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 thrown vertically into the air with a velocity \( v \), and reaches a maximum height \( h \). At what height was the baseball moving with one-half its original velocity? Assume air resistance is negligible.
Which of the following statements about the acceleration due to gravity is TRUE?
A stone is thrown vertically upwards with a speed of \( 20.0 \) \( \text{m/s} \). How fast is it moving when it reaches a height of \( 12.0 \) \( \text{m} \)?
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
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?
\( t_1 = 4.65 \, \text{s} \)
\( t_2 = 85.35 \, \text{s} \)
\( v = 60.5 \, \text{m/s} \)
By continuing you (1) agree to our Terms of 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.
