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# (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."
Which of the following graphs represent an object at rest?
Which of the following graphs represent an object having zero acceleration? (There could be more than one answer)
The figure shows a graph of the position x of two cars, C and D, as a function of time t. According to this graph, which statements about these cars must be true? (There could be more
than one correct choice.)
(a) 4 m
(b) 9 m
(c) 8 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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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