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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | Choose east as +x. Set the red car’s start at the origin. \[x_R(0)=0,\quad v_{R0}=0,\quad a_R=+3.5,\text{m/s}^2\] Blue car starts 600 m to the east: \[x_B(0)=600,\text{m},\quad v_{B0}=-15,\text{m/s},\quad a_B=-1.2,\text{m/s}^2\] |
Since the blue car moves west, its velocity is negative. Interpreting “accelerating at 1.2 m/s²” while traveling west means its acceleration is also westward (negative) in this coordinate system. |
| 2 | Constant-acceleration position functions: \[x_R(t)=x_R(0)+v_{R0}t+\tfrac{1}{2} a_R t^2=\tfrac{1}{2}(3.5)t^2=1.75t^2\] \[x_B(t)=x_B(0)+v_{B0}t+\tfrac{1}{2} a_B t^2=600-15t-0.6t^2\] |
Use \(x=x_0+v_0 t+\tfrac{1}{2} a t^2\) for each car. |
| 3 | They meet when \(x_R(t)=x_B(t)\): \[1.75t^2=600-15t-0.6t^2\] \[2.35t^2+15t-600=0\] |
Meeting means same position at the same time; rearranging gives a quadratic in \(t\). |
| 4 | Quadratic formula for \(at^2+bt+c=0\) with \(a=2.35\), \(b=15\), \(c=-600\): \[t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] Discriminant: \[D=b^2-4ac=15^2-4(2.35)(-600)=225+5640=5865\] |
Solve the quadratic exactly; keep sufficient precision to avoid inconsistent positions later. |
| 5 | Compute \(\sqrt{5865}\) accurately: \[\sqrt{5865}\approx 76.583\] Then \[t=\frac{-15+76.583}{2(2.35)}=\frac{61.583}{4.7}\approx 13.103,\text{s}\] |
Choose the (+) root because time must be positive; the other root is negative and not physical. |
| 6 | Meeting position (use either car; results should match when using the same \(t\)): \[x= x_R(t)=1.75t^2\approx 1.75(13.103)^2\] \[(13.103)^2\approx 171.683\quad\Rightarrow\quad x\approx 1.75(171.683)\approx 300.45,\text{m}\] Check with blue car: \[x_B=600-15t-0.6t^2\approx 600-15(13.103)-0.6(171.683)\approx 300.45,\text{m}\] |
Using consistent precision eliminates the earlier mismatch; both equations must give the same meeting point. |
Just ask: "Help me solve this problem."
A spacecraft accelerates at a rate of \(20.0 \, \text{m/s}^2\).
A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of \( 1500 \, \text{m} \). The rocket is launched with an initial velocity of \( 180 \, \text{m/s} \). Find the following:
Given the graph of velocity versus time for a duck flying due south for the winter, at what labeled point did the duck stop its forward motion?
In which of the following is the particle’s acceleration constant?
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?
\(t\approx 13.10\,\text{s}\), \(x\approx 300.45\,\text{m}\) east of the red car’s starting point (i.e., \(600-300.45\approx 299.55\,\text{m}\) west of the blue car’s starting point).
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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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