| 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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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?
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:
An airplane is traveling \( 900. \) \( \text{km/h} \) in a direction \( 38.5^\circ \) west of north.
Vector \( A \) is \( 44.0 \) units and \( 28.0^\circ \) above the \( +x \) axis, vector \( B \) is \( 26.5 \) units and \( 56.0^\circ \) above the \( -x \) axis, and vector \( C \) is \( 31.0 \) units along the \( -y \) axis. Determine the resultant (sum) of the three vectors.
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.
While Santa was delivering presents to the children of Nashville, Tennessee he experienced a strong wind perpendicular to his motion.
When we refer to an object’s speed, we’re talking about:
A seagull first flies \( 160 \, \text{m} \) North, then heads \( 120.65 \, \text{m} \) at \( 18.43^\circ \) North of West. After it lands:
\(24.0\,\hat\imath+11.6\,\hat j\) 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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