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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. |
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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.
A car moves forward at a steady \( 10 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). The driver slams the brakes and brings it to rest in \( 2 \) \( \text{s} \). Without waiting, the driver immediately accelerates backward (negative velocity) for \( 3 \) \( \text{s} \) until reaching \( 8 \) \( \text{m/s} \) in reverse. Draw the velocity vs. time graph.
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 \(t = 0\)?
A large beach ball is dropped from the ceiling of a school gymnasium to the floor about 10 meters below. Which of the following graphs would best represent its velocity as a function of time? (do not neglect air resistance)
In which of the following is the particle’s acceleration constant?
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.
In which of the following is the rate of change of the particle’s momentum zero?
Which statement is true about the distances the two objects have traveled at time \( t_f \)?
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.
(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] |
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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