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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_b = 5 \, \text{km/hr}\) | Identify the maximum speed of the boat in still water. |
| 2 | \(v_c = 4 \, \text{km/hr}\) | Identify the speed of the current flowing east. |
| 3 | \(v_{\text{net}} = \sqrt{v_b^2 – v_c^2}\) | Calculate the effective speed of the boat perpendicular to the current. Use Pythagorean theorem to determine the component of the speed that allows the boat to cross the river directly. |
| 4 | \(v_{\text{net}} = \sqrt{5^2 – 4^2}\) | Substitute given speeds into the formula for net velocity across the river. |
| 5 | \(v_{\text{net}} = \sqrt{25 – 16}\) | Simplify the expression within the square root. |
| 6 | \(v_{\text{net}} = \sqrt{9}\) | Calculate the difference. |
| 7 | \(v_{\text{net}} = 3 \, \text{km/hr}\) | Determine the net speed of the boat across the river. |
| 8 | Time to cross = \(\frac{\Delta x}{v_{\text{net}}} = \frac{1 \, \text{km}}{3 \, \text{km/hr}}\) | Calculate the time required to cross by dividing the river width by the net speed. |
| 9 | Time to cross = \(\frac{1}{3} \, \text{hours}\) | Simplify the expression to find the time taken in hours. |
| 10 | Time to cross = 20 \, \text{minutes} | Convert the time taken to minutes by multiplying by 60. |
Just ask: "Help me solve this problem."
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 \( h = 1.8 \, \text{m} \), and the radius of the Earth is \( R = 6.38 \times 10^6 \, \text{m} \). 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 \( 90^\circ \).)
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?
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
While Santa was delivering presents to the children of Nashville, Tennessee he experienced a strong wind perpendicular to his motion.
A seagull first flies \( 160 \, \text{m} \) North, then heads \( 120.65 \, \text{m} \) at \( 18.43^\circ \) North of West. After it lands:
\( \frac{1}{3} \, \text{hours} \) or \( 20 \, \text{minutes} \)
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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] |
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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