| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[a = -g\] | Choose the upward direction as positive; the ball is constantly accelerated downward by gravity, so acceleration is the negative constant \(g\). |
| 2 | \[v(t) = v_i + at\] | Use the linear kinematic relation that ties velocity, constant acceleration, and time. |
| 3 | \[v(t) = v_i – g t\] | Substitute \(a = -g\) from Step 1 to obtain the explicit velocity–time function. |
| 4 | \[0 = v_i – g t_{\text{peak}} \quad\Rightarrow\quad t_{\text{peak}} = \frac{v_i}{g}\] | At the peak, the ball’s velocity is zero. Solve \(v(t)=0\) for the time to peak height. |
| 5 | \[v_i = 20\,\text{m/s},\; g = 9.8\,\text{m/s}^2\;\Rightarrow\; t_{\text{peak}} \approx 2.0\,\text{s}\] | Insert the given initial speed and standard gravitational acceleration to obtain the numerical peak time. |
| 6 | \[-v_i = v_i – g t_{\text{final}} \quad\Rightarrow\quad t_{\text{final}} = \frac{2v_i}{g}\] | The ball returns with speed \(-v_i\) (same magnitude, opposite sign). Solve for the total flight time until it regains that speed. |
| 7 | \[t_{\text{final}} = \frac{2(20)}{9.8} \approx 4.1\,\text{s}\] | Compute the numeric total time from launch to the instant it again has \(-20\,\text{m/s}.\) |
| 8 | \[v(t) = v_i – g t\quad (0 \le t \le t_{\text{final}})\] | The velocity–time graph is a straight line with slope \(-g\) that starts at \((0,\, +20)\), crosses the axis at \(t_{\text{peak}}\approx2\,\text{s}\), and ends at \((4.1\,\text{s},\,-20)\). The line’s constant slope visually represents uniform acceleration. |
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A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard \( 3.4 \) \( \text{s} \) later. If the speed of sound is \( 340 \) \( \text{m/s} \), how high is the cliff?
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
A projectile has the least speed at what point in its path?
A \( 1.5 \) \( \text{kg} \) block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is \( 0.300 \) and the coefficient of static friction is \( 0.400 \), calculate the amount of time that passes from when the force is applied to when the block stops.
A block starts from rest at the top of a \(50^\circ\) incline. The coefficient of kinetic friction between the block and the incline is \(0.4\). If the block reaches a velocity of \(7 \, \text{m/s}\) at the bottom of the incline, what is the length of the incline?
A cart begins to move from rest on a horizontal track. Which of the following correctly indicates the magnitude of the average velocity of the cart during the interval shown and provides a valid explanation?
Hint: when solving this, its consider that the area of the acceleration vs time graph tells you the change in velocity.
Two balls are dropped off a cliff, 3 seconds apart. The first ball dropped is twice as heavy as the second ball dropped. Air resistance is negligible. While both balls are falling, the distance between the two balls is
An object is thrown straight upward at 64 m/s.
A 100-pound rock and a 1-pound metal arrow pointed downwards are dropped from height \( h \). Assuming there is no air resistance, which one hits the ground first and why?
\(v(t)=v_i-g t\)
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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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