| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x_A = v_{iA} t\] | Runner A moves at constant speed, so displacement \( \Delta x_A \) equals speed \( v_{iA} \) times time \( t \) measured from the instant A starts. |
| 2 | \[\Delta x_B = v_{iB} (t-3)\] | Runner B starts \(3\text{ s}\) later, therefore his running time is \( t-3 \) whenever \( t \ge 3\text{ s} \). |
| 3 | \[v_{iA} t = v_{iB}(t-3)\] | Catch-up occurs when both runners have the same displacement, i.e. \( \Delta x_A = \Delta x_B \). |
| 4 | \[t = \frac{3 v_{iB}}{v_{iB}-v_{iA}}\] | Algebraically solving the equality for time \( t \). |
| 5 | \[t = \frac{3\times9.0}{9.0-6.0}=9.0\text{ s}\] | Substitute \( v_{iA}=6.0\text{ m/s} \) and \( v_{iB}=9.0\text{ m/s} \). The resulting answer is the time at which the runners meet. |
| 6 | \[\Delta x = v_{iA} t = 6.0\times9.0 = 54\text{ m}\] | Using either equation from step 1 or 2, calculate displacement (meet up position) at the meeting time of \(9\) seconds. |
| 7 | \[\boxed{\text{Yes}}\] | Because \( 54\text{ m} < 100\text{ m} \), Runner B catches Runner A before the finish line. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\boxed{t = 9.0\text{ s}}\] | The catch-up time relative to Runner A’s start. |
| 2 | \[\boxed{\Delta x = 54\text{ m}}\] | Distance from the start line where they meet. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[t_A = \frac{100}{v_{iA}} = \frac{100}{6.0} = 16.7\text{ s}\] | Time for Runner A to cover \(100\text{ m}\). |
| 2 | \[t_B = 3.0 + \frac{100}{v_{iB}} = 3.0 + \frac{100}{9.0} = 14.1\text{ s}\] | Runner B starts \(3\text{ s}\) late but runs faster; add start delay to running time. |
| 3 | \[\Delta t = t_A – t_B = 16.7 – 14.1 = 2.6\text{ s}\] | Difference in finishing times. |
| 4 | \[\boxed{\text{Runner B wins by }2.6\text{ s}}\] | Runner B’s smaller total time shows he finishes first. |
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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:
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
A stone is thrown vertically upward with a speed of \( 24.0 \) \( \text{m/s} \).
The motion of a particle is described in the velocity vs. time graph shown above. Over the nine-second interval shown, we can say that the speed of the particle…
A car is heading rightward but accelerating to the left. This means the car is:
Which of these scenarios involve accelerated motion? (Select all that apply)
A car is driving to the right at \( 20 \) \( \text{m/s} \). A motorcycle starts \( 30 \) \( \text{m} \) behind the car and is moving at \( 30 \) \( \text{m/s} \) in the same direction.
In a 4.0-kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6.0 minutes. What is the average speed of the runner? Use standard units: m/s.
\(\text{Yes}\)
\(t = 9\,\text{s},\; \Delta x = 54\,\text{m}\)
\(\text{Runner B, }2.56\,\text{s}\)
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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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