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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 seagull first flies \( 160 \, \text{m} \) North, then heads \( 120.65 \, \text{m} \) at \( 18.43^\circ \) North of West. After it lands:
While Santa was delivering presents to the children of Nashville, Tennessee he experienced a strong wind perpendicular to his motion.
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?
Seo-Jun throws a ball to her friend Zuri. The ball leaves Seo-Jun’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground with an initial speed \( \vec{v}_{s,0} = 12 \) \( \text{m/s} \) at an angle of \( \theta = 25^\circ \) with respect to the horizontal. Zuri catches the ball at a height of \( h = 1.5 \) \( \text{m} \) above the ground, as shown in the figure.
After catching the ball, Zuri throws it back to Seo-Jun. The ball leaves Zuri’s hand from a height \( h = 1.5 \) \( \text{m} \) above the ground. The ball is moving with a speed of \( 15 \) \( \text{m/s} \) when it reaches a maximum height of \( 5.8 \) \( \text{m} \) above the ground.
At what height \( h’ \) above the ground will the ball be when the return throw reaches Seo-Jun?
Does the odometer of a car measure a scalar or a vector quantity? What about the speedometer?
\( \frac{1}{3} \, \text{hours} \) or \( 20 \, \text{minutes} \)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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