0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x_A = v_{iA} t\] | Runner A moves at constant speed, so displacement \( \Delta x_A \) equals speed \( v_{iA} \) times time \( t \) measured from the instant A starts. |
| 2 | \[\Delta x_B = v_{iB} (t-3)\] | Runner B starts \(3\text{ s}\) later, therefore his running time is \( t-3 \) whenever \( t \ge 3\text{ s} \). |
| 3 | \[v_{iA} t = v_{iB}(t-3)\] | Catch-up occurs when both runners have the same displacement, i.e. \( \Delta x_A = \Delta x_B \). |
| 4 | \[t = \frac{3 v_{iB}}{v_{iB}-v_{iA}}\] | Algebraically solving the equality for time \( t \). |
| 5 | \[t = \frac{3\times9.0}{9.0-6.0}=9.0\text{ s}\] | Substitute \( v_{iA}=6.0\text{ m/s} \) and \( v_{iB}=9.0\text{ m/s} \). |
| 6 | \[\Delta x = v_{iA} t = 6.0\times9.0 = 54\text{ m}\] | Compute the common displacement at the meeting time. |
| 7 | \[\boxed{\text{Yes}}\] | Because \( 54\text{ m} < 100\text{ m} \), Runner B catches Runner A before the finish line. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\boxed{t = 9.0\text{ s}}\] | The catch-up time relative to Runner A’s start. |
| 2 | \[\boxed{\Delta x = 54\text{ m}}\] | Distance from the start line where they meet. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t_A = \frac{100}{v_{iA}} = \frac{100}{6.0} = 16.7\text{ s}\] | Time for Runner A to cover \(100\text{ m}\). |
| 2 | \[t_B = 3.0 + \frac{100}{v_{iB}} = 3.0 + \frac{100}{9.0} = 14.1\text{ s}\] | Runner B starts \(3\text{ s}\) late but runs faster; add start delay to running time. |
| 3 | \[\Delta t = t_A – t_B = 16.7 – 14.1 = 2.6\text{ s}\] | Difference in finishing times. |
| 4 | \[\boxed{\text{Runner B wins by }2.6\text{ s}}\] | Runner B’s smaller total time shows he finishes first. |
Just ask: "Help me solve this problem."
On Saturday, Ashley rode her bicycle to visit Maria. Maria’s house is directly east of Ashley’s. The graph shows how far Ashley was from her house after each minute of her trip.(Hint – Use the standard units of velocity (m/s) for all parts)
A block is projected up a ramp with an initial speed \( v_0 \). It travels along the surface of the ramp with constant acceleration \( a \). Take the positive direction of motion to be up the ramp. If the acceleration vector points opposite the initial velocity vector, which of the following MUST be true?
You are a bungee jumping fanatic and want to be the first bungee jumper on Jupiter. The length of your bungee cord is \( 45.0 \) \( \text{m} \). Free fall acceleration on Jupiter is \( 23.1 \) \( \text{m/s}^2 \). What is the ratio of your speed on Jupiter to your speed on Earth when you have dropped \( 45 \) \( \text{m} \)? Ignore the effects of air resistance and assume that you start at rest.
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance,
The first \(10 \, \text{meters}\) of a \(100 \, \text{meter}\) dash are covered in \(2 \, \text{seconds}\) by a sprinter who starts from rest and accelerates with a constant acceleration. The remaining \(90 \, \text{meters}\) are run with the same velocity the sprinter had after \(2 \, \text{seconds}\).
\(\text{Yes}\)
\(t = 9\,\text{s},\; \Delta x = 54\,\text{m}\)
\(\text{Runner B, }2.56\,\text{s}\)
By continuing you (1) agree to our Terms of Use and Terms of Sale 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
We created THE ultimate A.P Physics 1 course by simplifying everything so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
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.
