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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| (a) How fast is the rocket traveling when the engine cuts off? | ||
| 1 | Use the kinematic equation:
\( v^2 = u^2 + 2 a s \) |
Relates final velocity, initial velocity, acceleration, and distance. |
| 2 | Substitute values:
\( v^2 = 0 + 2 \times 12.0 \times 1,000 \) |
Calculated \( v^2 \) using given values. |
| 3 | Solve for \( v \):
\( v = \sqrt{24,000} \approx 154.92 \, \text{m/s} \) |
Found the rocket’s speed at engine cutoff. |
| (b) What maximum height relative to the ground does the rocket reach? | ||
| 4 | After engine cutoff, use \( v^2 = u^2 + 2 a s \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \):
\( 0 = (154.92)^2 + 2 (-9.8) s \) |
Used kinematic equation for upward motion until velocity is zero. |
| 5 | Solve for \( s \):
\( s = \dfrac{(154.92)^2}{2 \times 9.8} \) |
Calculated additional height after engine cutoff. |
| 6 | Total maximum height:
\( h_{\text{total}} = 1,000 + 1,224.49 = 2,224.49 \, \text{m} \) |
Added initial altitude to additional height for total height. |
| (c) Velocity just before the rocket hits the earth. | ||
| 7 | Use \( v^2 = u^2 + 2 a s \) for free fall from maximum height with \( u = 0 \, \text{m/s} \), \( a = 9.8 \, \text{m/s}^2 \), \( s = 2,224.49 \, \text{m} \):
\( v^2 = 0 + 2 \times 9.8 \times 2,224.49 \) |
Calculated final velocity during descent. |
| 8 | Compute \( v^2 \):
\( v^2 = 43,598.00 \, \text{m}^2/\text{s}^2 \) |
Found velocity just before impact. |
| (d) Total amount of time the rocket was in the air. | ||
| 9 | **Time during powered ascent:**
Use \( s = u t + \dfrac{1}{2} a t^2 \) with \( s = 1,000 \, \text{m} \), \( u = 0 \, \text{m/s} \), \( a = 12.0 \, \text{m/s}^2 \): |
Calculated time for powered ascent. |
| 10 | Solve for \( t \):
\( t^2 = \dfrac{1,000}{6.0} \approx 166.67 \) |
Found time during powered ascent. |
| 11 | **Time during coasting ascent:**
Use \( v = u + a t \) with \( v = 0 \, \text{m/s} \), \( u = 154.92 \, \text{m/s} \), \( a = -9.8 \, \text{m/s}^2 \): |
Calculated time from engine cutoff to maximum height. |
| 12 | Solve for \( t \):
\( t = \dfrac{154.92}{9.8} \approx 15.81 \, \text{s} \) |
Found time during coasting ascent. |
| 13 | **Time during free fall descent:**
Use \( s = \dfrac{1}{2} a t^2 \) with \( s = 2,224.49 \, \text{m} \), \( a = 9.8 \, \text{m/s}^2 \): |
Calculated time for descent back to earth. |
| 14 | Solve for \( t \):
\( t^2 = \dfrac{2,224.49}{4.9} \approx 454.9996 \) |
Found time during free fall descent. |
| 15 | **Total time in the air:**
\( t_{\text{total}} = t_{\text{ascent}} + t_{\text{coasting}} + t_{\text{descent}} \) |
Summed all time intervals for total flight time. |
Just ask: "Help me solve this problem."
A cart starts from rest and accelerates uniformly at 4.0 m/s2Â for 5.0 s. It next maintains the velocity it has reached for 10 s. Then it slows down at a steady rate of 2.0 m/s2Â for 4.0 s. What is the final speed of the car?
A Corvette is traveling at a constant velocity \( 30 \, \text{m/s} \) when it passes a stationary supped up Civic. At that moment, the Civic puts the pedal to the floor and accelerates at \( 6 \, \text{m/s}^2 \). The Civic eventually catches up to the Corvette.
Toy car W travels across a horizontal surface with an acceleration of \( a_w \) after starting from rest. Toy car Z travels across the same surface toward car W with an acceleration of \( a_z \), after starting from rest. Car W is separated from car Z by a distance \( d \). Which of the following pairs of equations could be used to determine the location on the horizontal surface where the two cars will meet, and why?
In which of these cases is the rate of change of the particle’s displacement constant?
Note: Answers may be off by \( \pm 0.2 \).
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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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