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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( T_{up} = \frac{T}{2} \) | The total time in the air is \( T \). Therefore, the time taken to reach the maximum height is half the total time, \( \frac{T}{2} \). |
| 2 | \(H = v_i \left( \frac{T}{2} \right) – \frac{1}{2} g \left( \frac{T}{2} \right)^2\) | Using the kinematic equation for maximum height \( H \), where final velocity at the top is 0, initial velocity is \(v_i\), and \( g \) is acceleration due to gravity. |
| 3 | \( H = v_i \left( \frac{T}{2} \right) – \frac{1}{8} g T^2 \) | Simplifying the maximum height equation. |
| 4 | \( H = \left( \frac{v_i T}{2} \right) – \frac{1}{8}g T^2 \) | Simplifying same expression for better understanding. |
| 5 | \(\Delta x_{T/4} = v_i \left( \frac{T}{4} \right) – \frac{1}{2} g \left( \frac{T}{4} \right)^2 \) | Calculate the height after \(T/4\) time using the same kinematic equation. |
| 6 | \(\Delta x_{T/4} = v_i \left( \frac{T}{4} \right) – \frac{1}{32} g T^2 \) | Simplify the height equation for \(T/4\). |
| 7 | \(2 \Delta x_{T/4} = H + \frac{1}{8}g T^2 \) | Rewriting \(\Delta x_{T/4}\) equation and multiply both sides by 2 to normalize expression |
| 8 | \( 4 \Delta x_{T/4} = \left( \frac{v_i T}{2} \right) – \frac{1}{8}g T^2 + \frac{1}{8}g T^2 \) | Simplifying both sides and reordering variables |
| 9 | \( 4 \Delta x_{T/4} = H + H = 2H \) | Solving for \(H\), and then dividing right side by 2 to isolate \( \Delta(T/4) \). |
| 10 | \(\frac{3H}{4}\) | Thus the height after \(T/4\) seconds |
Just ask: "Help me solve this problem."
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
In which of the following is the rate of change of the particle’s momentum zero?
An object is thrown upward at \( 65 \, \text{m/s} \) from the top of a \( 800 \, \text{m} \) tall building.
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
Two balls are dropped off a cliff, 3 seconds apart. The first ball dropped is twice as heavy as the second ball dropped. Air resistance is negligible. While both balls are falling, the distance between the two balls is
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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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