| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[A_x=-A\cos56^\circ\] | Vector \(\vec A\) points \(56^\circ\) north of west, so its horizontal component is westward (negative). |
| 2 | \[A_y=A\sin56^\circ\] | The vertical component of \(\vec A\) is northward (positive). |
| 3 | \[A_x\;{=}\;-14.8\,\text{m},\;\;A_y\;{=}\;22.0\,\text{m}\] | Substituted \(A=26.5\,\text{m}\) and the trigonometric values \(\cos56^\circ=0.559\) and \(\sin56^\circ=0.829\). |
| 4 | \[B_x=B\cos28^\circ,\;\;B_y=B\sin28^\circ\] | \(\vec B\) is \(28^\circ\) north of east, giving positive \(x\) and \(y\) components. |
| 5 | \[B_x\;{=}\;38.9\,\text{m},\;\;B_y\;{=}\;20.7\,\text{m}\] | Inserted \(B=44\,\text{m}\) with \(\cos28^\circ=0.883\) and \(\sin28^\circ=0.469\). |
| 6 | \[C_x=0,\;\;C_y=-31\,\text{m}\] | \(\vec C\) points straight south, so only a negative \(y\) component exists. |
| 7 | \[R_x=A_x+B_x+C_x\] | Horizontal components add algebraically. |
| 8 | \[R_x=-14.8+38.9+0=24.0\,\text{m}\] | Computed the net \(x\)-component. |
| 9 | \[R_y=A_y+B_y+C_y\] | Vertical components add algebraically. |
| 10 | \[R_y=22.0+20.7-31=11.6\,\text{m}\] | Computed the net \(y\)-component. |
| 11 | \[\boxed{\vec R = 24.0\,\hat\imath + 11.6\,\hat j\;\text{m}}\] | Expressed \(\vec R\) in component form. |
| 12 | \[\boxed{R_x = 24,\; R_y = 11.6}\] | Alternative way of writing final answer |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[R=\sqrt{R_x^2+R_y^2}\] | Pythagorean theorem gives the magnitude of the resultant in 2-D. |
| 2 | \[R=\sqrt{(24.0)^2+(11.6)^2}=26.7\,\text{m}\] | Inserted the component values from part (a). |
| 3 | \[\theta=\tan^{-1}\!\left(\frac{R_y}{R_x}\right)\] | Angle measured from the positive \(x\)-axis toward the positive \(y\)-axis. |
| 4 | \[\theta=\tan^{-1}\!\left(\frac{11.6}{24.0}\right)=25.8^\circ\] | Computed the arctangent; both components are positive, placing \(\vec R\) in the NE quadrant. |
| 5 | \[\boxed{R=26.7\,\text{m}\;@\;25.8^\circ\;\text{NE}}\] | Final magnitude and compass direction (degrees north of east). |
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What does displacement mean in the context of motion?
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?
An airplane is traveling \( 900. \) \( \text{km/h} \) in a direction \( 38.5^\circ \) west of north.
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 \(20.1\ \text{m}\)?
Two racing boats set out from the same dock and speed away at the same constant speed of \(101 \, \text{km/h}\) for half an hour (\(0.5 \, \text{hr}\)). Boat 1 is headed \(27.6^\circ\) south of west, and Boat 2 is headed \(35.3^\circ\) south of west, as shown in the graph above. During this half-hour calculate:
Vector \( V_1 \) is \( 6.0 \) units long and points along the negative \( y \) axis. Vector \( V_2 \) is \( 4.5 \) units long and points at \( +45^\circ \) to the positive \( x \) axis.
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
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 \( 90^\circ \).)
\(24.0\,\hat\imath+11.6\,\hat j\) or \(\boxed{R_x = 24,\; R_y = 11.6}\)
\(26.7\,\text{m}\text{ at }25.8^\circ\text{ NE}\)
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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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