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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[V_{1x}=0\] | \(V_1\) points strictly along the negative \(y\)-axis, so it has no \(x\)-component. |
| 2 | \[V_{1y}=-6.0\] | The entire magnitude of \(V_1\) lies on the \(y\)-axis in the negative direction. |
| 3 | \[V_{2x}=V_2\cos45^{\circ}=4.5\left(\frac{\sqrt{2}}{2}\right)=3.18\] | Resolve \(V_2\) along the \(x\)-axis using the cosine of \(45^{\circ}\). |
| 4 | \[V_{2y}=V_2\sin45^{\circ}=4.5\left(\frac{\sqrt{2}}{2}\right)=3.18\] | Resolve \(V_2\) along the \(y\)-axis using the sine of \(45^{\circ}\). (Same value because \(\sin45^{\circ}=\cos45^{\circ}\).) |
| 5 | \[\boxed{(V_{1x},V_{1y})=(0,-6.0)}\quad\boxed{(V_{2x},V_{2y})=(3.18,3.18)}\] | Components of each vector boxed for clarity. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[R_x = V_{1x}+V_{2x}=0+3.18 = 3.18\] | Add the \(x\)-components to get the resultant \(x\)-component. |
| 2 | \[R_y = V_{1y}+V_{2y}=-6.0+3.18 = -2.82\] | Add the \(y\)-components to get the resultant \(y\)-component. |
| 3 | \[|\mathbf{R}| = \sqrt{R_x^2+R_y^2}=\sqrt{(3.18)^2+(-2.82)^2}=4.25\] | Pythagorean theorem yields the magnitude of the resultant vector. |
| 4 | \[\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)=\tan^{-1}\left(\frac{-2.82}{3.18}\right)=-42^{\circ}\] | The angle \(\theta\) is measured from the positive \(x\)-axis; the negative value indicates it lies below the axis. |
| 5 | \[\boxed{|\mathbf{R}|=4.25\text{ units at }-42^{\circ}}\] | Final expression for the sum’s magnitude and direction (\(42^{\circ}\) below the +\(x\)-axis). |
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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:
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?
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:
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 \(20.1\ \text{m}\)?
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?
An airplane is traveling \( 900. \) \( \text{km/h} \) in a direction \( 38.5^\circ \) west of north.
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
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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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