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# Part (a)
Calculate Ashley’s constant speed for the first 4 minutes.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \Delta x = 800 [/katex] m | Determine the displacement for the first 4 minutes from the graph; Ashley traveled 800 meters from 0 to 4 minutes. |
| 2 | [katex] \Delta t = 4 [/katex] min = [katex]4 \times 60[/katex] s = 240 s | Convert the time duration from minutes to seconds, as the standard unit of velocity is m/s. |
| 3 | [katex] v = \frac{\Delta x}{\Delta t} = \frac{800 \text{ m}}{240 \text{ s}} [/katex] | Use the formula for speed: speed equals displacement divided by time. |
| 4 | [katex] v = \frac{800}{240} \approx 3.33 \text{ m/s} [/katex] | Calculate the speed by dividing distance by time. |
# Part (b)
Calculate Ashley’s average speed for the entire trip.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \Delta x_{\text{total}} = 1600 [/katex] m | Determine the total distance from the graph; Ashley traveled 1600 meters in total. |
| 2 | [katex] \Delta t_{\text{total}} = 10 [/katex] min = [katex]10 \times 60[/katex] s = 600 s | Convert the total time duration from minutes to seconds. |
| 3 | [katex] v_{\text{avg}} = \frac{\Delta x_{\text{total}}}{\Delta t_{\text{total}}} = \frac{1600 \text{ m}}{600 \text{ s}} [/katex] | Use the formula for average speed: total distance divided by total time. |
| 4 | [katex] v_{\text{avg}} = \frac{1400}{600} \approx 2.67 \text{ m/s} [/katex] | Calculate the average speed by dividing total distance by total time. |
# Part (c)
Calculate Ashley’s average velocity for the entire trip.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \Delta x_{\text{net}} = 1600 [/katex] m to the east. | Determine the net displacement from the graph; In this case, both her distance and displacement are numerically the same. |
| 2 | [katex] \Delta t_{\text{total}} = 10 [/katex] min = [katex]10 \times 60[/katex] s = 600 s | Convert the total time duration from minutes to seconds. |
| 3 | [katex] v_{\text{avg}} = \frac{\Delta x_{\text{net}}}{\Delta t_{\text{total}}} = \frac{1600 \text{ m}}{600 \text{ s}} [/katex] | Use the formula for average velocity: total displacement divided by total time. |
| 4 | [katex] v_{\text{avg}} = \frac{1600}{600} \approx 2.67 \text{ m/s} \, \text{East} [/katex] | Calculate the average velocity by dividing net displacement by total time. |
# Part (d)
Determine how far Ashley was from her house when she stopped.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Read from Graph: [katex] \Delta x_{\text{stop}} = 800 [/katex] m | Ashley stopped at 4 minutes, as shown by the horizontal line on the graph. The distance at this point was 800 meters. |
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You throw a ball straight upward. It leaves your hand at \( 20 \) \( \text{m/s} \) and slows at a steady rate until it stops at the peak. The ball then comes back down, speeding up steadily until it hits the ground with the same speed it left your hand. Draw the velocity vs. time graph or explain it in terms of functions.
In which of the following is the rate of change of the particle’s momentum zero?
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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] |
General Metric Conversion Chart
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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