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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_i = 0 \, \text{m/s} \) | The car starts from rest, so the initial velocity is zero. |
| 2 | \( a_1 = 4.3 \, \text{m/s}^2 \) | The car accelerates with an acceleration of \(4.3 \, \text{m/s}^2\). |
| 3 | \( t_1 = 6.8 \, \text{s} \) | The car accelerates for a time of \(6.8 \, \text{s}\). |
| 4 | \( v_x = v_i + a_1 t_1 \) | Use the formula for final velocity after a time interval, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time. |
| 5 | \( v_x = 0 + (4.3 \, \text{m/s}^2)(6.8 \, \text{s}) \) | Plug in the values for the initial velocity, acceleration, and time. |
| 6 | \( v_x = 29.24 \, \text{m/s} \) | Calculate the final velocity after the acceleration phase. |
| 7 | \( d_1 = v_i t_1 + \frac{1}{2} a_1 t_1^2 \) | Use the formula for distance traveled during the acceleration phase, where \( v_i \) is the initial velocity, \( a_1 \) is the acceleration, and \( t_1 \) is the time. |
| 8 | \( d_1 = 0 \cdot 6.8 + \frac{1}{2} (4.3) (6.8)^2 \) | Set initial velocity term to zero and plug in the values for acceleration and time. |
| 9 | \( d_1 = 99.292 \, \text{m} \) | Calculate the distance traveled during the acceleration phase. |
| 10 | \( a_2 = -5.1 \, \text{m/s}^2 \) | The car decelerates with an acceleration of \( -5.1 \, \text{m/s}^2 \) (since the car is slowing down). |
| 11 | \( v_{x2} = 0 \, \text{m/s} \) | Final velocity when the car comes to rest again. |
| 12 | \( v_x^2 = v_i^2 + 2 a_2 \Delta x_2 \) | Use the kinematic equation relating initial velocity, final velocity, acceleration, and distance traveled. |
| 13 | \( 0 = (29.24)^2 + 2(-5.1) \Delta x_2 \) | Plug in the values for initial velocity during deceleration, final velocity, and acceleration. |
| 14 | \( 0 = 855.2976 – 10.2 \Delta x_2 \) | Calculate the term \( (29.24)^2 \) and simplify the equation. |
| 15 | \( \Delta x_2 = \frac{855.2976}{10.2} \) | Rearrange the equation to solve for the distance traveled during deceleration \( \Delta x_2 \). |
| 16 | \( \Delta x_2 = 83.95 \, \text{m} \) | Calculate the distance traveled during the deceleration phase. |
| 17 | \( d_{\text{total}} = d_1 + \Delta x_2 \) | Sum the distances traveled during the acceleration phase and deceleration phase to find the total distance. |
| 18 | \( d_{\text{total}} = 99.292 + 83.95 \) | Plug in the values for the distances calculated in both phases. |
| 19 | \( d_{\text{total}} = 183.242 \, \text{m} \) | Calculate the total distance traveled by the car and box the final answer. |
Just ask: "Help me solve this problem."
The figure shows the velocity-versus-time graph for a basketball player traveling up and down the court in a straight-line path. Find the displacement of the player…
Which of the following statements about the acceleration due to gravity is TRUE?
A blue sphere and a red sphere with the same diameter are released from rest at the top of a ramp. The red sphere takes a longer time to reach the bottom of the ramp. The spheres are then rolled off a horizontal table at the same time with the same speed and fall freely to the floor. Which sphere reaches the floor first?
A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
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?
\(183.24 \,\text{m}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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