# (a) Displacement during the first two seconds
The displacement is the area under the velocity-time graph from \(t = 0\) to \(t = 2\) seconds.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}\) | The graph from \(t = 0\) to \(t = 2\) forms a right triangle. Calculate its area. |
| 2 | \(\text{Area} = \frac{1}{2} \times 2 \, \text{s} \times 4 \, \text{m/s}\) | Substitute the base (time interval) and height (velocity) into the area formula. |
| 3 | \(\text{Displacement} = 4 \,\text{meters}\) | The area (in square units) represents the player’s displacement in meters. |
# (b) Displacement between \(t = 4 \, \text{s}\) and \(t = 9 \, \text{s}\)
Calculate the area under the velocity-time graph between \(t = 4 \, \text{s}\) and \(t = 9 \, \text{s}\).
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(A_{\text{rectangle}} = \text{base} \times \text{height}\) | Calculate the area of the rectangle from \(t = 4\) to \(t = 8\). |
| 2 | \(A_{\text{rectangle}} = (8 – 4) \, \text{s} \times 2 \, \text{m/s} = 8 \, \text{meters}\) | Substitute the base (4 seconds) and height (2 m/s) into the equation. |
| 3 | \(A_{\text{triangle}} = \frac{1}{2} \times (9 – 8) \, \text{s} \times 2 \, \text{m/s} = 1 \, \text{meter}\) | Calculate the area of the triangle from \(t = 8\) to \(t = 9\). |
| 4 | \(\text{Total displacement} = 8 \, \text{meters} + 1 \, \text{meter} = 9 \, \text{meters}\) | Sum of the rectangle and triangle areas give the total displacement. |
# (c) Displacement between \(t = 4 \, \text{s}\) and \(t = 10 \, \text{s}\)
Calculate the area under the velocity-time graph between \(t = 4 \, \text{s}\) and \(t = 10 \, \text{s}\), including the negative area.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(A_{\text{rectangle}} = \text{base} \times \text{height} = 8 \, \text{meters}\) | Area calculated previously for rectangle from \(t = 4\) to \(t = 8\). |
| 2 | \(A_{\text{triangle1}} = \frac{1}{2} \times (9 – 8) \, \text{s} \times 2 \, \text{m/s} = 1 \, \text{meter}\) | Area calculated previously for triangle from \(t = 8\) to \(t = 9\). |
| 3 | \(A_{\text{triangle2}} = \frac{1}{2} \times (10 – 9) \, \text{s} \times (-2) \, \text{m/s} = -1 \, \text{meter}\) | Calculate the area (negative) for the triangle from \(t = 9\) to \(t = 10\). |
| 4 | \(\text{Total displacement} = 8 \, \text{meters} + 1 \, \text{meter} – 1 \, \text{meter} = 8 \, \text{meters}\) | Sum the areas of the rectangle and the two triangles. |
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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.)
The graph above shows velocity \( v \) versus time \( t \) for an object in linear motion. Which of the following is a possible graph of position (\( x \)) versus time (\( t \)) for this object?
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 graph below is a plot of position versus time. For which labeled segments is the velocity positive and the acceleration negative?
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.
Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
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.
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 motion of a particle is described in the velocity vs. time graph shown above. Over the nine-second interval shown, we can say that the speed of the particle…
(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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