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| Step | Formula / Calculation | Reasoning |
|---|---|---|
| 1 | [katex]v = 101 , \text{km/h}[/katex], [katex]t = 0.500 , \text{hr}[/katex] | Defining the constant speed and time for both boats. |
| 2 | [katex]v_{\text{m/s}} = \frac{101 \times 1000}{3600}[/katex] | Converting the speed from km/h to m/s. |
| 3 | [katex]v_{\text{west, boat 1}} = v_{\text{m/s}} \cos(\theta_{\text{boat 1}})[/katex], [katex]v_{\text{south, boat 1}} = v_{\text{m/s}} \sin(\theta_{\text{boat 1}})[/katex] | Decomposing Boat 1’s speed into westward and southward components. |
| 4 | [katex]v_{\text{west, boat 2}} = v_{\text{m/s}} \cos(\theta_{\text{boat 2}})[/katex], [katex]v_{\text{south, boat 2}} = v_{\text{m/s}} \sin(\theta_{\text{boat 2}})[/katex] | Decomposing Boat 2’s speed into westward and southward components. |
| 5 | [katex]\text{Distance west, boat 1} = v_{\text{west, boat 1}} \times t[/katex], [katex]\text{Distance south, boat 1} = v_{\text{south, boat 1}} \times t[/katex] | Calculating the westward and southward distances for Boat 1. |
| 6 | [katex]\text{Distance west, boat 2} = v_{\text{west, boat 2}} \times t[/katex], [katex]\text{Distance south, boat 2} = v_{\text{south, boat 2}} \times t[/katex] | Calculating the westward and southward distances for Boat 2. |
| 7 | [katex]\text{Difference west} = \text{Distance west, boat 1} – \text{Distance west, boat 2}[/katex] | Calculating the difference in the westward distance between the two boats. |
| 8 | [katex]\text{Difference south} = \text{Distance south, boat 2} – \text{Distance south, boat 1}[/katex] | Calculating the difference in the southward distance between the two boats. |
| 9 | [katex]\text{Difference west} \approx 0.98[/katex] km, [katex]\text{Difference south} \approx 1.61[/katex] km | Evaluating the differences in distances. |
During the half-hour journey, Boat 1 travels approximately 0.98 km farther west compared to Boat 2, and Boat 2 travels approximately 1.61 km farther south compared to Boat 1.
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Vector \( A \) is \( 44.0 \) units and \( 28.0^\circ \) above the \( +x \) axis, vector \( B \) is \( 26.5 \) units and \( 56.0^\circ \) above the \( -x \) axis, and vector \( C \) is \( 31.0 \) units along the \( -y \) axis. Determine the resultant (sum) of the three vectors.
You are adding vectors of length \( 20 \) and \( 40 \) units. Which of the following choices is a possible resultant magnitude?
A boat can row across a still \( 1 \, \text{km} \) wide river at a maximum speed of \( 5 \, \text{km/hr} \). If a current of \( 4 \, \text{km/hr} \) flows east as you try to directly cross the river, how long would it take?
What does displacement mean in the context of motion?
When we refer to an object’s speed, we’re talking about:
Gregory was walking through the halls of the school when he realized that he was walking in perpendicular directions and he could easily calculate his displacement using the incredibly useful techniques he learned in physics. He recognized that he walked \(12.5\ \text{m}\) left and then \(18.9\ \text{m}\) down. How far must he walk to the right so that his resultant displacement is \(20.1\ \text{m}\)?
A seagull first flies \( 160 \, \text{m} \) North, then heads \( 120.65 \, \text{m} \) at \( 18.43^\circ \) North of West. After it lands:
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
A skier is accelerating down a \( 30.0^{\circ} \) hill at \( 3.80 \) \( \text{m/s}^2 \).
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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] |
Metric Prefixes
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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