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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."
The International Space Station travels at \( 7660 \, \text{m/s} \). Find the average velocity of the space station if it takes \( 90 \, \text{minutes} \) to make one full orbit around Earth.
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 decelerates from \( 25 \, \text{m/s} \) to \( 5 \, \text{m/s} \) at \( 10 \, \text{m/s}^2 \). How far does the car travel during this deceleration?
A car accelerates from rest with an acceleration of \( 3.5 \, \text{m/s}^2 \) for \( 10 \, \text{s} \). After this, it continues at a constant speed for an unknown amount of time. The driver notices a ramp \( 50 \, \text{m} \) ahead and takes \( 0.6 \, \text{s} \) to react. After reacting, the driver hits the brakes, which slow the car with an acceleration of \( 7.2 \, \text{m/s}^2 \). Unfortunately, the driver does not stop in time and goes off the \( 3 \, \text{m} \) high ramp that is angled at \( 27^\circ \).
A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m. What is its acceleration during this time?
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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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