| Derivation or Formula | Reasoning |
|---|---|
| \[\vec v_{\text{sled}}=\vec v_{\text{Santa (air)}}+\vec v_{\text{wind}}\] | Ground (actual) velocity is the vector sum of Santa’s velocity relative to the air and the wind velocity. |
| \[\vec v_{\text{Santa (air)}}=\langle -78.0,\,0\rangle\,\text{m/s},\qquad \vec v_{\text{wind}}=\langle 0,\,-20.3\rangle\,\text{m/s}\] | Take \(\hat x\) positive east and \(\hat y\) positive north. West is negative \(x\), south is negative \(y\). |
| \[\vec v_{\text{sled}}=\langle -78.0,\,0\rangle+\langle 0,\,-20.3\rangle=\langle -78.0,\,-20.3\rangle\,\text{m/s}\] | Add components to get the actual velocity vector. |
| \[|\vec v_{\text{sled}}|=\sqrt{(-78.0)^2+(-20.3)^2}\] | The speed is the magnitude of the velocity vector using the Pythagorean theorem. |
| \[|\vec v_{\text{sled}}|=\sqrt{6084+412.09}=\sqrt{6496.09}=80.6\,\text{m/s}\] | Compute the numerical magnitude. |
| \[\theta=\tan^{-1}\left(\frac{|v_y|}{|v_x|}\right)=\tan^{-1}\left(\frac{20.3}{78.0}\right)=14.6^\circ\] | The direction relative to west is found from the ratio of southward to westward components. |
| \[\boxed{\vec v_{\text{sled}}=\langle -78.0,\,-20.3\rangle\,\text{m/s}}\] | Final actual velocity vector (west and south components). |
| \[\boxed{|\vec v_{\text{sled}}|=80.6\,\text{m/s at }14.6^\circ\text{ south of west}}\] | Final speed and direction description. |
| Derivation or Formula | Reasoning |
|---|---|
| \[\vec v_{\text{sled}}=\vec v_{\text{Santa (air)}}+\vec v_{\text{wind}}\] | Same vector-addition relationship as in part (a). |
| \[\text{To go due west: }(v_{\text{sled}})_y=0\] | Due west means no north/south component in the ground velocity. |
| \[(v_{\text{sled}})_y=(v_{\text{Santa}})_y+(v_{\text{wind}})_y=0\] | Apply the condition \((v_{\text{sled}})_y=0\) using component addition. |
| \[(v_{\text{Santa}})_y+(-20.3)=0\Rightarrow (v_{\text{Santa}})_y=20.3\,\text{m/s}\] | Santa must aim northward so his north component cancels the wind’s south component. |
| \[|\vec v_{\text{Santa}}|=78.0\,\text{m/s},\quad (v_{\text{Santa}})_y=|\vec v_{\text{Santa}}|\sin\theta\] | If \(\theta\) is the angle north of west, the north component is \(78.0\sin\theta\). |
| \[78.0\sin\theta=20.3\Rightarrow \sin\theta=\frac{20.3}{78.0}=0.260\] | Set the required north component equal to \(20.3\,\text{m/s}\) and solve for \(\sin\theta\). |
| \[\theta=\sin^{-1}(0.260)=15.1^\circ\] | Compute the needed heading angle. |
| \[\boxed{\theta\approx 15.1^\circ\text{ north of west}}\] | Santa must point slightly north of west so the wind pushes him back to due west overall. |
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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)
\[\vec{A} = 26.5 \, \text{m} \ @ \ 56^\circ \, \text{NW}\]
\[\vec{B} = 44 \, \text{m} \ @ \ 28^\circ \, \text{NE}\]
\[\vec{C} = 31 \, \text{m} \, \text{South}\]
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:
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.
What does displacement mean in the context of motion?
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
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.
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
An object is moving to the west at a constant speed. Three forces are exerted on the object. One force is \( 10 \) \( \text{N} \) directed due north, and another is \( 10 \) \( \text{N} \) directed due west. What is the magnitude and direction of the third force if the object is to continue moving to the west at a constant speed?
When we refer to an object’s speed, we’re talking about:
\(\boxed{|\vec v_{\text{sled}}|=80.6\,\text{m/s at }14.6^\circ\text{ south of west},\quad \theta=15.1^\circ\text{ north of west}}\)
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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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