| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[v = a t\] | The coaster starts from rest and its speed increases steadily (constant acceleration \(a\)) for the first \(6\;\text{s}\). |
| 2 | \[v_x = a \times 6\] | At \(t = 6\;\text{s}\) the speed has reached some value \(v_x\), given by the product of the acceleration and the time interval. |
| 3 | \[v = v_x\] | For the next \(4\;\text{s}\) (from \(t = 6\;\text{s}\) to \(t = 10\;\text{s}\)) the coaster travels at a constant speed, so the velocity–time graph is a horizontal line. |
| 4 | \[v = v_x + a_b (t-10)\] | When the brakes are applied (\(t = 10\;\text{s}\)) the coaster undergoes a constant negative acceleration \(a_b\) for \(3\;\text{s}\). |
| 5 | \[0 = v_x + a_b (13-10)\;\Rightarrow\;a_b = -\dfrac{v_x}{3}\] | Setting the final speed to zero at \(t = 13\;\text{s}\) gives the braking acceleration. |
| 6 | \[v(t)=\left\{\begin{array}{ll} a\,t & 0 \le t \le 6 \\ 6a & 6 < t \le 10 \\ 2a(13 – t) & 10 < t \le 13 \end{array}\right.\quad (a>0)\] | Piecewise form summarises the entire velocity–time graph: linear rise, flat section, and linear fall back to zero. |
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A coin is dropped from a hot air-balloon that is \(250 \, \text{m}\) above the ground rising at \(11 \, \text{m/s}\) upwards. For the coin, assume up is positive and find the following:
A kangaroo jumps straight up to a vertical height of \( 1.45 \) \( \text{m} \). How long was it in the air before returning to Earth?
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.
Divers in Acapulco jump from a cliff that is \( 36 \, \text{m} \) above the water with an initial vertical velocity of \( 2 \, \text{m/s} \).
A ball is tossed directly upward. Its total time in the air is \( T \). Its maximum height is \( H \). What is its height after it has been in the air a time \( T/4 \)? Air resistance is negligible.
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.
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 driver is traveling at a speed of \( 18.0 \) \( \text{m/s} \) when she sees a red light ahead. Her car is capable of decelerating at a rate of \( 3.65 \) \( \text{m/s}^2 \). If it takes her \( 0.350 \) \( \text{s} \) to get the brakes on and she is \( 20.0 \) \( \text{m} \) from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in what direction?
A car’s velocity increases as follows each second: \( 2 \) \( \text{m/s} \), \( 4 \) \( \text{m/s} \), \( 6 \) \( \text{m/s} \), \( 8 \) \( \text{m/s} \). This pattern shows that the car is:
A projectile of mass 0.750 kg is shot straight up with an initial speed of 18.0 m/s.
\[v(t)=\left\{\begin{array}{ll} a\,t & 0 \le t \le 6 \\ 6a & 6 < t \le 10 \\ 2a(13 – t) & 10 < t \le 13 \end{array}\right.\quad (a>0)\]
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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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