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| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for displacement. |
| 2 | Set \( y = 0 \), \( y_0 = 19.6 \) m, \( a = -9.8 \) m/s². | Define the variables. |
| 3 | For Ball A (downward): \( v_i = -14.7 \) m/s | Initial velocity downward is negative. |
| 4 | \( 0 = 19.6 -14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 5 | \( 4.9 t^2 +14.7 t -19.6 = 0 \) | Rearrange into standard quadratic form. |
| 6 | Solve for \( t \): \( t = 1 \) s | Find the positive root of the quadratic equation. |
| 7 | For Ball B (upward): \( v_i = +14.7 \) m/s | Initial velocity upward is positive. |
| 8 | \( 0 = 19.6 +14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 9 | \( -4.9 t^2 +14.7 t +19.6 = 0 \) | Simplify equation. |
| 10 | \( 4.9 t^2 -14.7 t -19.6 = 0 \) | Multiply both sides by -1. |
| 11 | Solve for \( t \): \( t = 4 \) s | Find the positive root of the quadratic equation. |
| 12 | \( \Delta t = t_{\text{Ball B}} – t_{\text{Ball A}} = 4 \, \text{s} – 1 \, \text{s} = 3 \, \text{s} \) | Calculate the difference in time. |
Answer: The difference in time the balls spend in the air is 3 seconds.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( v = v_i + a t \) | Use the kinematic equation for velocity. |
| 2 | For Ball A: \( v = -14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(1 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball A. |
| 3 | For Ball B: \( v = +14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(4 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball B. |
Answer: Each ball strikes the ground with a velocity of -24.5 m/s downward.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for position. |
| 2 | For Ball A: \( y_{\text{A}} = 19.6 + (-14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball A at \( t = 0.8 \) s. |
| 3 | \( y_{\text{A}} = 19.6 -11.76 -3.136 = 4.704 \, \text{m} \) | Simplify to find \( y_{\text{A}} \). |
| 4 | For Ball B: \( y_{\text{B}} = 19.6 + (+14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball B at \( t = 0.8 \) s. |
| 5 | \( y_{\text{B}} = 19.6 +11.76 -3.136 = 28.224 \, \text{m} \) | Simplify to find \( y_{\text{B}} \). |
| 6 | \( \Delta y = y_{\text{B}} – y_{\text{A}} = 28.224 – 4.704 = 23.52 \, \text{m} \) | Calculate the distance between the balls. |
Answer: The balls are 23.52 meters apart 0.800 seconds after they are thrown.
Just ask: "Help me solve this problem."
The alarm at a fire station rings and a 79.34-kg fireman, starting from rest, slides down a pole to the floor below (a distance of 4.20 m). Just before landing, his speed is 1.36 m/s. What is the magnitude of the kinetic frictional force exerted on the fireman as he slides down the pole?
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?
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:
Why is the stopping distance of a truck much shorter than for a train going the same speed? Hint: try deriving a formula or stopping distance.
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
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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] |
General Metric Conversion Chart
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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