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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{x}^{2}=v_{i}^{2}+2(-g)\Delta x\] | Use the kinematic relation \(v_{x}^{2}=v_{i}^{2}+2a\Delta x\) with upward positive (so \(a=-g\)). \(v_{x}=14\,\text{m/s}\) at the window, \(\Delta x=18\,\text{m}\). |
| 2 | \[v_{i}^{2}=v_{x}^{2}+2g\Delta x\] | Algebraically solve for \(v_{i}^{2}\). |
| 3 | \[v_{i}^{2}=14^{2}+2(9.8)(18)=196+352.8=548.8\] | Substitute the numerical values. |
| 4 | \[\boxed{v_{i}=23.43\,\text{m/s}}\] | Take the square root to obtain the initial speed. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[0=v_{i}^{2}+2(-g)\Delta x_{\text{max}}\] | At the peak, the final velocity is zero, so set \(v_{x}=0\). |
| 2 | \[\Delta x_{\text{max}}=\frac{v_{i}^{2}}{2g}\] | Re-arrange for the upward displacement from the street. |
| 3 | \[\Delta x_{\text{max}}=\frac{548.8}{19.6}=28\,\text{m}\] | Insert \(v_{i}^{2}=548.8\) and \(g=9.8\,\text{m/s}^{2}\). |
| 4 | \[\boxed{28\,\text{m}}\] | The ball rises 28 m above the street. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v_{x}=v_{i}-g t_{1}\] | Use \(v_{x}=v_{i}+at\) with \(a=-g\) to relate velocities and time. |
| 2 | \[t_{1}=\frac{v_{i}-v_{x}}{g}\] | Solve for \(t_{1}\), the interval from the throw to passing the window upward. |
| 3 | \[t_{1}=\frac{23.5-14}{9.8}=0.964\,\text{s}\] | Insert the numerical values. |
| 4 | \[\boxed{0.96\,\text{s}}\] | The ball was thrown roughly one second before being seen at the window. |
| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[T_{\text{total}}=\frac{2v_{i}}{g}\] | For motion that starts and ends at the same height, total flight time is twice the time to the peak, \(v_{i}/g\). |
| 2 | \[T_{\text{total}}=\frac{2(23.5)}{9.8}=4.79\,\text{s}\] | Insert \(v_{i}=23.5\,\text{m/s}\) and \(g=9.8\,\text{m/s}^{2}\). |
| 3 | \[t_{\text{after window}}=T_{\text{total}}-t_{1}=4.79-0.96=3.83\,\text{s}\] | Subtract the elapsed time before passing the window to find the interval after it. |
| 4 | \[\boxed{4.8\,\text{s}\;\text{(total from throw)}}\] | The ball returns to the street 4.8 s after being thrown, i.e., about 3.8 s after passing the window. |
Just ask: "Help me solve this problem."
An object’s velocity \(v\) as a function of time \(t\) is given in the graph. Which of the following statements is true about the motion of the object?
A horizontal spring with spring constant 162 N/m is compressed 50 cm and used to launch a 3 kg box across a frictionless, horizontal surface. After the box travels some distance, the surface becomes rough. The coefficient of kinetic friction of the box on the rough surface is 0.2. Find the total distance the box travels before stopping.
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
A Corvette is traveling at a constant velocity \( 30 \, \text{m/s} \) when it passes a stationary supped up Civic. At that moment, the Civic puts the pedal to the floor and accelerates at \( 6 \, \text{m/s}^2 \). The Civic eventually catches up to the Corvette.
Which of these scenarios involve accelerated motion? (Select all that apply)
\(23.43\,\text{m/s}\)
\(28\,\text{m}\)
\(0.96\,\text{s}\)
\(4.8\,\text{s}\)
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| 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] |
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