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| Step | Formula / Calculation | Reasoning |
|---|---|---|
| 1 | [katex]v = 101 , \text{km/h}[/katex], [katex]t = 0.500 , \text{hr}[/katex] | Defining the constant speed and time for both boats. |
| 2 | [katex]v_{\text{m/s}} = \frac{101 \times 1000}{3600}[/katex] | Converting the speed from km/h to m/s. |
| 3 | [katex]v_{\text{west, boat 1}} = v_{\text{m/s}} \cos(\theta_{\text{boat 1}})[/katex], [katex]v_{\text{south, boat 1}} = v_{\text{m/s}} \sin(\theta_{\text{boat 1}})[/katex] | Decomposing Boat 1’s speed into westward and southward components. |
| 4 | [katex]v_{\text{west, boat 2}} = v_{\text{m/s}} \cos(\theta_{\text{boat 2}})[/katex], [katex]v_{\text{south, boat 2}} = v_{\text{m/s}} \sin(\theta_{\text{boat 2}})[/katex] | Decomposing Boat 2’s speed into westward and southward components. |
| 5 | [katex]\text{Distance west, boat 1} = v_{\text{west, boat 1}} \times t[/katex], [katex]\text{Distance south, boat 1} = v_{\text{south, boat 1}} \times t[/katex] | Calculating the westward and southward distances for Boat 1. |
| 6 | [katex]\text{Distance west, boat 2} = v_{\text{west, boat 2}} \times t[/katex], [katex]\text{Distance south, boat 2} = v_{\text{south, boat 2}} \times t[/katex] | Calculating the westward and southward distances for Boat 2. |
| 7 | [katex]\text{Difference west} = \text{Distance west, boat 1} – \text{Distance west, boat 2}[/katex] | Calculating the difference in the westward distance between the two boats. |
| 8 | [katex]\text{Difference south} = \text{Distance south, boat 2} – \text{Distance south, boat 1}[/katex] | Calculating the difference in the southward distance between the two boats. |
| 9 | [katex]\text{Difference west} \approx 0.98[/katex] km, [katex]\text{Difference south} \approx 1.61[/katex] km | Evaluating the differences in distances. |
During the half-hour journey, Boat 1 travels approximately 0.98 km farther west compared to Boat 2, and Boat 2 travels approximately 1.61 km farther south compared to Boat 1.
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Gregory was walking through the halls of the school when he realized that he was walking in perpendicular directions and he could easily calculate his displacement using the incredibly useful techniques he learned in physics. He recognized that he walked 12.5 meters left and then 18.9 meters down. How far must he walk to the right so that his resultant displacement is 20.1 m?
While Santa was delivering presents to the children of Nashville, Tennessee he experienced a strong wind perpendicular to his motion.
A person is standing at the edge of the water and looking out at the ocean. The height of the person’s eyes above the water is h = 1.8 m, and the radius of the Earth is R = 6.38 × 106 m. How far is it to the horizon (in meters)? In other words, find the distance d from the person’s eyes to the horizon.
(Note: At the horizon the angle between the line of sight and the radius of the earth is 90°)
A seagull first flies 160 m North, the heads up 120.65 m 18.43° North of West. After you land:
Determine the sum of the 3 vectors given below. Give the resultant (R) in terms of (a) vector components (b) resultant vector.
Vectors:
[katex] \vec{A} = 26.5 m [/katex] @ at 56° NW
[katex] \vec{B} = 44 m [/katex] @ at 28° NE
[katex] \vec{C} = 31 m [/katex] South
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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