| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x_{\text{react}} = v_i t_r\] | The car travels at its initial speed \( v_i = 18.0\,\text{m/s} \) for the driver’s reaction time \( t_r = 0.350\,\text{s} \). |
| 2 | \[\Delta x_{\text{react}} = (18.0)(0.350) = 6.3\,\text{m}\] | Calculate the distance covered during the reaction interval. |
| 3 | \[\Delta x_{\text{left}} = 20.0 – 6.3 = 13.7\,\text{m}\] | Subtract the reaction distance from the original \( 20.0\,\text{m} \) gap to find how much roadway remains before the intersection when braking begins. |
| 4 | \[v_x^{2} = v_i^{2} + 2(-a)\,\Delta x_{\text{brake}}\] | Use the kinematic relation with final speed \( v_x = 0 \) and deceleration magnitude \( a = 3.65\,\text{m/s}^2 \). |
| 5 | \[\Delta x_{\text{brake}} = \frac{v_i^{2}}{2a} = \frac{(18.0)^2}{2(3.65)} = 44.4\,\text{m}\] | This is the distance actually required to stop once the brakes engage. |
| 6 | \[\Delta x_{\text{brake}} > \Delta x_{\text{left}}\] | Since \( 44.4\,\text{m} > 13.7\,\text{m} \), the car cannot stop before reaching the intersection. |
| 7 | \[\Delta x_{\text{beyond}} = 44.4 – 13.7 = 30.7\,\text{m}\] | The driver will still need this extra distance beyond the intersection to come to rest. |
| 8 | \[\boxed{30.7\,\text{m forward}}\] | The car will stop \( 30.7\,\text{m} \) past the start of the intersection; therefore, she will not be able to stop in time. |
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Can an object’s average velocity equal zero when object’s speed is greater than zero? Explain using the formula for average velocity vs speed.
Which of the following graphs represent an object having zero acceleration?
You throw a ball straight upward. It leaves your hand at \( 20 \) \( \text{m/s} \) and slows at a steady rate until it stops at the peak. The ball then comes back down, speeding up steadily until it hits the ground with the same speed it left your hand. Draw the velocity vs. time graph or explain it in terms of functions.
An object can move upward while having a downward acceleration.
Priscilla the Penguin stands at the edge of a rock ledge and tosses a small ice cube directly upward with an initial velocity of \( v_0 \). The ice cube’s initial height above the ground is \( 3.25 \, \text{m} \), and it reaches its maximum height above the ground \( 0.586 \, \text{s} \) after being thrown. The ice cube then plummets to the ground, missing the edge of the rock ledge on its way down.
Which of these scenarios involve accelerated motion? (Select all that apply)
A large beach ball is dropped from the ceiling of a school gymnasium to the floor about 10 meters below. Which of the following graphs would best represent its velocity as a function of time? (do not neglect air resistance)
A car accelerates from rest with an acceleration of \( 4.3 \, \text{m/s}^2 \) for a time of \( 6.8 \, \text{s} \). The car then slows to a stop with an acceleration of \( 5.1 \, \text{m/s}^2 \). What is the total distance traveled by the car?
A ball is dropped off a high cliff, and \( 2 \) \( \text{s} \) later another ball is thrown vertically downward with an initial speed of \( 30 \) \( \text{m/s} \). How long will it take the second ball to overtake the first?
A \( 1.5 \) \( \text{kg} \) block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is \( 0.300 \) and the coefficient of static friction is \( 0.400 \), calculate the amount of time that passes from when the force is applied to when the block stops.
\(\text{No}\)
\(30.7\,\text{m forward}\)
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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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