Part a:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(y = y_0 + v_0 t – \frac{1}{2} g t^2\) | This is the kinematic equation for vertical position, where \(y_0\) is the initial position, \(v_0\) is the initial velocity, \(g\) is the acceleration due to gravity, and \(t\) is the time. |
| 2 | Substitute \(y = 0\), \(y_0 = 800\ \text{m}\), \(v_0 = 65\ \text{m/s}\), \(g = 9.8\ \text{m/s}^2\) | The object hits the ground when \(y=0\). It is thrown upward from a height of 800 m and the initial velocity is upward, hence positive. |
| 3 | \(0 = 800 + 65t – 4.9t^2\) | Rearrange the substituted equation and simplify \(g/2\) from \(9.8/2\) to \(4.9\). |
| 4 | \(4.9t^2 – 65t – 800 = 0\) | This is a quadratic equation in the form of \(at^2 + bt + c = 0\). |
| 5 | \(t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\) | Apply the quadratic formula to solve for \(t\). Here, \(a = 4.9\), \(b = -65\), and \(c = -800\). |
| 6 | \(t = \frac{-(-65) \pm \sqrt{(-65)^2 – 4 \cdot 4.9 \cdot (-800)}}{2 \cdot 4.9}\) | Plug in the values of \(a\), \(b\), and \(c\). |
| 10 | \(t \approx 21.03\ \text{s} \textbf{ (positive root)}\) | 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 | \(v = v_0 – gt\) | Use the kinematic equation for velocity in vertical motion, taking downward as the negative direction. |
| 2 | \(v = 65 – 9.8 \times 21.04\) | Substitute the values of \(g\) and \(t\) into the equation. |
| 3 | \(v \approx -141.14\ \text{m/s}\) | Subtract to find the final velocity, where negative indicates direction downwards. |
Part c:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(v \propto v_0\) | The final velocity is proportional to the initial velocity when all other factors remain the same. |
| 2 | \(v_{\text{final, new}} = 2v_0 – gt\) | 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 | \(v_{\text{final, new}} = {2v_0}^2 + 2g\Delta \theta\) | Use this kinematic equation instead. |
| 4 | \({v_{\text{final, new}}}^2 = = 180.54 \, m/s\) | Plug in values and solve for the new final speed. |
| 5 | \( \frac{v_{\text{final, new}}}{v_{\text{final, orginal}}\) | Find the factor the final velocity has increased by. |
| 6 | \( \frac{180.54}{141.14} \approx 1.28 \) | 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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Divers in Acapulco jump from a cliff that is \( 36 \, \text{m} \) above the water with an initial vertical velocity of \( 2 \, \text{m/s} \).
A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard \( 3.4 \) \( \text{s} \) later. If the speed of sound is \( 340 \) \( \text{m/s} \), how high is the cliff?
A car is driving to the right at \( 20 \) \( \text{m/s} \). A motorcycle starts \( 30 \) \( \text{m} \) behind the car and is moving at \( 30 \) \( \text{m/s} \) in the same direction.
A kangaroo jumps straight up to a vertical height of \( 1.45 \) \( \text{m} \). How long was it in the air before returning to Earth?
A particle moves along the x-axis with an acceleration of \( a = 18t \), where \( a \) has units of \( \text{m/s}^2 \). If the particle at time \( t = 0 \) is at the origin with a velocity of \( -12 \, \text{m/s} \), what is its position at \( t = 4.0 \, \text{s} \)? Note this requires calculus to solve.
A stone is thrown vertically upwards with a speed of \( 20.0 \) \( \text{m/s} \). How fast is it moving when it reaches a height of \( 12.0 \) \( \text{m} \)?
A large beach ball is dropped from the ceiling of a school gymnasium to the floor about 10 meters below. Which of the following graphs would best represent its velocity as a function of time? (do not neglect air resistance)
An object is moving in the \( +x \) direction and begins to slow down. What must be true about its acceleration?
Traveling at a speed of 15.9 m/s, the driver of an automobile suddenly locks the wheels by slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.659. What is the speed of the automobile after 1.59 s have elapsed? Ignore the effects of air resistance.
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
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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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