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| Step | Derivation / Formula | Reasoning |
|---|---|---|
| 1 | \[v(t)=6\quad(0\le t\le4)\] | The skater glides with a constant speed of \(6\,\text{m/s}\) for the first \(4\,\text{s}\); hence the velocity–time graph is a horizontal line at \(+6\,\text{m/s}\). |
| 2 | \[a_1=\frac{0-6}{6}=-1\,\text{m/s}^2\] | Using \(a=\frac{v_x-v_i}{\Delta t}\), the skater slows from \(+6\,\text{m/s}\) to rest in \(6\,\text{s}\); the constant (negative) acceleration is \(-1\,\text{m/s}^2\). |
| 3 | \[v(t)=6-1\,(t-4)\quad(4< t\le10)\] | With slope \(a_1=-1\,\text{m/s}^2\), velocity decreases linearly from \(+6\,\text{m/s}\) at \(t=4\,\text{s}\) to \(0\,\text{m/s}\) at \(t=10\,\text{s}\). |
| 4 | \[v(t)=0\quad(10< t\le12)\] | The skater is motionless for \(2\,\text{s}\); the graph lies on the time axis between \(t=10\,\text{s}\) and \(t=12\,\text{s}\). |
| 5 | \[a_2=\frac{-5-0}{5}=-1\,\text{m/s}^2\] | They now push off the opposite way, gaining speed steadily for \(5\,\text{s}\). Choosing the same magnitude acceleration gives a final speed of \(5\,\text{m/s}\) in the negative direction, so \(a_2=-1\,\text{m/s}^2\). |
| 6 | \[v(t)=-1\,(t-12)\quad(12< t\le17)\] | Starting from rest at \(t=12\,\text{s}\) and accelerating at \(-1\,\text{m/s}^2\), velocity decreases linearly to \(-5\,\text{m/s}\) at \(t=17\,\text{s}\). |
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Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
Will Clark throws a baseball with a horizontal component of velocity of \(25 \, \text{m/s}\). It takes \(3 \, \text{s}\) to come back to its original height. Calculate the baseball’s:
An object undergoes constant acceleration. Starting from rest, the object travels \( 5 \, \text{m} \) in the first second. Then it travels \( 15 \, \text{m} \) in the next second. What additional distance will be covered in the third second?
At time \( t = 0 \) an object is traveling to the right along the \( +x \) axis at a speed of \( 10.0 \) \( \text{m/s} \) with acceleration \( -2.0 \) \( \text{m/s}^2 \). Which statement is true?
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge that is \( 235 \) \( \text{m} \) below. The plane is traveling horizontally with a speed of \( 250 \) \( \text{km/h} \). How far in advance of the recipients (horizontal distance) must the goods be dropped?
A mine shaft is known to be 57.8 m deep. If you dropped a rock down the shaft, how long would it take for you to hear the sound of the rock hitting the bottom of the shaft knowing that sound travels at a constant velocity of 345 m/s?
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
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.
\[v(t)=\begin{cases}6 & 0 \le t \le 4 \\ 6-(t-4) & 4 < t \le 10 \\ 0 & 10 < t \le 12 \\ -(t-12) & 12 < t \le 17\end{cases}\].
View the fully drawn graph by clicking here.
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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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