0 attempts
0% avg
UBQ Credits
# 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. |
Just ask: "Help me solve this problem."
The displacement \( x \) of an object moving in one dimension is shown above as a function of time \( t \). The acceleration of this object must be
The graph above represents the motion of an object traveling in a straight line as a function of time. What is the average speed of the object during the first four seconds? Note the displacemnt from 0 to 4 seconds is 2 meters
Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
The graph above shows velocity as a function of time for an object moving along a straight line. For which of the following sections of the graph is the acceleration constant and nonzero?
Which of the following graphs shows runners moving at the same speed? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
