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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 rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance,
Which pair of quantities will always have the same magnitude if motion is in a straight line and in one direction?
Two students start \( 100 \) \( \text{m} \) apart.
• Student A walks to the right at \( 2 \) \( \text{m/s} \).
• Student B walks to the left at \( 3 \) \( \text{m/s} \).
At what time do the students meet, and how far has each student walked when they collide?
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.
A car travels at \( 20 \, \text{m/s} \) for \( 5 \, \text{mins} \) and then travels another \( 2 \, \text{km} \) at \( 40 \, \text{m/s} \). What is the total distance traveled and time of travel for the car?
\(183.24 \,\text{m}\)
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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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