0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(0 \, \text{m/s} = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot t \) | At maximum height, the final velocity \(v_f\) is zero, the initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity). |
| 3 | \( t = \frac{5.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 0.51 \, \text{s} \) | Calculate the time interval for the rock to reach its maximum height. This is only the time to reach the maximum height, so the total time to fall back to the original location is double this time. |
| 4 | \( t_{\text{total}} = 2 \cdot 0.51 \, \text{s} \approx 1.02 \, \text{s} \) | Multiply by 2 to get the total time interval for the rock to return to its original location. |
| 5 | \(\boxed{1.02 \, \text{s}}\) | Final answer for part (a) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(v_f = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot 1.02 \, \text{s} \) | The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity), and the total time \(t\) is \(1.02 \, \text{s}\). |
| 3 | \(v_f \approx 5.0 \, \text{m/s} – 10.0 \, \text{m/s} = -5.0 \, \text{m/s} \) | Calculate the final velocity when the rock returns to the same height. Note the negative sign indicates the direction is downward. |
| 4 | \(\boxed{-5.0 \, \text{m/s}}\) | Final answer for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(a = -g\) | Acceleration due to gravity is always acting downward. |
| 2 | \(a = -9.8 \, \text{m/s}^2\) | Even at maximum height, the acceleration due to gravity remains \(-9.8 \, \text{m/s}^2\). |
| 3 | \(\boxed{-9.8 \, \text{m/s}^2}\) | Final answer for part (c). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = 0 \, \text{m/s}\) | At maximum height, the velocity of the rock is zero as it changes direction. |
| 2 | \(\boxed{0 \, \text{m/s}}\) | Final answer for part (d) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f^2 = v_i^2 + 2a\Delta x\) | Use the third equation of motion to relate initial velocity, final velocity, acceleration, and displacement. |
| 2 | \(0 = (5.0 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)\Delta x\) | At maximum height, the final velocity \(v_f\) is zero. The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and acceleration \(a\) is \(-9.8 \, \text{m/s}^2\). |
| 3 | \(0 = 25 – 19.6 \Delta x \) | Simplify the equation. |
| 4 | \( 19.6 \Delta x = 25 \) | Rearrange the equation to solve for height \( \Delta x \). |
| 5 | \( \Delta x = \frac{25}{19.6} \approx 1.28 \, \text{m} \) | Calculate the maximum height using the final value obtained from the rearranged equation. |
| 6 | \( \boxed{1.28 \, \text{m}} \) | Final answer for part (e). |
Just ask: "Help me solve this problem."
A car travels east at a steady \( 30 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). What is its acceleration during this motion?
A driver is traveling at a speed of \( 18.0 \) \( \text{m/s} \) when she sees a red light ahead. Her car is capable of decelerating at a rate of \( 3.65 \) \( \text{m/s}^2 \). If it takes her \( 0.350 \) \( \text{s} \) to get the brakes on and she is \( 20.0 \) \( \text{m} \) from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction?
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from the ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magnitudes of their acceleration:
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?
A) \( 1.02 \, \text{s} \)
B) \( -5 \, \text{m/s} \)
C) \( -9.8 \, \text{m/s}^2 \)
D) \( 0 \, \text{m/s} \)
E) \( 1.28 \, \text{m} \)
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
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.
