| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v(t) = \text{Area under velocity-time graph}\) | Object passes through its initial position when the net area under the velocity-time graph (Displacement, \( \Delta x \)) is zero. |
| 2 | N/A (Graph Inspection) | From \( t = 0 \, \text{s} \) to \( t = 5 \, \text{s} \), the area under the curve is a triangle with base 5 s and height 20 m/s giving an area of \( \frac{1}{2} \times 5 \, \text{s} \times 20 \, \text{m/s} = 50 \, \text{m} \). |
| 3 | N/A (Graph Inspection) | From \( t = 5 \, \text{s} \) to \( t = 11 \, \text{s} \), the area under the curve is a large negative area as shown in the graph. Calculate areas to compensate for positive 50 m. |
| 4 | N/A (Graph Inspection) | The area from \( t = 5 \, \text{s} \) to \( t = 8 \, \text{s} \) is -40, while the area from \( t = 5 \, \text{s} \) to \( t = 9 \, \text{s} \) is -60 m. |
| 5 | Conclusion | Between 8 to 9 seconds the area, representing displacement, reaches \( -50 \, \text{m} \), which would cancel out the \( +50 \, \text{m} \) covered in the first 5 seconds. Thus, between 8 and 9 seconds, displacement is 0, which means the particle is back to its original starting point. |
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\text{Average acceleration} = \frac{\Delta v}{\Delta t}\) | The average acceleration is the change in velocity divided by the change in time (the slope of the graph) |
| 2 | \(\Delta v = v_f – v_i = 0 – 0 \, \text{m/s} = 0 \, \text{m/s}\) | The change in velocity for \(12 \, \text{s} < t < 14 \, \text{s}\) is \( \Delta v = 0 \, \text{m/s}\). |
| 3 | \(\Delta t = 14 \, \text{s} – 12 \, \text{s} = 2 \, \text{s}\) | The time interval is \( \Delta t = 2 \, \text{s}\). |
| 4 | \(\text{Average acceleration} = \frac{0 \, \text{m/s}}{2 \, \text{s}}\) | Substitute the values obtained for \( \Delta v \) and \( \Delta t \). |
| 5 | \(\text{Average acceleration} = 0 \, \text{m/s}^2\) | The average acceleration for \(12 \, \text{s} < t < 14 \, \text{s}\) is \( 0 \, \text{m/s}^2\). |
| 6 | Find the identical slope on a different part of of the graph. | The interval from \( 7 \, \text{s} \) to \( 10 \, \text{s} \) also shows 0 slope resulting in same acceleration. |
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The graph above represents the motion of an object traveling in a straight line as a function of time. What is the average speed of the object during the first four seconds?
Above is the graph of an object’s velocity as a function of time. Which of the following is true about the motion?
A rollercoaster leaves the station at rest. Its speed increases steadily for \( 6 \) \( \text{s} \) as it heads down the first drop. The ride then levels out and it moves at a constant speed for \( 4 \) \( \text{s} \) before hitting the brakes and stopping in \( 3 \) \( \text{s} \). Draw the velocity vs. time graph or explain it in terms of functions.
The graph in the figure shows the position of a particle as it travels along the x-axis. At what value of \(t\) is the speed of the particle equal to \(0 \, \text{m/s}\)?
note that the slope of position vs time is velocity. And the graph most closely reemsbles a flat or 0 slope at 3 seconds
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
On Saturday, Ashley rode her bicycle to visit Maria. Maria’s house is directly east of Ashley’s. The graph shows how far Ashley was from her house after each minute of her trip. (Hint – Use the standard units of velocity \(\text{m/s}\) for all parts)
The graph above shows velocity as a function of time for an object moving along a straight line. For which of the following sections of the graph is the acceleration constant and nonzero?
In which of these cases is the rate of change of the particle’s displacement constant?
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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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