{m/s}$
- Interval 3 ( $2.00\,\text{s}$ to $4.00\,\text{s}$ ): $\bar{v} = \frac{+5.00\,\text{m} - (+10.0\,\text{m})}{4.00\,\text{s} - 2.00\,\text{s}} = \frac{-5.00\,\text{m}}{2.00\,\text{s}} = -2.50\,\text{m/s}$
- Interval 4 ( $4.00\,\text{s}$ to $7.00\,\text{s}$ ): $\bar{v} = \frac{-5.00\,\text{m} - (+5.00\,\text{m})}{7.00\,\text{s} - 4.00\,\text{s}} = \frac{-10.0\,\text{m}}{3.00\,\text{s}} = -3.33\,\text{m/s}$
- Interval 5 ( $0\,\text{s}$ to $8.00\,\text{s}$ ): $\bar{v} = \frac{0.00\,\text{m} - 0.00\,\text{m}}{8.00\,\text{s} - 0.00\,\text{s}} = 0.00\,\text{m/s}$
Nonlinear Position Versus Time Graph Interpretation:

For a graph starting deep in the negative $x$ region and curving asymptotically towards $x = 0$ as time advances:
- Statement Selection: The object is moving in the positive direction and slowing down (Option E).
- Method 1 (Instantaneous Velocity Analysis): As $t$ increases, $x$ values become less negative, indicating displacement $\Delta x > 0$ (motion is in the positive $x$ -direction). Tangent lines drawn to the curve display positive slopes ( $v > 0$ ), but their steepness systematically decreases toward zero, demonstrating that the object is slowing down.
- Method 2 (Average Velocity Over Equal Intervals): Partitioning motion into equal, consecutive time intervals $\Delta t$ reveals that displacements decrease across successive intervals ( $\Delta x_2 < \Delta x_1$ ). Because $\bar{v}_1 = \frac{\Delta x_1}{\Delta t}$ and $\bar{v}_2 = \frac{\Delta x_2}{\Delta t}$ , it follows that $\bar{v}_2 < \bar{v}_1$ . The progressive decline in average velocity confirms that the object is decelerating.
- Matching Velocity-Time Graphs to Acceleration-Time Graphs:
The slope of a velocity versus time graph defines acceleration ( $\bar{a} = \frac{\Delta v}{\Delta t}$ ).
Linear $v$ vs. $t$ graph with constant non-zero slope corresponds to a horizontal line on the $a$ vs. $t$ graph ( $a = \text{constant}$ , matching Graph a with Graph e).
Parabolic/quadratic $v$ vs. $t$ graph with continuously steepening slope corresponds to a linearly increasing line on the $a$ vs. $t$ graph (matching Graph b with Graph d).
Segmented $v$ vs. $t$ graph with a linearly increasing segment followed by a plateau ( $v = \text{constant}$ ) corresponds to a constant positive acceleration that drops abruptly to zero ( $a = 0$ , matching Graph c with Graph f).

Matching velocity-time graphs to acceleration-time graphs

One-Dimensional Motion with Constant Acceleration

Physical Context and Properties:
- Constant acceleration occurs when velocity changes at a steady rate throughout motion.
- Applies accurately to physical systems including objects in free fall (under negligible air resistance) and systems subject to a constant net force.
- Because acceleration is uniform, the average acceleration equals instantaneous acceleration at all times: $\bar{a} = a = \text{constant}$
Standard Notational Definitions:
- Initial time: $t_i = 0$
- Initial velocity at $t = 0$ : $v_i = v_0$
- Elapsed time: $t_f = t$
- Velocity at elapsed time $t$ : $v_f = v$
Derivation of the Kinematic Equations:
- Kinematic Equation 1 (Velocity as a Function of Time):
- From the fundamental definition of acceleration: $a = \frac{v_f - v_i}{t_f - t_i} = \frac{v - v_0}{t - 0} = \frac{v - v_0}{t}$
- Multiplying by $t$ and isolating $v$ : $v - v_0 = at \implies v = v_0 + at$
- Kinematic Equation 2 (Displacement as a Function of Average Velocity):
- On a velocity-time graph, the area under the curve equals displacement ( $\Delta x$ ).

Velocity versus time graph showing discrete increments under constant acceleration

* When velocity changes linearly, the average velocity is the arithmetic mean of initial and final velocities:

$\bar{v} = \frac{v_0 + v}{2}$ * Since displacement is average velocity multiplied by time: $\Delta x = \bar{v}t = \frac{1}{2}(v_0 + v)t$

Kinematic Equation 3 (Displacement as a Function of Time and Acceleration):
- Substituting Kinematic Equation 1 ( $v = v_0 + at$ ) into Kinematic Equation 2: $\Delta x = \frac{1}{2}\left(v_0 + (v_0 + at)\right)t = \frac{1}{2}(2v_0 + at)t$ $\Delta x = v_0 t + \frac{1}{2}at^2$
Kinematic Equation 4 (Velocity as a Function of Displacement):
- Solving Kinematic Equation 1 for time: $t = \frac{v - v_0}{a}$
- Substituting this into Kinematic Equation 2: $\Delta x = \left(\frac{v + v_0}{2}\right)\left(\frac{v - v_0}{a}\right) = \frac{v^2 - v_0^2}{2a}$
- Rearranging algebraically: $v^2 - v_0^2 = 2a\Delta x \implies v^2 = v_0^2 + 2a\Delta x$
- Sign Conventions for Acceleration and Velocity (PollEverywhere Q9):
Question: If a coordinate system results in a vehicle having both negative velocity ( $v < 0$ ) and negative acceleration ( $a < 0$ ), which statement describes its motion?
Determination: A negative velocity signifies that the vehicle travels in the negative coordinate direction. When acceleration shares the same algebraic sign as velocity, the acceleration reinforces the motion, causing the object's speed to increase.
Correct Conclusion: The car is moving in the negative direction with increasing speed (Option A).
- Comprehensive Overtake Problem (PollEverywhere Q10):
Scenario: A motorcycle waiting at a traffic light accelerates from rest at $4.20\,\text{m/s}^2$ the instant the signal turns green. At that same moment, a car traveling at a constant speed of $15.0\,\text{m/s}$ passes the motorcycle and maintains that speed.
Initial Parameters at $t = 0\,\text{s}$ :
- Motorcycle ( $b$ ): $v_{0b} = 0\,\text{m/s}$ , $a_b = +4.20\,\text{m/s}^2$
- Car ( $c$ ): $v_{0c} = +15.0\,\text{m/s}$ , $a_c = 0\,\text{m/s}^2$
Part (a): Elapsed Time Before the Motorcycle Overtakes the Car:
- The overtake occurs when both vehicles have covered identical displacements from the traffic light ( $\Delta x_b = \Delta x_c$ ).
- Setting kinematic equations equal: $v_{0b}t + \frac{1}{2}a_b t^2 = v_{0c}t + \frac{1}{2}a_c t^2$ $0 + \frac{1}{2}a_b t^2 = v_{0c}t + 0$
- Dividing by $t$ ( $t \neq 0$ ): $\frac{1}{2}a_b t = v_{0c} \implies t = \frac{2v_{0c}}{a_b}$
- Evaluating numerically: $t = \frac{2(+15.0\,\text{m/s})}{+4.20\,\text{m/s}^2} = 7.1428\,\text{s} \approx 7.14\,\text{s}$
Part (b): Speed of the Motorcycle at Overtake:
- Applying the first kinematic equation: $v_b = v_{0b} + a_b t = 0 + (+4.20\,\text{m/s}^2)(7.1428\,\text{s}) = +30.0\,\text{m/s}$
Part (c): Distance Traveled From the Light When Overtake Occurs:
- Calculating displacement using the car's motion: $\Delta x_c = v_{0c}t = (+15.0\,\text{m/s})(7.1428\,\text{s}) = 107.14\,\text{m} \approx 107\,\text{m}$
- Cross-verification using the motorcycle's motion: $\Delta x_b = v_{0b}t + \frac{1}{2}a_b t^2 = 0 + \frac{1}{2}(+4.20\,\text{m/s}^2)(7.1428\,\text{s})^2 = 107.14\,\text{m} \approx 107\,\text{m}$