0 attempts
0% avg
UBQ Credits
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for displacement. |
| 2 | Set \( y = 0 \), \( y_0 = 19.6 \) m, \( a = -9.8 \) m/s². | Define the variables. |
| 3 | For Ball A (downward): \( v_i = -14.7 \) m/s | Initial velocity downward is negative. |
| 4 | \( 0 = 19.6 -14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 5 | \( 4.9 t^2 +14.7 t -19.6 = 0 \) | Rearrange into standard quadratic form. |
| 6 | Solve for \( t \): \( t = 1 \) s | Find the positive root of the quadratic equation. |
| 7 | For Ball B (upward): \( v_i = +14.7 \) m/s | Initial velocity upward is positive. |
| 8 | \( 0 = 19.6 +14.7 t – 4.9 t^2 \) | Substitute values into the equation. |
| 9 | \( -4.9 t^2 +14.7 t +19.6 = 0 \) | Simplify equation. |
| 10 | \( 4.9 t^2 -14.7 t -19.6 = 0 \) | Multiply both sides by -1. |
| 11 | Solve for \( t \): \( t = 4 \) s | Find the positive root of the quadratic equation. |
| 12 | \( \Delta t = t_{\text{Ball B}} – t_{\text{Ball A}} = 4 \, \text{s} – 1 \, \text{s} = 3 \, \text{s} \) | Calculate the difference in time. |
Answer: The difference in time the balls spend in the air is 3 seconds.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( v = v_i + a t \) | Use the kinematic equation for velocity. |
| 2 | For Ball A: \( v = -14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(1 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball A. |
| 3 | For Ball B: \( v = +14.7 \, \text{m/s} + (-9.8 \, \text{m/s}^2)(4 \, \text{s}) = -24.5 \, \text{m/s} \) | Compute final velocity for Ball B. |
Answer: Each ball strikes the ground with a velocity of -24.5 m/s downward.
| Step | Derivation/Formula | Explanation |
|---|---|---|
| 1 | \( y = y_0 + v_i t + \frac{1}{2} a t^2 \) | Use the kinematic equation for position. |
| 2 | For Ball A: \( y_{\text{A}} = 19.6 + (-14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball A at \( t = 0.8 \) s. |
| 3 | \( y_{\text{A}} = 19.6 -11.76 -3.136 = 4.704 \, \text{m} \) | Simplify to find \( y_{\text{A}} \). |
| 4 | For Ball B: \( y_{\text{B}} = 19.6 + (+14.7)(0.8) + \frac{1}{2}(-9.8)(0.8)^2 \) | Compute position of Ball B at \( t = 0.8 \) s. |
| 5 | \( y_{\text{B}} = 19.6 +11.76 -3.136 = 28.224 \, \text{m} \) | Simplify to find \( y_{\text{B}} \). |
| 6 | \( \Delta y = y_{\text{B}} – y_{\text{A}} = 28.224 – 4.704 = 23.52 \, \text{m} \) | Calculate the distance between the balls. |
Answer: The balls are 23.52 meters apart 0.800 seconds after they are thrown.
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
Above is the graph of the velocity vs. time of a duck flying due south for the winter. At what point might the duck begin reversing directions?
A tennis ball is thrown straight up with an initial speed of \( 22.5 \, \text{m/s} \). It is caught at the same distance above ground.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
You drive \( 4 \) \( \text{km} \) at \( 30 \) \( \text{km/h} \) and then another \( 4 \) \( \text{km} \) at \( 50 \) \( \text{km/h} \). What is your average speed for the whole \( 8 \) \( \text{km} \) trip?
A \( 1.5 \) \( \text{kg} \) block is pushed to the right with just enough force to get it to move. The block is pushed for five seconds with this constant force, then the force is released and the block slides to a stop. If the coefficient of kinetic friction is \( 0.300 \) and the coefficient of static friction is \( 0.400 \), calculate the amount of time that passes from when the force is applied to when the block stops.
A projectile has the least speed at what point in its path?
A car moves forward at a steady \( 10 \) \( \text{m/s} \) for \( 5 \) \( \text{s} \). The driver slams the brakes and brings it to rest in \( 2 \) \( \text{s} \). Without waiting, the driver immediately accelerates backward (negative velocity) for \( 3 \) \( \text{s} \) until reaching \( 8 \) \( \text{m/s} \) in reverse. Draw the velocity vs. time graph.
A rocket, initially at rest, is fired vertically upward with an acceleration of \( 12.0 \, \text{m/s}^2 \). At an altitude of \( 1.00 \, \text{km} \), the rocket engine cuts off. Drag is negligible.
A boat is rowed directly upriver at a speed of \(2.5 \, \text{m/s}\) relative to the water. Viewers on the shore find that it is moving at only \(0.5 \, \text{m/s}\) relative to the shore. What is the speed of the river? Is it moving with or against the boat?
By continuing you (1) agree to our Terms of Use and Terms of Sale 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] |
Metric Prefixes
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] |
One price to unlock most advanced version of Phy across all our tools.
per month
Billed Monthly. Cancel Anytime.
We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.
Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.
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.
Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed
