| Derivation/Formula | Reasoning |
|---|---|
| \[A_x = 44.0\cos 28.0^{\circ}\] | Horizontal component of \(A\) lies along the \(+x\) axis. |
| \[A_y = 44.0\sin 28.0^{\circ}\] | Vertical component of \(A\); upward is positive. |
| \[B_x = -26.5\cos 56.0^{\circ}\] | \(B\) is above the \(-x\) axis; the \(x\)-component is leftward (negative). |
| \[B_y = 26.5\sin 56.0^{\circ}\] | \(y\)-component of \(B\) is upward (positive). |
| \[C_x = 0\] | Vector \(C\) has no horizontal part. |
| \[C_y = -31.0\] | \(C\) is directed straight down the \(-y\) axis. |
| \[A_x \approx 38.9,\; A_y \approx 20.7\] | Numerical values for \(A\). |
| \[B_x \approx -14.8,\; B_y \approx 22.0\] | Numerical values for \(B\). |
| \[R_x = A_x + B_x + C_x\] | Add horizontal components to obtain resultant \(x\)-component. |
| \[R_y = A_y + B_y + C_y\] | Add vertical components to obtain resultant \(y\)-component. |
| \[R_x \approx 24.0,\; R_y \approx 11.6\] | Numeric sum of components. |
| \[\boxed{\mathbf{R}_{\text{components}} = (24.0,\; 11.6)}\] | Resultant written in component form \((R_x, R_y)\). |
| Derivation/Formula | Reasoning |
|---|---|
| \[R = \sqrt{R_x^2 + R_y^2}\] | Pythagorean theorem for the magnitude of a 2-D vector. |
| \[R \approx \sqrt{(24.0)^2 + (11.6)^2} \approx 26.7\] | Insert values from part (a). |
| \[\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)\] | Inverse tangent yields the angle above the \(+x\) axis. |
| \[\theta \approx \tan^{-1}\left(\frac{11.6}{24.0}\right) \approx 25.8^{\circ}\] | Numerical evaluation of the direction. |
| \[\boxed{R \approx 26.7,\; \theta \approx 25.8^{\circ}\;\text{(above +x)}}\] | Magnitude and orientation of the resultant vector. |
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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?
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 \(20.1\ \text{m}\)?
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
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?
What does displacement mean in the context of 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)
\[\vec{A} = 26.5 \, \text{m} \ @ \ 56^\circ \, \text{NW}\]
\[\vec{B} = 44 \, \text{m} \ @ \ 28^\circ \, \text{NE}\]
\[\vec{C} = 31 \, \text{m} \, \text{South}\]
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.
An airplane is traveling \( 900. \) \( \text{km/h} \) in a direction \( 38.5^\circ \) west of north.
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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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