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. |
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Which graph below shows that one of the runners started 10 meters further ahead of the other? Assume the y-axis is measured in meters and the x-axis is measured in seconds.
A teacher walks the following path in \( 10 \) \( \text{s} \): \( 2 \) \( \text{m} \) south, \( 4 \) \( \text{m} \) east, \( 2 \) \( \text{m} \) north, \( 4 \) \( \text{m} \) west. What is the teacher’s average velocity?
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
You throw a rock straight up with an initial velocity of \( 5.0 \, \text{m/s} \).
A helicopter is ascending vertically with a speed of \( 5.40 \) \( \text{m/s} \). At a height of \( 105 \) \( \text{m} \) above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground?
An elevator of height \(h\) ascends with constant acceleration \(a\). When it crosses a platform, it has acquired a velocity \(u\). At this instant a bolt drops from the top of the elevator. Find the time for the bolt to hit the floor of the elevator. Give your answer in terms of \(h\), \(a\), and any constant.
The graph below is a plot of position versus time. For which labeled segments is the velocity positive and the acceleration negative?
A rubber ball bounces on the ground. After each bounce, the ball reaches one-half the height of the bounce before it. If the time the ball was in the air between the first and second bounce was 1 second. What would be the time between the second and third bounce?
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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] |
Metric Prefixes
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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