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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."
Which of the following graphs represent an object having zero acceleration? (There could be more than one answer)
A student is running at her top speed of \( 5.0 \, \text{m/s} \) to catch a bus, which is stopped at the bus stop. When the student is still \( 40.0 \, \text{m} \) from the bus, it starts to pull away, moving with a constant acceleration of \( 0.170 \, \text{m/s}^2 \).
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge that is \( 235 \) \( \text{m} \) below. The plane is traveling horizontally with a speed of \( 250 \) \( \text{km/h} \). How far in advance of the recipients (horizontal distance) must the goods be dropped?
The graph in the figure shows the position of a particle as it travels along the x-axis. At what value of \(t\) is the speed of the particle equal to \(0 \, \text{m/s}\)?
note that the slope of position vs time is velocity. And the graph most closely reemsbles a flat or 0 slope at 3 seconds
You are a bungee jumping fanatic and want to be the first bungee jumper on Jupiter. The length of your bungee cord is \( 45.0 \) \( \text{m} \). Free fall acceleration on Jupiter is \( 23.1 \) \( \text{m/s}^2 \). What is the ratio of your speed on Jupiter to your speed on Earth when you have dropped \( 45 \) \( \text{m} \)? Ignore the effects of air resistance and assume that you start at rest.
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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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