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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\text{total displacement}^2 = (\Delta x_{\text{left}} + \Delta x_{\text{right}})^2 + \Delta x_{\text{down}}^2 [/katex] | Apply the Pythagorean theorem to calculate total displacement where you have to walk a certain distance to the right to complete the diagonal hypotenuse. |
| 2 | [katex](20.1)^2 = (12.5 – x)^2 + (18.9)^2[/katex] | Substitute given values: [katex]\text{total displacement}=20.1 \, \text{m}[/katex], [katex]\Delta x_{\text{left}}=12.5\, \text{m}[/katex], [katex]\Delta x_{\text{down}}=18.9 \,\text{m}[/katex]. [katex]x[/katex] is the variable we need to solve. |
| 3 | [katex]404.01 = (12.5 – x)^2 + 357.21[/katex] | Simplify the squares, where [katex]404.01 = 20.1^2[/katex] and [katex]357.21 = 18.9^2 [/katex] |
| 4 | [katex]404.01 – 357.21 = (12.5 – x)^2[/katex] | Isolate the term [katex](12.5 – x)^2[/katex] by subtracting [katex]357.21[/katex] from both sides. |
| 5 | [katex]46.8 = (12.5 – x)^2[/katex] | Simplify further to isolate the squared term completely. |
| 6 | [katex]\sqrt{46.8} = 12.5 – x [/katex] | Take the square root of both sides to solve for [katex] x [/katex]. |
| 7 | [katex]6.84 = 12.5 – x [/katex] | Calculate the square root to approximate value. |
| 8 | [katex] x = 12.5 – 6.84 [/katex] | Rearrange the equation to isolate [katex] x [/katex]. |
| 9 | [katex]\boxed{x = 5.66 \, \text{m}} [/katex] | Final answer is boxed: Gregory must walk [katex]5.66 \, \text{m}[/katex] to the right to achieve the resultant displacement. |
Note – Gregory can also walk (12.5 + 5.66) meters, which is 19.34 meters. In this case his horizontal resultant displacement would be to the right, whereas, in the table above, it is to the left. In either case as long as the magnitude of his horizontal resultant displacement is 6.84 meters, then his total displacement will be 20.1 meters.
Just ask: "Help me solve this problem."
Determine the sum of the three vectors given below. Calculate the resultant \( \vec{R} \) expressed as:
(a) Vector components
(b) Resultant vector (its total magnitude and direction)
\[\vec{A} = 26.5 \, \text{m} \ @ \ 56^\circ \, \text{NW}\]
\[\vec{B} = 44 \, \text{m} \ @ \ 28^\circ \, \text{NE}\]
\[\vec{C} = 31 \, \text{m} \, \text{South}\]
A student walks \( 3 \) \( \text{m} \) east, then \( 4 \) \( \text{m} \) west in \( 7 \) \( \text{s} \). What is their displacement and average velocity?
What does displacement mean in the context of motion?
An object is moving to the west at a constant speed. Three forces are exerted on the object. One force is \( 10 \) \( \text{N} \) directed due north, and another is \( 10 \) \( \text{N} \) directed due west. What is the magnitude and direction of the third force if the object is to continue moving to the west at a constant speed?
\(5.67\ \text{m}\) or \(19.34\ \text{m}\)
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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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