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."
A coin is dropped from a hot air-balloon that is 250 m above the ground rising at 11 m/s upwards. For the coin, assume up is positive and find the following:
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 \( 90.0 \, \text{s} \).
A car travels at \( 20 \, \text{m/s} \) for 5 minutes and then travels another 2 km at \( 40 \, \text{m/s} \). What is the total distance traveled and time of travel for the car?
The driver of a car makes an emergency stop by slamming on the car’s brakes and skidding to a stop. How far would the car have skidded if it had been traveling twice as fast?
Ball 1 is dropped from rest at time \( t = 0 \) from a tower of height \( h \). At the same instant, ball 2 is launched upward from the ground with the initial speed \( v_0 \). If air resistance is negligible, at what time \( t \) will the two balls pass each other?
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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.
NEW! PHY instantly solves any question
🔥 Elite Members get up to 30% off Physics Tutoring
🧠 Learning Physics this summer? Try our free course.
🎯 Need exam style practice questions? We’ve got over 2000.