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# Part (a): At what times is the object 20 m above the ground?
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]y = v_0 t + \frac{1}{2} a t^2[/katex] | Use the kinematic equation for vertical displacement where [katex]y[/katex] is the height, [katex]v_0[/katex] is the initial velocity, [katex]t[/katex] is the time, and [katex]a[/katex] is the acceleration. |
| 2 | [katex]20 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2[/katex] | Substitute [katex]y = 20 \text{ m}[/katex], [katex]v_0 = 64 \text{ m/s}[/katex], and [katex]a = -9.8 \text{ m/s}^2[/katex]. Note the negative sign of acceleration due to gravity. |
| 3 | [katex]4.9 t^2 – 64 t + 20 = 0[/katex] | Rearrange the equation into standard quadratic form [katex]a t^2 + b t + c = 0[/katex] where [katex]a = 4.9[/katex], [katex]b = -64[/katex], and [katex]c = 20[/katex]. |
| 4 | [katex]t = \frac{64 \pm \sqrt{(-64)^2 – 4 \cdot 4.9 \cdot 20}}{2 \cdot 4.9}[/katex] | Use the quadratic formula [katex]t = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}[/katex] to solve for [katex]t[/katex]. |
| 5 | [katex]t_1 = \frac{64 + 60.88}{9.8} \approx 12.74 \text{ s}, \\ t_2 = \frac{64 – 60.88}{9.8} \approx 0.32 \text{ s}[/katex] | Calculate the two possible values for [katex]t[/katex]. The results are [katex]t_1 \approx 12.74 \text{ s}[/katex] and [katex]t_2 \approx 0.32 \text{ s}[/katex]. |
| 6 | [katex]t = 0.32 \text{ s}, 12.74 \text{ s}[/katex] | Final answer: There are two times when the object is 20 m above the ground, at [katex]0.32 \text{ s}[/katex] and [katex]12.74 \text{ s}[/katex]. |
# Part (b): Why are there two answers for part (a)?
| Step | Explanation | Reasoning |
|---|---|---|
| 1 | Two Answers Explanation | There are two answers because the object passes 20 m twice: once while going up and once while coming back down. |
# Part (c): How long does it take to come back to the ground?
| Step | Derivation/Formula/Answer | Reasoning |
|---|---|---|
| 1 | [katex]y = v_0 t + \frac{1}{2} a t^2[/katex] | Use the kinematic equation for vertical displacement where [katex]y[/katex] is zero as it returns to the ground. |
| 2 | [katex]0 = 64 t – \frac{1}{2} \cdot 9.8 \cdot t^2[/katex] | Substitute [katex]y = 0 \text{ m}[/katex], [katex]v_0 = 64 \text{ m/s}[/katex], and [katex]a = -9.8 \text{ m/s}^2[/katex]. |
| 3 | [katex]0 = 64 t – 4.9 t^2[/katex] | Simplify the equation by calculating [katex]\frac{1}{2} \cdot 9.8 = 4.9[/katex]. |
| 4 | [katex]0 = t (64 – 4.9t)[/katex] | Factor out [katex]t[/katex] to solve for the time at which the object returns to the ground. |
| 5 | [katex]t = 0 \quad \text{or} \quad 64 = 4.9t \Rightarrow t = \frac{64}{4.9} \approx 13.06 \text{ s}[/katex] | The first solution [katex]t = 0[/katex] corresponds to the initial throw. Solving [katex]64 – 4.9t = 0[/katex] gives the time it takes to reach the ground. |
| 6 | [katex]t = 13.06 \text{ s}[/katex] | Final answer: The object takes [katex]13.06 \text{ s}[/katex] to come back to the ground. |
Just ask: "Help me solve this problem."
A rock is dropped from the top of a tall tower. Half a second later another rock, twice as massive as the first, is dropped. Ignoring air resistance and using ONLY simple kinematics (DO NOT use energy to explain this). Explain it like you would to a 5th grader and select the correct choice:
A car slows down uniformly from a speed of \( 28.0 \) \( \text{m/s} \) to rest in \( 8.00 \) \( \text{s} \). How far did it travel in that time?
A projectile is launched at \( 25 \) \( \text{m/s} \) at an angle of \( 45^\circ \). It lands on a slope \( 5 \) \( \text{m} \) below the launch height. On landing, it rebounds vertically with \( 80\% \) of its speed and falls straight down from there. Find the total time from launch to final impact at the base of the slope.
You are standing on a bathroom scale in an elevator. The elevator starts from rest on the first floor and accelerates up to the third floor, \(12 \, \text{m}\) above, in a time of \(6 \, \text{s}\). The scale reads \(800 \, \text{N}\). What is the mass of the person?
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at \( +13.0 \, \text{m/s}^2 \). At \( t_1 \), the rocket engine is shut down and the sled moves with constant velocity \( v \) until \( t_2 \). The total distance traveled by the sled is \( 5.30 \times 10^3 \, \text{m} \) and the total time is \( 90.0 \, \text{s} \).
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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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