| Derivation/Formula | Reasoning |
|---|---|
| \[\Delta x_1 = \frac{1}{2} a_1 t_1^2\] | For motion starting from rest under constant acceleration, the displacement is \( \Delta x = \frac{1}{2} a t^2 \). |
| \[\Delta x_1 = \frac{1}{2} (4.3)(6.8)^2 = 99.4\,\text{m}\] | Substitute \( a_1 = 4.3\,\text{m/s}^2 \) and \( t_1 = 6.8\,\text{s} \) to obtain the distance covered while speeding up. |
| \[v_x = a_1 t_1\] | The final speed after the first interval is given by \( v = a t \) since \( v_i = 0 \). |
| \[v_x = (4.3)(6.8) = 29.24\,\text{m/s}\] | Insert the known values to compute the speed just before braking. |
| \[\Delta x_2 = \frac{v_x^2}{2 a_2}\] | Re-arrange \( 0 = v_x^2 + 2(-a_2)\Delta x_2 \) to find the stopping distance with constant deceleration \( a_2 \). |
| \[\Delta x_2 = \frac{(29.24)^2}{2(5.1)} = 83.8\,\text{m}\] | Plug in \( v_x = 29.24\,\text{m/s} \) and \( a_2 = 5.1\,\text{m/s}^2 \). |
| \[\Delta x_{\text{total}} = \Delta x_1 + \Delta x_2\] | Total distance is the sum of the distances during acceleration and deceleration. |
| \[\Delta x_{\text{total}} = 99.4\,\text{m} + 83.8\,\text{m} = 183\,\text{m}\] | Add the two segments to obtain the total displacement. |
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A truck is traveling at \(35 \, \text{m/s}\) when the driver realizes the truck has no brakes. He sees a ramp off the road, inclined at \(20^\circ\), and decides to go up it to help the truck come to a stop. How far does the truck travel before coming to a stop (assume no friction)?
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?
A skateboarder, with an initial speed of \( 20.0 \, \text{m/s} \), rolls to the end of friction-free incline of length \( 25 \, \text{m} \). At what angle is the incline oriented above the horizontal?
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.
Two identical metal balls are being held side by side at the top of a ramp. Alex lets one ball, \( A \), start rolling down the hill. A few seconds later, Alex’s partner, Bob, starts the second ball, \( B \), down the hill by giving it a push. Ball \( B \) rolls down an identical, parallel path to the first ball and passes it. At the instant ball \( B \) passes ball \( A \) (select all that apply):
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?
A car is heading rightward but accelerating to the left. This means the car is:
A cart starts from rest and accelerates uniformly at \(4.0 \, \text{m/s}^2\) for \(5.0 \, \text{s}\). It next maintains the velocity it has reached for \(10 \, \text{s}\). Then it slows down at a steady rate of \(2.0 \, \text{m/s}^2\) for \(4.0 \, \text{s}\). What is the final speed of the car?
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?
At time \( t = 0 \), a cart is at \( x = 10 \, \text{m} \) and has a velocity of \( 3 \, \text{m/s} \) in the \( -x \) direction. The cart has a constant acceleration in the \( +x \) direction with magnitude \( 3 \, \text{m/s}^2 < a < 6 \, \text{m/s}^2 \). Which of the following gives the possible range of the position of the cart at \( t = 1 \, \text{s} \)?
\(183\,\text{m}\)
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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] |
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