| Derivation/Formula | Reasoning |
|---|---|
| \[a = \frac{v_x – v_i}{\Delta t}\] | The average acceleration is defined as the change in velocity \(v_x – v_i\) divided by the time interval \(\Delta t\). |
| \[a = \frac{30\,\text{m/s} – 30\,\text{m/s}}{5\,\text{s}}\] | Since the car keeps a steady speed, its initial velocity \(v_i = 30\,\text{m/s}\) and final velocity \(v_x = 30\,\text{m/s}\) over \(\Delta t = 5\,\text{s}\). |
| \[a = 0\,\text{m/s}^2\] | The change in velocity is zero, so the acceleration is \(0\,\text{m/s}^2\). \(\boxed{0\,\text{m/s}^2}\) |
| \[\text{Choice\ analysis}\] | Option \((a)\,0\) is correct because the velocity does not change. Options \((b)\,6\), \((c)\,30\), and \((d)\,150\) require a non-zero change in velocity, which is absent. Option \((e)\) is incorrect since a correct choice already exists. |
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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.)
A baseball is tossed from street level by a student straight up at a speed of \(25.3 \text{ m/s}\). After reaching maximum height, it is caught by another student on the roof of a building, \(17.4 \text{ m}\) above the street. How long did this take?
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at \( +13.0 \, \text{m/s}^2 \). At \( t_1 \), the rocket engine is shut down and the sled moves with constant velocity \( v \) until \( t_2 \). The total distance traveled by the sled is \( 5.30 \times 10^3 \, \text{m} \) and the total time is \( 90.0 \, \text{s} \).
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?
On another planet, a ball is in free fall after being released from rest at time \( t = 0 \). A graph of the height of the ball above the planet’s surface as a function of time \( t \) is shown. The acceleration due to gravity on the planet is most nearly
An elevator of height \(h\) ascends with constant acceleration \(a\). When it crosses a platform, it has acquired a velocity \(u\). At this instant a bolt drops from the top of the elevator. Find the time for the bolt to hit the floor of the elevator. Give your answer in terms of \(h\), \(a\), and any constant.
A red car, initially at rest, travels east with an acceleration of \( 3.5 \, \text{m/s}^2 \). At the same time as the red car starts to move, a blue car is traveling west at \( 15 \, \text{m/s} \) and accelerating at \( 1.2 \, \text{m/s}^2 \). If they are \( 600 \, \text{m} \) apart the moment the red car starts to move and they are traveling towards each other, where and when will they meet?
The figure shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the displacement of the player…
A truck is traveling at \(35 \, \text{m/s}\) when the driver realizes the truck has no brakes. He sees a ramp off the road, inclined at \(20^\circ\), and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction)?
A \(30 \, \text{g}\) bullet is fired with a speed of \(500 \, \text{m/s}\) into a wall.
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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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