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UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v(t) = \text{Area under velocity-time graph}\) | Object passes through its initial position when the net area under the velocity-time graph (Displacement, \( \Delta x \)) is zero. |
| 2 | N/A (Graph Inspection) | From \( t = 0 \, \text{s} \) to \( t = 5 \, \text{s} \), the area under the curve is a triangle with base 5 s and height 20 m/s giving an area of \( \frac{1}{2} \times 5 \, \text{s} \times 20 \, \text{m/s} = 50 \, \text{m} \). |
| 3 | N/A (Graph Inspection) | From \( t = 5 \, \text{s} \) to \( t = 11 \, \text{s} \), the area under the curve is a large negative area as shown in the graph. Calculate areas to compensate for positive 50 m. |
| 4 | N/A (Graph Inspection) | The area from \( t = 5 \, \text{s} \) to \( t = 8 \, \text{s} \) is -40, while the area from \( t = 5 \, \text{s} \) to \( t = 9 \, \text{s} \) is -60 m. |
| 5 | Conclusion | Between 8 to 9 seconds the area, representing displacement, reaches \( -50 \, \text{m} \), which would cancel out the \( +50 \, \text{m} \) covered in the first 5 seconds. Thus, between 8 and 9 seconds, displacement is 0, which means the particle is back to its original starting point. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\text{Average acceleration} = \frac{\Delta v}{\Delta t}\) | The average acceleration is the change in velocity divided by the change in time (the slope of the graph) |
| 2 | \(\Delta v = v_f – v_i = 0 – 0 \, \text{m/s} = 0 \, \text{m/s}\) | The change in velocity for \(12 \, \text{s} < t < 14 \, \text{s}\) is \( \Delta v = 0 \, \text{m/s}\). |
| 3 | \(\Delta t = 14 \, \text{s} – 12 \, \text{s} = 2 \, \text{s}\) | The time interval is \( \Delta t = 2 \, \text{s}\). |
| 4 | \(\text{Average acceleration} = \frac{0 \, \text{m/s}}{2 \, \text{s}}\) | Substitute the values obtained for \( \Delta v \) and \( \Delta t \). |
| 5 | \(\text{Average acceleration} = 0 \, \text{m/s}^2\) | The average acceleration for \(12 \, \text{s} < t < 14 \, \text{s}\) is \( 0 \, \text{m/s}^2\). |
| 6 | Find the identical slope on a different part of of the graph. | The interval from \( 7 \, \text{s} \) to \( 10 \, \text{s} \) also shows 0 slope resulting in same acceleration. |
Just ask: "Help me solve this problem."
In which of the following is the rate of change of the particle’s momentum zero?
Which of the following graphs shows runners moving at the same speed? Assume the \(y\)-axis is measured in meters and the \(x\)-axis is measured in seconds.
Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
The displacement \(x\) of an object moving in one dimension is shown above as a function of time \(t\). The velocity of this object must be
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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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