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UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_i = 40 \, \text{m/s} \) | Initial velocity of the car before the driver reacts to the red light. |
| 2 | \( t_{\text{reaction}} = 0.9 \, \text{s} \) | Time taken for the driver to react and hit the brakes. |
| 3 | \( v_x = 40 \, \text{m/s} \) | Car continues to travel with initial velocity during the reaction time. |
| 4 | \( \Delta x_{\text{reaction}} = v_i \cdot t_{\text{reaction}} \) | Distance traveled during the driver’s reaction time. |
| 5 | \( \Delta x_{\text{reaction}} = 40 \, \text{m/s} \times 0.9 \, \text{s} \) | Substituting the values for initial velocity and reaction time. |
| 6 | \( \Delta x_{\text{reaction}} = 36 \, \text{m} \) | Calculate the distance traveled during the reaction time. |
| 7 | \( a = -3.5 \, \text{m/s}^2 \) | Determine the acceleration (deceleration) after the driver hits the brakes. |
| 8 | \( v_f = 0 \, \text{m/s} \) | Final velocity of the car when it comes to a stop. |
| 9 | \( v_f^2 = v_i^2 + 2a \Delta x_{\text{braking}} \) | Using the kinematic equation to solve for the braking distance \( \Delta x_{\text{braking}} \). |
| 10 | \( 0 = (40 \, \text{m/s})^2 + 2(-3.5 \, \text{m/s}^2) \Delta x_{\text{braking}} \) | Substitute the values of initial velocity, acceleration, and final velocity into the kinematic equation. |
| 11 | \( 0 = 1600 \, \text{m}^2/\text{s}^2 – 7 \, \text{m/s}^2 \Delta x_{\text{braking}} \) | Simplify the equation by performing the multiplications and adding/subtracting. |
| 12 | \( 7 \, \text{m/s}^2 \Delta x_{\text{braking}} = 1600 \, \text{m}^2/\text{s}^2 \) | Rearrange the equation to isolate \( \Delta x_{\text{braking}} \). |
| 13 | \( \Delta x_{\text{braking}} = \frac{1600 \, \text{m}^2/\text{s}^2}{7 \, \text{m/s}^2} \) | Divide both sides by the coefficient of \( \Delta x_{\text{braking}} \). |
| 14 | \( \Delta x_{\text{braking}} = 228.57 \, \text{m} \) | Calculate the braking distance. |
| 15 | \( \Delta x_{\text{total}} = \Delta x_{\text{reaction}} + \Delta x_{\text{braking}} \) | Total distance traveled is the sum of the reaction distance and the braking distance. |
| 16 | \( \Delta x_{\text{total}} = 36 \, \text{m} + 228.57 \, \text{m} \) | Substitute the values into the total distance equation. |
| 17 | \( \Delta x_{\text{total}} = 264.57 \, \text{m} \) | Final distance traveled by the car before coming to a complete stop. |
Just ask: "Help me solve this problem."
An object is thrown straight upward at 64 m/s.
A car decelerates from \( 25 \, \text{m/s} \) to \( 5 \, \text{m/s} \) at \( 10 \, \text{m/s}^2 \). How far does the car travel during this deceleration?
A rock is thrown vertically upward with a velocity of \( 20 \, \text{m/s} \) from the edge of a bridge \( 42 \, \text{m} \) above a river.
In a 4.0-kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6.0 minutes. What is the average speed of the runner? Use standard units: m/s.
\( \Delta x_{\text{total}} = 264.57 \, \text{m} \)
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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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