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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(0 \, \text{m/s} = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot t \) | At maximum height, the final velocity \(v_f\) is zero, the initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity). |
| 3 | \( t = \frac{5.0 \, \text{m/s}}{9.8 \, \text{m/s}^2} \approx 0.51 \, \text{s} \) | Calculate the time interval for the rock to reach its maximum height. This is only the time to reach the maximum height, so the total time to fall back to the original location is double this time. |
| 4 | \( t_{\text{total}} = 2 \cdot 0.51 \, \text{s} \approx 1.02 \, \text{s} \) | Multiply by 2 to get the total time interval for the rock to return to its original location. |
| 5 | \(\boxed{1.02 \, \text{s}}\) | Final answer for part (a) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = v_i + at \) | Use the first equation of motion to relate the initial velocity, final velocity, and time. |
| 2 | \(v_f = 5.0 \, \text{m/s} – 9.8 \, \text{m/s}^2 \cdot 1.02 \, \text{s} \) | The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), the acceleration \(a\) is \(-9.8 \, \text{m/s}^2\) (due to gravity), and the total time \(t\) is \(1.02 \, \text{s}\). |
| 3 | \(v_f \approx 5.0 \, \text{m/s} – 10.0 \, \text{m/s} = -5.0 \, \text{m/s} \) | Calculate the final velocity when the rock returns to the same height. Note the negative sign indicates the direction is downward. |
| 4 | \(\boxed{-5.0 \, \text{m/s}}\) | Final answer for part (b). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(a = -g\) | Acceleration due to gravity is always acting downward. |
| 2 | \(a = -9.8 \, \text{m/s}^2\) | Even at maximum height, the acceleration due to gravity remains \(-9.8 \, \text{m/s}^2\). |
| 3 | \(\boxed{-9.8 \, \text{m/s}^2}\) | Final answer for part (c). |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f = 0 \, \text{m/s}\) | At maximum height, the velocity of the rock is zero as it changes direction. |
| 2 | \(\boxed{0 \, \text{m/s}}\) | Final answer for part (d) |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v_f^2 = v_i^2 + 2a\Delta x\) | Use the third equation of motion to relate initial velocity, final velocity, acceleration, and displacement. |
| 2 | \(0 = (5.0 \, \text{m/s})^2 + 2(-9.8 \, \text{m/s}^2)\Delta x\) | At maximum height, the final velocity \(v_f\) is zero. The initial velocity \(v_i\) is \(5.0 \, \text{m/s}\), and acceleration \(a\) is \(-9.8 \, \text{m/s}^2\). |
| 3 | \(0 = 25 – 19.6 \Delta x \) | Simplify the equation. |
| 4 | \( 19.6 \Delta x = 25 \) | Rearrange the equation to solve for height \( \Delta x \). |
| 5 | \( \Delta x = \frac{25}{19.6} \approx 1.28 \, \text{m} \) | Calculate the maximum height using the final value obtained from the rearranged equation. |
| 6 | \( \boxed{1.28 \, \text{m}} \) | Final answer for part (e). |
Just ask: "Help me solve this problem."
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
A rocket, initially at rest, is fired vertically upward with an acceleration of \( 12.0 \, \text{m/s}^2 \). At an altitude of \( 1.00 \, \text{km} \), the rocket engine cuts off. Drag is negligible.
Can an object have \( 0 \) velocity and nonzero acceleration at the same time? Give two examples.
Which of the following statements about the acceleration due to gravity is TRUE?
A) \( 1.02 \, \text{s} \)
B) \( -5 \, \text{m/s} \)
C) \( -9.8 \, \text{m/s}^2 \)
D) \( 0 \, \text{m/s} \)
E) \( 1.28 \, \text{m} \)
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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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