0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_x = v_i + at_1\) | First, use the equation of motion to express final velocity \(v_x\) at \(t_1\). Since the sled starts from rest, \(v_i = 0\), so \(v_x = at_1\). |
| 2 | \(v_x = 13.0 \, \text{m/s}^2 \times t_1\) | Substitute the given acceleration \(a = 13.0 \, \text{m/s}^2\). |
| 3 | \(d_1 = \frac{1}{2} a t_1^2\) | Use the equation for displacement under constant acceleration to find \(d_1\), the distance covered during the acceleration phase. |
| 4 | \(d_1 = \frac{1}{2} \times 13.0 \, \text{m/s}^2 \times t_1^2\) | Substitute the value of \(a\). |
| 5 | \(d_2 = v_x(t_2 – t_1)\) | Use the equation for displacement under constant velocity for the distance traveled from \(t_1\) to \(t_2\). |
| 6 | \(d_2 = \left(13.0 \, \text{m/s}^2 \times t_1\right)\left(90.0\, \text{s} – t_1\right)\) | Substitute \(v_x\) and the time intervals into the equation for the second part of the journey. |
| 7 | \(d_1 + d_2 = 5300 \, \text{m}\) | Use the total distance formula, combining both displacements to equal the total given distance. |
| 8 | \(\frac{1}{2} \times 13.0 \, \text{m/s}^2 \times t_1^2 + (13.0 \, \text{m/s}^2 \times t_1)(90.0 \, \text{s} – t_1) = 5300 \, \text{m}\) | Combine the equations for \(d_1\) and \(d_2\). |
| 9 | \(6.5 t_1^2 + 1170 t_1 – 13 t_1^2 = 5300\) | Simplify the equation by combining like terms. |
| 10 | \(-6.5 t_1^2 + 1170 t_1 – 5300 = 0\) | Combine like terms and solve the quadratic equation via the quadratic formula or graphing. |
| 11 | \(t_1 = 4.65 \, \text{s}\) | Final answer. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(t_2 + t_1 = 90.0 \, \text{s}\) | The total time of travel is the time of the \(t_1\) and \(t_2\) added together. |
| 2 | \(t_2 = 90.0 \, \text{s} – t_1 \, \text{s}\) | Rearrange and solve for \(t_2\). |
| 3 | \(t_2 = 90.0 \, \text{s} – 4.58 \, \text{s}\) | Substitute the calculated value of \(t_1 \) found in part a. |
| 4 | \(t_2 = 85.35 \, \text{s}\) | Final \(t_2\) value. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v = a t_1\) | Use the equation for final velocity under constant acceleration. This comes from the kinematic formula \(v = v_0 + a t_1\), where \(v_0\) is zero. |
| 2 | \(v = 13.0 \, \text{m/s}^2 \times 4.65 \, \text{s}\) | Substitute the value of \(a\) and \(t_1\) into the equation. |
| 3 | \(v = 60.5 \, \text{m/s}\) | Calculate the velocity. |
– (a) \( \boxed{t_1 = 4.58 \, \text{s}} \)
– (b) \( \boxed{t_2 = 85.42\, \text{s}} \)
– (c) \( \boxed{v = 59.54 \, \text{m/s}} \)
Just ask: "Help me solve this problem."
A baseball is seen to pass upward by a window with a vertical speed of \( 14 \) \( \text{m/s} \). If the ball was thrown by a person \( 18 \) \( \text{m} \) below on the street, determine the following.
You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, \(12 \, \text{m}\) above, in a time of \(6 \, \text{s}\). The scale reads \(800 \, \text{N}\). What is the mass of the person?
A car increases its forward velocity uniformly from \(40 ~ \text{m/s}\) to \(80 ~ \text{m/s}\) while traveling a distance of \(200 ~ \text{m}\). What is its acceleration during this time?
Two students start \( 100 \) \( \text{m} \) apart.
• Student A walks to the right at \( 2 \) \( \text{m/s} \).
• Student B walks to the left at \( 3 \) \( \text{m/s} \).
At what time do the students meet, and how far has each student walked when they collide?
What does displacement mean in the context of motion?
\( t_1 = 4.65 \, \text{s} \)
\( t_2 = 85.35 \, \text{s} \)
\( v = 60.5 \, \text{m/s} \)
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 crafted THE Ultimate A.P Physics 1 course 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.
