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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_i = 0 \, \text{m/s} \) | The car starts from rest, so the initial velocity is zero. |
| 2 | \( a_1 = 4.3 \, \text{m/s}^2 \) | The car accelerates with an acceleration of \(4.3 \, \text{m/s}^2\). |
| 3 | \( t_1 = 6.8 \, \text{s} \) | The car accelerates for a time of \(6.8 \, \text{s}\). |
| 4 | \( v_x = v_i + a_1 t_1 \) | Use the formula for final velocity after a time interval, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time. |
| 5 | \( v_x = 0 + (4.3 \, \text{m/s}^2)(6.8 \, \text{s}) \) | Plug in the values for the initial velocity, acceleration, and time. |
| 6 | \( v_x = 29.24 \, \text{m/s} \) | Calculate the final velocity after the acceleration phase. |
| 7 | \( d_1 = v_i t_1 + \frac{1}{2} a_1 t_1^2 \) | Use the formula for distance traveled during the acceleration phase, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time. |
| 8 | \( d_1 = 0 \cdot 6.8 + \frac{1}{2} (4.3) (6.8)^2 \) | Set initial velocity term to zero and plug in the values for acceleration and time. |
| 9 | \( d_1 = 99.292 \, \text{m} \) | Calculate the distance traveled during the acceleration phase. |
| 10 | \( a_2 = -5.1 \, \text{m/s}^2 \) | The car decelerates with an acceleration of \( -5.1 \, \text{m/s}^2 \) (since the car is slowing down). |
| 11 | \( v_{x2} = 0 \, \text{m/s} \) | Final velocity when the car comes to rest again. |
| 12 | \( v_x^2 = v_i^2 + 2 a_2 \Delta x_2 \) | Use the kinematic equation relating initial velocity, final velocity, acceleration, and distance traveled. |
| 13 | \( 0 = (29.24)^2 + 2(-5.1) \Delta x_2 \) | Plug in the values for initial velocity during deceleration, final velocity, and acceleration. |
| 14 | \( 0 = 855.2976 – 10.2 \Delta x_2 \) | Calculate the term \( (29.24)^2 \) and simplify the equation. |
| 15 | \( \Delta x_2 = \frac{855.2976}{10.2} \) | Rearrange the equation to solve for the distance traveled during deceleration \( \Delta x_2 \). |
| 16 | \( \Delta x_2 = 83.95 \, \text{m} \) | Calculate the distance traveled during the deceleration phase. |
| 17 | \( d_{\text{total}} = d_1 + \Delta x_2 \) | Sum the distances traveled during the acceleration phase and deceleration phase to find the total distance. |
| 18 | \( d_{\text{total}} = 99.292 + 83.95 \) | Plug in the values for the distances calculated in both phases. |
| 19 | \( d_{\text{total}} = 183.242 \, \text{m} \) | Calculate the total distance traveled by the car and box the final answer. |
Just ask: "Help me solve this problem."
A 100-pound rock and a 1-pound metal arrow pointed downwards are dropped from height \( h \). Assuming there is no air resistance, which one hits the ground first and why?
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
An object is released from rest near the surface of a planet. The velocity of the object as a function of time is expressed in the following equation. \( v_y = (-3) \, \text{m/s}^2 \, t \) All frictional forces are considered to be negligible. What distance does the object fall \( 10 \) \( \text{s} \) after it is released from rest?
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.
Which pair of graphs represents the same 1- dimensional motion?
\(183.24 \,\text{m}\)
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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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