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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] v_f^2 = v_i^2 + 2ad [/katex] | Use the kinematic equation relating final velocity [katex] v_f [/katex], initial velocity [katex] v_i [/katex], acceleration [katex] a [/katex], and distance [katex] d [/katex]. |
| 2 | [katex] v_f = 0 [/katex] | Since the car skids to a stop, the final velocity [katex] v_f [/katex] is 0. |
| 3 | [katex] 0 = v_i^2 + 2a d [/katex] | Substitute [katex] v_f = 0 [/katex] into the kinematic equation. |
| 4 | [katex] d = \frac{-v_i^2}{2a} [/katex] | Solve the equation for distance [katex] d [/katex]. |
| 5 | [katex] d_1 = \frac{-v_i^2}{2a} [/katex] | Express the initial distance [katex] d_1 [/katex] using the initial speed [katex] v_i [/katex]. |
| 6 | [katex] v_f = 0, v_i = 2v_i \rightarrow (2v_i)^2 [/katex] | Consider the scenario where the initial speed is twice as fast, i.e.,[katex] v_i = 2v_i [/katex]. Substituting this in the same equation. |
| 7 | [katex] d_2 = \frac{-(2v_i)^2}{2a} [/katex] | Solve for the new distance [katex] d_2 [/katex] using the new initial speed [katex] 2v_i [/katex]. |
| 8 | [katex] d_2 = \frac{-4v_i^2}{2a} = 4 \left( \frac{-v_i^2}{2a} \right) = 4d_1 [/katex] | By simplifying the resulting equation, we see that the new stopping distance is 4 times the initial stopping distance. |
| 9 | 4 times as far: (a) | Therefore, if the car was traveling twice as fast, it would skid 4 times as far. |
| 10 | (b) The same distance | Incorrect: This contradicts the earlier derivation that shows the distance scales quadratically with speed. |
| 11 | (c) 2 times as far | Incorrect: This would be true if distance scaled linearly with speed, which it does not. |
| 12 | (d) The mass of the car must be known | Incorrect: The stopping distance in the given context is independent of the mass of the car. |
Just ask: "Help me solve this problem."
A ball is tossed directly upward. Its total time in the air is \( T \). Its maximum height is \( H \). What is its height after it has been in the air a time \( T/4 \)? Air resistance is negligible.
A car travels east at a steady \( 30 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). What is its acceleration during this motion?
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 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
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?
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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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