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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(d = \frac{1}{2} g t^2\) | Use the formula for the distance traveled by an object under free fall where \( g \) is the gravitational acceleration and \( t \) is the time elapsed. |
| 2 | \(d_1 = \frac{1}{2} g t^2\) | Distance traveled by the first ball after time \( t \). |
| 3 | \(d_2 = \frac{1}{2} g (t-3)^2\) | Distance traveled by the second ball, which is dropped 3 seconds later. So, after time \( t \), it has been falling for \( t-3 \) seconds. |
| 4 | \(\Delta d = d_1 – d_2\) | The distance between the two balls is the difference between \( d_1 \) and \( d_2 \). |
| 5 | \(\Delta d = \frac{1}{2} g t^2 – \frac{1}{2} g (t-3)^2\) | Substitute the expressions for \( d_1 \) and \( d_2 \). |
| 6 | \(\Delta d = \frac{1}{2} g \left[ t^2 – (t-3)^2 \right]\) | Factor out \(\frac{1}{2} g\). |
| 7 | \(\Delta d = \frac{1}{2} g \left[ t^2 – (t^2 – 6t + 9) \right]\) | Expand the term \((t-3)^2\) to get \(t^2 – 6t + 9\). |
| 8 | \(\Delta d = \frac{1}{2} g \left[ t^2 – t^2 + 6t – 9 \right]\) | Distribute the negative sign across the terms inside the parenthesis. |
| 9 | \(\Delta d = \frac{1}{2} g \left[ 6t – 9 \right]\) | Simplify the expression by combining like terms (the \(t^2\) terms cancel each other out). |
| 10 | \(\Delta d = 3g \left( t – \frac{3}{2} \right)\) | Further simplify the expression by factoring out constants. |
| 11 | \(\Delta d \propto t\) | The expression \( \Delta d = 3g \left( t – \frac{3}{2} \right) \) shows that the distance between the balls \( \Delta d \) increases with time \( t \). |
| 12 | (b) increasing | Therefore, the correct answer is that the distance between the two balls is increasing with time. |
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A ranger in a national park is driving at \( 56 \, \text{km/h} \) when a deer jumps onto the road \( 65 \, \text{m} \) ahead of the vehicle. After a reaction time of \( t \, \text{s} \), the ranger applies the brakes to produce an acceleration of \( -3 \, \text{m/s}^2 \). What is the maximum reaction time allowed if the ranger is to avoid hitting the deer?
A person whose weight is \(4.92 \times 10^2 \, \text{N}\) is being pulled up vertically by a rope from the bottom of a cave that is \(35.2 \, \text{m}\) deep. The maximum tension that the rope can withstand without breaking is \(592 \, \text{N}\). What is the shortest time, starting from rest, in which the person can be brought out of the cave?
Which pair of quantities will always have the same magnitude if motion is in a straight line and in one direction?
A car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What additional distance will be covered in the third second?
Given the graph of velocity versus time for a duck flying due south for the winter, at what labeled point did the duck stop its forward motion?
Which pair of graphs represents the same 1-dimensional motion?
A helicopter is ascending vertically with a speed of \( 5.40 \) \( \text{m/s} \). At a height of \( 105 \) \( \text{m} \) above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
An object is dropped from the top of a 45 m tall building
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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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