| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x=800\,\text{m}-0\,\text{m}=800\,\text{m}\] | From the graph, Ashley travels from \(0\,\text{m}\) to \(800\,\text{m}\) during the first \(4\,\text{min}\). |
| 2 | \[\Delta t=4\,\text{min}=4(60)\,\text{s}=240\,\text{s}\] | The time must be converted to seconds because the standard velocity unit is \(\text{m/s}\). |
| 3 | \[\text{speed}=\frac{\Delta x}{\Delta t}=\frac{800\,\text{m}}{240\,\text{s}}=3.33\,\text{m/s}\] | Speed is distance traveled divided by elapsed time. |
| 4 | \[\boxed{3.33\,\text{m/s}}\] | Ashley’s speed for the first \(4\,\text{min}\) is \(3.33\,\text{m/s}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[d_{\text{total}}=800\,\text{m}+0\,\text{m}+800\,\text{m}=1600\,\text{m}\] | From \(0\) to \(4\,\text{min}\), Ashley travels \(800\,\text{m}\). From \(4\) to \(8\,\text{min}\), she is stopped, so she travels \(0\,\text{m}\). From \(8\) to \(10\,\text{min}\), she travels from \(800\,\text{m}\) to \(1600\,\text{m}\), which is another \(800\,\text{m}\). |
| 2 | \[\Delta t_{\text{total}}=10\,\text{min}=10(60)\,\text{s}=600\,\text{s}\] | The total trip time is \(10\,\text{min}\), converted to seconds. |
| 3 | \[\text{average speed}=\frac{d_{\text{total}}}{\Delta t_{\text{total}}}=\frac{1600\,\text{m}}{600\,\text{s}}=2.67\,\text{m/s}\] | Average speed depends on total distance traveled, not just final displacement. |
| 4 | \[\boxed{2.67\,\text{m/s}}\] | Ashley’s average speed for the entire trip is \(2.67\,\text{m/s}\). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x=x_f-x_i=1600\,\text{m}-0\,\text{m}=1600\,\text{m}\] | Ashley starts at her house and ends \(1600\,\text{m}\) east of her house, so her displacement is \(1600\,\text{m}\) east. |
| 2 | \[\Delta t_{\text{total}}=10\,\text{min}=600\,\text{s}\] | The total elapsed time for the trip is \(600\,\text{s}\). |
| 3 | \[v_{\text{avg}}=\frac{\Delta x}{\Delta t_{\text{total}}}=\frac{1600\,\text{m}}{600\,\text{s}}=2.67\,\text{m/s}\] | Average velocity is displacement divided by total time. Since Maria’s house is directly east of Ashley’s, the direction is east. |
| 4 | \[\boxed{2.67\,\text{m/s east}}\] | Ashley’s average velocity for the entire trip is \(2.67\,\text{m/s}\) east. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t=4\,\text{min}\text{ to }8\,\text{min}\] | The graph is horizontal from \(4\,\text{min}\) to \(8\,\text{min}\), which means Ashley’s distance from home is not changing. Therefore, she is stopped during this interval. |
| 2 | \[x=800\,\text{m}\] | During the horizontal section, the graph stays at \(800\,\text{m}\). |
| 3 | \[\boxed{800\,\text{m}}\] | Ashley was \(800\,\text{m}\) from her house when she stopped to talk to Maria. |
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The graph below is a plot of position versus time. For which labeled segments is the velocity positive and the acceleration negative?
Which of the following graphs represent an object at rest?
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The graph in the figure shows the position of a particle as it travels along the x-axis. At what value of \(t\) is the speed of the particle equal to \(0 \, \text{m/s}\)?
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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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