0 attempts
0% avg
UBQ Credits
| 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."
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
A ball is dropped from a window [katex]10 \, [/katex] above the sidewalk. Determine the time it takes for the ball to fall to the sidewalk.
A car increases its forward velocity uniformly from 40 m/s to 80 m/s while traveling a distance of 200 m. What is its acceleration during this time?
A 100-pound rock and a 1-pound metal arrow pointed downwards are dropped from height \( h \). Assuming there is no air resistance, which one hits the ground first and why?
A mine shaft is known to be 57.8 m deep. If you dropped a rock down the shaft, how long would it take for you to hear the sound of the rock hitting the bottom of the shaft knowing that sound travels at a constant velocity of 345 m/s?
Note: Answers may be off by \( \pm 0.2 \).
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
