| Derivation / Formula | Reasoning |
|---|---|
| \[(R + h)^2 = R^2 + d^2\] | Construct the right triangle whose hypotenuse is from Earth’s center to the observer’s eyes, with the right angle at the point on the surface where the line of sight is tangent; thus \(CT = R\), \(CP = R + h\), and \(PT = d\). |
| \[d^2 = (R + h)^2 – R^2\] | Rearrange the Pythagorean equation to isolate \(d^2\). |
| \[d = \sqrt{(R + h)^2 – R^2}\] | Take the square root of both sides to solve for \(d\). |
| \[d = \sqrt{2Rh + h^2}\] | Expand \((R + h)^2\) to \(R^2 + 2Rh + h^2\) and cancel the \(R^2\) terms. |
| \[d = \sqrt{2(6.38 \times 10^{6}\,\text{m})(1.8\,\text{m}) + (1.8\,\text{m})^{2}}\] | Insert the given values \(R = 6.38 \times 10^{6}\,\text{m}\) and \(h = 1.8\,\text{m}\). |
| \[d \approx \sqrt{2.2968 \times 10^{7}\,\text{m}^{2}}\] | Calculate \(2Rh = 2.2968 \times 10^{7}\,\text{m}^{2}\) and note that \(h^2 \approx 3.24\,\text{m}^{2}\) is negligible in comparison. |
| \[d \approx 4.79 \times 10^{3}\,\text{m}\] | Take the square root to find the distance, yielding about \(4.79\,\text{km}\). |
| \[\boxed{d \approx 4.8 \times 10^{3}\,\text{m}}\] | Present the final boxed result. |
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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 \(12.5\ \text{m}\) left and then \(18.9\ \text{m}\) down. How far must he walk to the right so that his resultant displacement is \(20.1\ \text{m}\)?
Two racing boats set out from the same dock and speed away at the same constant speed of \(101 \, \text{km/h}\) for half an hour (\(0.5 \, \text{hr}\)). Boat 1 is headed \(27.6^\circ\) south of west, and Boat 2 is headed \(35.3^\circ\) south of west, as shown in the graph above. During this half-hour calculate:
When we refer to an object’s speed, we’re talking about:
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
You are piloting a small plane, and you want to reach an airport \(450 \, \text{km}\) due south in \(3.0 \,\text{hours}\). A wind is blowing from the west at \(50.0 \,\text{km/h}\). What heading and airspeed should you choose to reach your destination in time?
A boat can row across a still \( 1 \, \text{km} \) wide river at a maximum speed of \( 5 \, \text{km/hr} \). If a current of \( 4 \, \text{km/hr} \) flows east as you try to directly cross the river, how long would it take?
While Santa was delivering presents to the children of Nashville, Tennessee he experienced a strong wind perpendicular to his motion.
Determine the sum of the three vectors given below. Calculate the resultant \( \vec{R} \) expressed as:
(a) Vector components
(b) Resultant vector (its total magnitude and direction)
\[\vec{A} = 26.5 \, \text{m} \ @ \ 56^\circ \, \text{NW}\]
\[\vec{B} = 44 \, \text{m} \ @ \ 28^\circ \, \text{NE}\]
\[\vec{C} = 31 \, \text{m} \, \text{South}\]
An object travels along a path shown above, with changing velocity as indicated by vectors \( A \) and \( B \). Which vector best represents the net acceleration of the object from time \( t_A \) to \( t_B \)? Vector A points north east and B points north west.
\(4.79 \times 10^{3}\,\text{m}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| 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] |
Metric Prefixes
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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