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Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \(\Delta x_{\text{Corvette}} = v_{\text{Corvette}} \cdot t\) | Write the formula for the displacement of the Corvette. Here, \(v_{\text{Corvette}} = 30 \, \text{m/s}\) and \(t\) is time. |
2 | \(\Delta x_{\text{Corvette}} = 30t\) | Substitute the given constant velocity of the Corvette into the displacement formula. |
3 | \(\Delta x_{\text{Civic}} = \frac{1}{2}a_{\text{Civic}} t^2\) | Write the formula for the displacement of the Civic starting from rest. Here, \(a_{\text{Civic}} = 6 \, \text{m/s}^2\). |
4 | \(\Delta x_{\text{Civic}} = \frac{1}{2} \cdot 6 \cdot t^2 = 3t^2\) | Substitute the given acceleration of the Civic into the displacement formula. |
5 | \(\Delta x_{\text{Corvette}} = \Delta x_{\text{Civic}}\) | Set the displacements equal to each other since the Civic catches up to the Corvette. |
6 | \(30t = 3t^2\) | Set the expressions we derived for displacement equal to each other. |
7 | \(t = \frac{30}{3} = 10 \, \text{s}\) | Solve for \(t\) by dividing both sides of the equation by 3. |
8 | \(\boxed{t = 10 \, \text{s}}\) | Final answer for the time it takes for the Civic to catch the Corvette. |
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \(\Delta x_{\text{Civic}} = \frac{1}{2}a t^2\) | Write the formula for the displacement of the Civic. Here, \(a = 6 \, \text{m/s}^2\) and \(t = 10 \, \text{s}\). |
2 | \(\Delta x_{\text{Civic}} = \frac{1}{2} \cdot 6 \cdot (10)^2\) | Substitute in the values for acceleration and time. |
3 | \(\Delta x_{\text{Civic}} = 3 \cdot 100 = 300 \, \text{m}\) | Calculate the displacement by multiplying. |
4 | \(\boxed{\Delta x_{\text{Civic}} = 300 \, \text{m}}\) | Final answer for the distance traveled by the Civic. |
Just ask: "Help me solve this problem."
A horizontal 300 N force pushes a 40 kg object across a horizontal 10 meter frictionless surface. After this, the block slides up a 20° incline. Assuming the incline has a coefficient of kinetic friction of 0.4, how far along the incline with the object slide?
A car travels at \( 20 \, \text{m/s} \) for 5 minutes and then travels another 2 km at \( 40 \, \text{m/s} \). What is the total distance traveled and time of travel for the car?
A coin is dropped from a hot air-balloon that is 250 m above the ground rising at 11 m/s upwards. For the coin, assume up is positive and find the following:
Ball 1 is dropped from rest at time \( t = 0 \) from a tower of height \( h \). At the same instant, ball 2 is launched upward from the ground with the initial speed \( v_0 \). If air resistance is negligible, at what time \( t \) will the two balls pass each other?
In which of the following is the rate of change of the particle’s momentum zero?
a) \( 10 \, \text{s} \)
b) \( 300 \, \text{m} \)
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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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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