0 attempts
0% avg
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[\Delta x_1 = v t\] | For \(0 \le t \le 1\text{ s}\), the body moves at constant speed \(v = 24\text{ m/s}\); displacement equals speed times time. |
| 2 | \[\Delta x_1 = 24 \times 1 = 24 \text{ m}\] | Substitute \(v = 24\text{ m/s}\) and \(t = 1\text{ s}\). |
| 3 | \[x(1) = 0 + \Delta x_1 = 24 \text{ m}\] | Position at \(t = 1\text{ s}\) is origin plus the displacement \(24\text{ m}\). |
| 4 | \[\Delta x_2 = v_i \Delta t + \frac{1}{2} a (\Delta t)^2\] | For \(1 \le t \le 11\text{ s}\), use kinematic displacement with constant acceleration \(a = -6\text{ m/s}^2\), initial velocity \(v_i = 24\text{ m/s}\), and time interval \(\Delta t = 10\text{ s}\). |
| 5 | \[\Delta x_2 = 24(10) + \frac{1}{2}(-6)(10)^2 = 240 – 300 = -60 \text{ m}\] | Plug in \(v_i = 24\text{ m/s}\), \(a = -6\text{ m/s}^2\), and \(\Delta t = 10\text{ s}\). |
| 6 | \[x_{11} = x(1) + \Delta x_2 = 24 + (-60) = -36 \text{ m}\] | Add displacement \(\Delta x_2\) to position at \(t = 1\text{ s}\) to get final position at \(t = 11\text{ s}\). |
| 7 | \[\boxed{-36 \text{ m}}\] | Numerical answer boxed. |
Just ask: "Help me solve this problem."
We'll help clarify entire units in one hour or less — guaranteed.
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
Police officers have measured the length of a car’s tire skid marks to be \( 23 \, \text{m} \). This particular car is known to decelerate at a constant \( 7.5 \, \text{m/s}^2 \). What was the car’s initial velocity?
An object is dropped from the top of a 45 m tall building
A car’s velocity increases as follows each second: \( 2 \) \( \text{m/s} \), \( 4 \) \( \text{m/s} \), \( 6 \) \( \text{m/s} \), \( 8 \) \( \text{m/s} \). This pattern shows that the car is:
A cart starts from rest and accelerates uniformly at 4.0 m/s2 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/s2 for 4.0 s. What is the final speed of the car?
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.
A ball rolls down a ramp and gains speed. Its velocity is increasing in the negative direction. What can be said about its acceleration?
An object is moving in the \( +x \) direction and begins to slow down. What must be true about its acceleration?
A large beach ball is dropped from the ceiling of a school gymnasium to the floor about 10 meters below. Which of the following graphs would best represent its velocity as a function of time? (do not neglect air resistance)
In which of these cases is the rate of change of the particle’s displacement constant?
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
