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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."
Mary and Sally are in a foot race. When Mary is \( 22 \) \( \text{m} \) from the finish line, she has a speed of \( 4.0 \) \( \text{m/s} \) and is \( 5.0 \) \( \text{m} \) behind Sally, who has a speed of \( 5.0 \) \( \text{m/s} \). Sally thinks she has an easy win and, during the remaining portion of the race, decelerates at a constant rate of \( 0.40 \) \( \text{m/s}^2 \) until she reaches the finish line. What constant acceleration must Mary maintain during the remaining portion of the race if she wishes to cross the finish line side-by-side with Sally?
A cart begins to move from rest on a horizontal track. Which of the following correctly indicates the magnitude of the average velocity of the cart during the interval shown and provides a valid explanation?
Hint: when solving this, its consider that the area of the acceleration vs time graph tells you the change in velocity.
A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from the ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magnitudes of their acceleration:
A cart is initially moving at 0.5 m/s along a track. The cart comes to rest after traveling 1 m. The experiment is repeated on the same track, but now the cart is initially moving at 1 m/s. How far does the cart travel before coming to rest?
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
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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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