0 attempts
0% avg
UBQ Credits
# (a) Displacement during the first two seconds
The displacement is the area under the velocity-time graph from [katex]t = 0[/katex] to [katex]t = 2[/katex] seconds.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}[/katex] | The graph from [katex]t = 0[/katex] to [katex]t = 2[/katex] forms a right triangle. Calculate its area. |
| 2 | [katex]\text{Area} = \frac{1}{2} \times 2 \, \text{s} \times 4 \, \text{m/s}[/katex] | Substitute the base (time interval) and height (velocity) into the area formula. |
| 3 | [katex]\text{Displacement} = 4 \,\text{meters}[/katex] | The area (in square units) represents the player’s displacement in meters. |
# (b) Displacement between [katex]t = 4 \, \text{s}[/katex] and [katex]t = 9 \, \text{s}[/katex]
Calculate the area under the velocity-time graph between [katex]t = 4 \, \text{s}[/katex] and [katex]t = 9 \, \text{s}[/katex].
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]A_{\text{rectangle}} = \text{base} \times \text{height}[/katex] | Calculate the area of the rectangle from [katex]t = 4[/katex] to [katex]t = 8[/katex]. |
| 2 | [katex]A_{\text{rectangle}} = (8 – 4) \, \text{s} \times 2 \, \text{m/s} = 8 \, \text{meters}[/katex] | Substitute the base (4 seconds) and height (2 m/s) into the equation. |
| 3 | [katex]A_{\text{triangle}} = \frac{1}{2} \times (9 – 8) \, \text{s} \times 2 \, \text{m/s} = 1 \, \text{meter}[/katex] | Calculate the area of the triangle from [katex]t = 8[/katex] to [katex]t = 9[/katex]. |
| 4 | [katex]\text{Total displacement} = 8 \, \text{meters} + 1 \, \text{meter} = 9 \, \text{meters}[/katex] | Sum of the rectangle and triangle areas give the total displacement. |
# (c) Displacement between [katex]t = 4 \, \text{s}[/katex] and [katex]t = 10 \, \text{s}[/katex]
Calculate the area under the velocity-time graph between [katex]t = 4 \, \text{s}[/katex] and [katex]t = 10 \, \text{s}[/katex], including the negative area.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]A_{\text{rectangle}} = \text{base} \times \text{height} = 8 \, \text{meters}[/katex] | Area calculated previously for rectangle from [katex]t = 4[/katex] to [katex]t = 8[/katex]. |
| 2 | [katex]A_{\text{triangle1}} = \frac{1}{2} \times (9 – 8) \, \text{s} \times 2 \, \text{m/s} = 1 \, \text{meter}[/katex] | Area calculated previously for triangle from [katex]t = 8[/katex] to [katex]t = 9[/katex]. |
| 3 | [katex]A_{\text{triangle2}} = \frac{1}{2} \times (10 – 9) \, \text{s} \times (-2) \, \text{m/s} = -1 \, \text{meter}[/katex] | Calculate the area (negative) for the triangle from [katex]t = 9[/katex] to [katex]t = 10[/katex]. |
| 4 | [katex]\text{Total displacement} = 8 \, \text{meters} + 1 \, \text{meter} – 1 \, \text{meter} = 8 \, \text{meters}[/katex] | Sum the areas of the rectangle and the two triangles. |
Just ask: "Help me solve this problem."
In which of these cases is the rate of change of the particle’s displacement constant?
The motions of a car and a truck along a straight road are represented by the velocity-time graphs in the figure. The two vehicles are initially alongside each other at time t = 0. At time T, what is true of the distances traveled by the vehicles since time t = 0?
The displacement \( x \) of an object moving in one dimension is shown above as a function of time \( t \). The acceleration of this object must be
Which of the following graphs shows runners moving at the same speed? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
The graph below is a plot of position versus time. For which labeled region is the velocity positive and the acceleration negative?
(a) 4 m
(b) 9 m
(c) 8 m
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
