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Part a:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = y_0 + v_0 t – \frac{1}{2} g t^2[/katex] | This is the kinematic equation for vertical position, where [katex]y_0[/katex] is the initial position, [katex]v_0[/katex] is the initial velocity, [katex]g[/katex] is the acceleration due to gravity, and [katex]t[/katex] is the time. |
| 2 | Substitute [katex]y = 0[/katex], [katex]y_0 = 800\ \text{m}[/katex], [katex]v_0 = 65\ \text{m/s}[/katex], [katex]g = 9.8\ \text{m/s}^2[/katex] | The object hits the ground when [katex]y=0[/katex]. It is thrown upward from a height of 800 m and the initial velocity is upward, hence positive. |
| 3 | [katex]0 = 800 + 65t – 4.9t^2[/katex] | Rearrange the substituted equation and simplify [katex]g/2[/katex] from [katex]9.8/2[/katex] to [katex]4.9[/katex]. |
| 4 | [katex]4.9t^2 – 65t – 800 = 0[/katex] | This is a quadratic equation in the form of [katex]at^2 + bt + c = 0[/katex]. |
| 5 | [katex]t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/katex] | Apply the quadratic formula to solve for [katex]t[/katex]. Here, [katex]a = 4.9[/katex], [katex]b = -65[/katex], and [katex]c = -800[/katex]. |
| 6 | [katex]t = \frac{-(-65) \pm \sqrt{(-65)^2 – 4 \cdot 4.9 \cdot (-800)}}{2 \cdot 4.9}[/katex] | Plug in the values of [katex]a[/katex], [katex]b[/katex], and [katex]c[/katex]. |
| 10 | [katex]t \approx 21.03\ \text{s} \textbf{ (positive root)}[/katex] | Discard negative time and keep the physically meaningful positive root, being the actual time the object takes to hit the ground. |
Part b:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = v_0 – gt[/katex] | Use the kinematic equation for velocity in vertical motion, taking downward as the negative direction. |
| 2 | [katex]v = 65 – 9.8 \times 21.04[/katex] | Substitute the values of [katex]g[/katex] and [katex]t[/katex] into the equation. |
| 3 | [katex]v \approx -141.14\ \text{m/s}[/katex] | Subtract to find the final velocity, where negative indicates direction downwards. |
Part c:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v \propto v_0[/katex] | The final velocity is proportional to the initial velocity when all other factors remain the same. |
| 2 | [katex]v_{\text{final, new}} = 2v_0 – gt[/katex] | Note this is the equation we used in part b to find the final speed. Time, however, is not the same in the two scenarios. So unless you have found time with the double the initial speed, do NOT use this equation again. |
| 3 | [katex]v_{\text{final, new}} = {2v_0}^2 + 2g\Delta \theta[/katex] | Use this kinematic equation instead. |
| 4 | [katex]{v_{\text{final, new}}}^2 = = 180.54 \, m/s[/katex] | Plug in values and solve for the new final speed. |
| 5 | [katex] \frac{v_{\text{final, new}}}{v_{\text{final, orginal}}[/katex] | Find the factor the final velocity has increased by. |
| 6 | [katex] \frac{180.54}{141.14} \approx 1.28 [/katex] | Plug in values and solve. |
| 7 | Factor = 1.28 | If the initial velocity doubles the final velocity of the ball increases by a factor of 1.28. |
Just ask: "Help me solve this problem."
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2 for 4.0 s. What is the final speed of the car?
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
A car is driving at \(25 \, \text{m/s}\) when a light turns red \(100 \, \text{m}\) ahead. The driver takes an unknown amount of time to react and hit the brakes, but manages to skid to a stop at the red light. If \(\mu_s = 0.9\) and \(\mu_k = 0.65\), what was the reaction time of the driver?
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are μs=0.6 and μk=0.2. Starting from rest, the object travels 10 meters in 4.5 seconds. What is the mass of the unknown object?
A car moving at 30 m/s makes a head-on collision with a stone wall. From what height would the car have to fall in order to make an equally hard collision with the ground?
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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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