| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t_{\text{total}} = t_{\text{fall}} + t_{\text{sound}}\] | The observed delay \(t_{\text{total}} = 3.4\,\text{s}\) is the sum of the rock’s fall time and the sound’s travel time. |
| 2 | \[t_{\text{fall}} = \sqrt{\frac{2\Delta x}{g}}\] | For free-fall with \(v_i = 0\) and constant gravity, kinematics gives \(\Delta x = \tfrac{1}{2} g t_{\text{fall}}^{2}\). |
| 3 | \[t_{\text{sound}} = \frac{\Delta x}{v_{\text{sound}}}\] | Sound travels the same distance back up at speed \(v_{\text{sound}} = 340\,\text{m/s}\). |
| 4 | \[\sqrt{\frac{2\Delta x}{g}} + \frac{\Delta x}{v_{\text{sound}}} = t_{\text{total}}\] | Substitute Steps 2 and 3 into Step 1. |
| 5 | \[\sqrt{\frac{2\Delta x}{9.8}} + \frac{\Delta x}{340} = 3.4\] | Insert \(g = 9.8\,\text{m/s}^2\) and the given numerical values. |
| 6 | \[y = \sqrt{\Delta x}\] | Let \(y\) simplify the square-root term, turning the equation into a quadratic in \(y\). |
| 7 | \[y^{2} + 340 \sqrt{\tfrac{2}{9.8}}\,y – 340(3.4) = 0\] | Multiply Step 5 by \(340\) and substitute \(y\). |
| 8 | \[y^{2} + 153.6\,y – 1156 = 0\] | Evaluate \(340\sqrt{\tfrac{2}{9.8}} \approx 153.6\). |
| 9 | \[y = \frac{-153.6 + \sqrt{(153.6)^{2} + 4(1156)}}{2}\] | Apply the quadratic formula and keep the positive root (time must be positive). |
| 10 | \[y \approx 7.19\] | Numerical evaluation of Step 9. |
| 11 | \[\Delta x = y^{2} \approx 51.7 \,\text{m}\] | Convert back to height with \(\Delta x = y^{2}\). |
| 12 | \[\boxed{\Delta x \approx 51.7\,\text{m}}\] | Final height of the sea cliff, expressed and boxed. |
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You drop a rock off a bridge. When the rock has fallen \( 4 \) \( \text{m} \), you drop a second rock. As the two rocks continue to fall, what happens to their velocities?
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
Person A throws a ball horizontally from a cliff \( 20 \) \( \text{m} \) tall at \( 12 \) \( \text{m/s} \). Person B is running to the right on the ground and catches the ball at the same height it would’ve landed after running \( 15 \) \( \text{m} \). How fast was Person B running?
Which pair of graphs represents the same 1-dimensional motion?
A cart starts from rest and accelerates uniformly at \(4.0 \, \text{m/s}^2\) for \(5.0 \, \text{s}\). It next maintains the velocity it has reached for \(10 \, \text{s}\). Then it slows down at a steady rate of \(2.0 \, \text{m/s}^2\) for \(4.0 \, \text{s}\). What is the final speed of the car?
A driver is traveling at a speed of \( 18.0 \) \( \text{m/s} \) when she sees a red light ahead. Her car is capable of decelerating at a rate of \( 3.65 \) \( \text{m/s}^2 \). If it takes her \( 0.350 \) \( \text{s} \) to get the brakes on and she is \( 20.0 \) \( \text{m} \) from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction?
A whiffle ball is tossed straight up, reaches a highest point, and falls back down. Air resistance is not negligible. Which of the following statements are true?
In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the:
\(51.7\,\text{m}\)
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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] |
Metric Prefixes
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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