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UBQ Credits
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) How high does the ball rise? | ||
| 1 | Use the kinematic equation: [katex]v_f^2 = v_i^2 + 2a d[/katex] | This equation relates the final velocity, initial velocity, acceleration, and distance traveled. |
| 2 | [katex]0 = (22.5 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)d[/katex] | At the highest point, the final velocity [katex]v_f = 0[/katex] m/s. The initial velocity [katex]v_i = 22.5[/katex] m/s and acceleration [katex]a = -9.8[/katex] m/s[katex]^2[/katex] (gravity acting downwards). |
| 3 | [katex]d = \frac{(22.5 \, \text{m/s})^2}{2 \cdot 9.8 \, \text{m/s}^2} \approx 25.8 \, \text{m}[/katex] | Solve for [katex]d[/katex], the height reached by the ball. |
| 4 | [katex]d \approx 25.8 \, \text{m}[/katex] | The maximum height the ball reaches. |
| (b) How long does it take for the ball to reach its highest point? | ||
| 1 | Use the kinematic equation: [katex]v_f = v_i + at[/katex] | This equation relates the final velocity, initial velocity, acceleration, and time. |
| 2 | 0 = 22.5 \, \text{m/s} + (-9.8 \, \text{m/s}^2) t | At the highest point, [katex]v_f = 0[/katex] m/s. The initial velocity [katex]v_i = 22.5[/katex] m/s and acceleration [katex]a = -9.8[/katex] m/s[katex]^2[/katex]. |
| 3 | t = \frac{22.5 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 2.30 \, \text{s} | Solve for [katex]t[/katex], the time taken to reach the highest point. |
| 4 | t \approx 2.30 \, \text{s} | The time taken to reach the maximum height. |
| (c) How long does the ball remain in the air? | ||
| 1 | \text{Total time} = 2 \times t_{\text{up}} | Time to go up is equal to time to come down because distances and accelerations are the same. |
| 2 | 2 \times 2.30 \, \text{s} \approx 4.60 \, \text{s} | Double the time to reach the highest point to find total time in the air. |
| 3 | \text{Total time} \approx 4.60 \, \text{s} | The total time the ball remains in the air. |
| (d) How fast was it going just before it is caught? | ||
| 1 | Use symmetry: [katex]v_{\text{final}} = -v_{\text{initial}}[/katex] | Due to symmetry, speed upon returning to original height equals initial speed but opposite in direction. |
| 2 | \text{Speed} = 22.5 \, \text{m/s} | The magnitude of the speed is the same. |
| 3 | \text{Speed} = 22.5 \, \text{m/s} | The speed just before being caught. |
| (e) What is the velocity and acceleration of the ball at the highest point? | ||
| 1 | \text{Velocity} = 0 \, \text{m/s} | At the highest point, the velocity is zero as the ball changes direction. |
| 2 | \text{Acceleration} = -9.8 \, \text{m/s}^2 | Acceleration due to gravity remains constant throughout the motion. |
| 3 | \text{Velocity} = 0 \, \text{m/s}, \text{Acceleration} = -9.8 \, \text{m/s}^2 | Velocity and acceleration at the highest point. |
Just ask: "Help me solve this problem."
Above is a graph of the \(distance\) vs. time for car moving along a road. According the graph, at which of the following times would the automobile have been accelerating positively?
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 student is running at her top speed of \( 5.0 \, \text{m/s} \) to catch a bus, which is stopped at the bus stop. When the student is still \( 40.0 \, \text{m} \) from the bus, it starts to pull away, moving with a constant acceleration of \( 0.170 \, \text{m/s}^2 \).
A baseball is tossed from street level by a student straight up at a speed of \(25.3 \text{ m/s}\). After reaching maximum height, it is caught by another student on the roof of a building, \(17.4 \text{ m}\) above the street. How long did this take?
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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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