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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."
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
A 135.0 N force is applied to a 30.0 kg box at 42 degree angle to the horizontal. If the force of friction is 85.0, what is the net force and acceleration? If the object starts from rest, how far has it traveled in 3.3 sec?
The motions of a car and a truck along a straight road are represented by the velocity-time graphs in the figure. The two vehicles are initially alongside each other at time t = 0. At time T, what is true of the distances traveled by the vehicles since time t = 0?
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?
In a 4.0-kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kilometer in 6.3 minutes, and the final kilometer in 6.0 minutes. What is the average speed of the runner? Use standard units: m/s.
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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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