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| Derivation / Formula | Reasoning |
|---|---|
| \[v_i = 25\,\text{m/s},\; v_x = 5\,\text{m/s},\; a = -10\,\text{m/s}^2\] | The car starts at \(v_i\), slows to \(v_x\) with constant deceleration \(a ~ <0)\). |
| \[v_x^2 = v_i^2 + 2 a \Delta x\] | Use the constant–acceleration kinematic relation connecting velocities, acceleration, and displacement. |
| \[\Delta x = \frac{v_x^2 – v_i^2}{2 a}\] | Algebraically solve the above equation for the unknown displacement \(\Delta x\). |
| \[\Delta x = \frac{5^2 – 25^2}{2(-10)} = \frac{25 – 625}{-20} = \frac{-600}{-20} = 30\,\text{m}\] | Substitute the given values and perform arithmetic to find the displacement during deceleration. |
| \[\boxed{\Delta x = 30\,\text{m}}\] | Distance traveled while slowing down. |
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Which pair of quantities will always have the same magnitude if motion is in a straight line and in one direction?
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 37^{\circ} \). It lands on a platform that is \( 5.0 \) \( \text{m} \) above the launch height.
A car increases its forward velocity uniformly from \(40 ~ \text{m/s}\) to \(80 ~ \text{m/s}\) while traveling a distance of \(200 ~ \text{m}\). What is its acceleration during this time?
The driver of a car makes an emergency stop by slamming on the car’s brakes and skidding to a stop. How far would the car have skidded if it had been traveling twice as fast?
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 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.
A car is driving to the right at \( 20 \) \( \text{m/s} \). A motorcycle starts \( 30 \) \( \text{m} \) behind the car and is moving at \( 30 \) \( \text{m/s} \) in the same direction.
On a strange, airless planet, a ball is thrown downward from a height of \( 17 \, \text{m} \). The ball initially travels at \( 15 \, \text{m/s} \). If the ball hits the ground in \( 1 \, \text{s} \), what is this planet’s gravitational acceleration?
An object is thrown upward at \( 65 \, \text{m/s} \) from the top of a \( 800 \, \text{m} \) tall building.
\(30\,\text{m}\)
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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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