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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\text{total displacement}^2 = (\Delta x_{\text{left}} + \Delta x_{\text{right}})^2 + \Delta x_{\text{down}}^2 [/katex] | Apply the Pythagorean theorem to calculate total displacement where you have to walk a certain distance to the right to complete the diagonal hypotenuse. |
| 2 | [katex](20.1)^2 = (12.5 – x)^2 + (18.9)^2[/katex] | Substitute given values: [katex]\text{total displacement}=20.1 \, \text{m}[/katex], [katex]\Delta x_{\text{left}}=12.5\, \text{m}[/katex], [katex]\Delta x_{\text{down}}=18.9 \,\text{m}[/katex]. [katex]x[/katex] is the variable we need to solve. |
| 3 | [katex]404.01 = (12.5 – x)^2 + 357.21[/katex] | Simplify the squares, where [katex]404.01 = 20.1^2[/katex] and [katex]357.21 = 18.9^2 [/katex] |
| 4 | [katex]404.01 – 357.21 = (12.5 – x)^2[/katex] | Isolate the term [katex](12.5 – x)^2[/katex] by subtracting [katex]357.21[/katex] from both sides. |
| 5 | [katex]46.8 = (12.5 – x)^2[/katex] | Simplify further to isolate the squared term completely. |
| 6 | [katex]\sqrt{46.8} = 12.5 – x [/katex] | Take the square root of both sides to solve for [katex] x [/katex]. |
| 7 | [katex]6.84 = 12.5 – x [/katex] | Calculate the square root to approximate value. |
| 8 | [katex] x = 12.5 – 6.84 [/katex] | Rearrange the equation to isolate [katex] x [/katex]. |
| 9 | [katex]\boxed{x = 5.66 \, \text{m}} [/katex] | Final answer is boxed: Gregory must walk [katex]5.66 \, \text{m}[/katex] to the right to achieve the resultant displacement. |
Note – Gregory can also walk (12.5 + 5.66) meters, which is 19.34 meters. In this case his horizontal resultant displacement would be to the right, whereas, in the table above, it is to the left. In either case as long as the magnitude of his horizontal resultant displacement is 6.84 meters, then his total displacement will be 20.1 meters.
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?
Two racing boats set out from the same dock and speed away at the same constant speed of 101 km/h for half an hour (0.5 hr). Boat 1 is headed 27.6° south of west, and Boat 2 is headed 35.3° south of west, as shown in the graph above. During this half-hour calculate:
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 \( h = 1.8 \, \text{m} \), 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 \( 90^\circ \).)
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
5.67 m OR 19.34 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] |
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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