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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \Delta x = 450 \, \text{km} \) | The total distance to the airport is \(450 \, \text{km}\). |
| 2 | \( t = 3.0 \, \text{h} \) | The total time to reach the airport is \(3.0 \, \text{hours}\). |
| 3 | \( v_x = \frac{\Delta x}{t} \) | The ground speed \( v_x \) required to reach the airport in the given time is the total distance divided by the total time. |
| 4 | \( v_x = \frac{450 \, \text{km}}{3.0 \, \text{h}} = 150 \, \text{km/h} \) | By calculation, the required ground speed is \(150 \, \text{km/h}\) due south. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_{\text{wind}} = 50.0 \, \text{km/h} \) | The wind speed blowing from the west is \(50.0 \, \text{km/h}\). |
| 2 | \( v_x = 150 \, \text{km/h} \) | The ground speed required to travel due south is \(150 \, \text{km/h}\). |
| 3 | \( v_{\text{plane}} \cos(\theta) = v_x \) | Horizontal component of the plane’s airspeed must counteract the wind to achieve the required southward ground speed. |
| 4 | \( v_{\text{plane}} \sin(\theta) = v_{\text{wind}} \) | Vertical component of the plane’s airspeed must counteract the wind speed blowing from the west. |
| 5 | \( v_{\text{plane}} = \sqrt{v_x^2 + v_{\text{wind}}^2} \) | Use Pythagorean theorem to find the magnitude of the plane’s airspeed vector. |
| 6 | \( v_{\text{plane}} = \sqrt{(150 \, \text{km/h})^2 + (50.0 \, \text{km/h})^2} \) | Substitute the known values into the equation. |
| 7 | \( v_{\text{plane}} = \sqrt{22500 \, \text{km}^2/\text{h}^2 + 2500 \, \text{km}^2/\text{h}^2} \) | Simplify the square terms inside the square root. |
| 8 | \( v_{\text{plane}} = \sqrt{25000 \, \text{km}^2/\text{h}^2} = 158.1 \, \text{km/h} \) | Evaluate the square root to find the plane’s airspeed. |
| 9 | \( \theta = \arctan\left(\frac{v_{\text{wind}}}{v_x}\right) \) | Use the inverse tangent function to find the heading angle \( \theta \) east of south. |
| 10 | \( \theta = \arctan\left(\frac{50.0}{150}\right) \) | Substitute the values for wind speed and ground speed. |
| 11 | \( \theta = \arctan\left(\frac{1}{3}\right) = 18.4^\circ \) | Evaluate the inverse tangent to find the heading angle. |
| 12 | \( v_{\text{plane}} = 158.1 \, \text{km/h}, \theta = 18.4^\circ \text{ east of south} \) | The required airspeed is \(158.1 \, \text{km/h}\) and the heading is \(18.4^\circ \) east of south. |
Just ask: "Help me solve this problem."
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
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?
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
\( v_{\text{plane}} = 158.1 \, \text{km/h}, \theta = 18.4^\circ \text{ east of south} \).
Note you could have also used the angle South of East, which would have been \(\theta = 18.4^\circ \text{ south of east} \). Either direction is acceptable.
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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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