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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] \Delta x = 100 \, \text{m} [/katex] | The total distance from the car’s position when the light turns red to the stop point is 100 meters. |
| 2 | [katex] v_i = 25 \, \text{m/s} [/katex] | The initial velocity of the car is 25 meters per second. |
| 3 | [katex] v_x = 0 \, \text{m/s} [/katex] | The final velocity of the car is 0 meters per second (the car comes to a halt). |
| 4 | [katex] \mu_k = 0.65 [/katex] | The coefficient of kinetic friction between the car’s tires and the road is 0.65. |
| 5 | [katex] f_k = \mu_k \cdot m \cdot g [/katex] | The kinetic friction force is equal to the coefficient of kinetic friction multiplied by the car’s mass and gravitational acceleration. |
| 6 | [katex] a = -\mu_k \cdot g [/katex] | The acceleration due to friction is the friction force divided by mass (mass cancels out). Here [katex] g [/katex] is the acceleration due to gravity. Using [katex] g = 9.8 \, \text{m/s}^2 [/katex]. |
| 7 | [katex] a = -0.65 \cdot 9.8 \, \text{m/s}^2 = -6.37 \, \text{m/s}^2 [/katex] | Substitute the values into the acceleration formula to find the deceleration. |
| 8 | [katex] v_x^2 = v_i^2 + 2a \Delta x_{braking} [/katex] | Use the kinematic equation to relate the distances and velocities during braking. |
| 9 | [katex] 0 = (25 \, \text{m/s})^2 + 2(-6.37 \, \text{m/s}^2) \Delta x_{braking} [/katex] | Set the final velocity to zero and substitute the initial velocity and acceleration to solve for [katex]\Delta x_{braking}[/katex]. |
| 10 | [katex] 0 = 625 \, \text{m}^2/\text{s}^2 – 12.74 \, \text{m/s}^2 \Delta x_{braking} [/katex] | Simplify the equation. |
| 11 | [katex] \Delta x_{braking} = \frac{625}{12.74} \approx 49.06 \, \text{m} [/katex] | Solve for the braking distance [katex]\Delta x_{braking}[/katex]. |
| 12 | [katex] \Delta x_{reaction} = 100 \, \text{m} – 49.06 \, \text{m} = 50.94 \, \text{m} [/katex] | Subtract the braking distance from the total distance to find the distance covered during reaction time. |
| 13 | [katex] \Delta x_{reaction} = v_i t_{reaction} [/katex] | During the reaction time, the car travels with a constant velocity of 25 m/s. |
| 14 | [katex] 50.94 \, \text{m} = 25 \, \text{m/s} \cdot t_{reaction} [/katex] | Substitute the known values into the reaction time equation. |
| 15 | [katex] t_{reaction} = \frac{50.94 \, \text{m}}{25 \, \text{m/s}} \approx 2.04 \, \text{s} [/katex] | Solve for the reaction time. |
| 16 | [katex] t_{reaction} \approx 2.04 \, \text{s} [/katex] | The reaction time of the driver is approximately [katex]\boxed{2.04 \, \text{s}}[/katex]. |
Just ask: "Help me solve this problem."
Will Clark throws a baseball with a horizontal component of velocity of \(25 \, \text{m/s}\). It takes \(3 \, \text{s}\) to come back to its original height. Calculate the baseball’s:
A child slides down a slide with a \( 34^\circ \) incline, and at the bottom her speed is precisely half what it would have been if the slide had been frictionless. Calculate the coefficient of kinetic friction between the slide and the child.
If the coefficient of static friction is \( \mu_s = 0.5 \), how much force must be applied to a spring (spring constant of \( 0.8 \) \( \text{N/m} \)) which is attached to a block of wood (mass \( 4.0 \) \( \text{kg} \)) in order to just begin to move the block?
A spring stretches \( 8.0 \) \( \text{cm} \) when a \( 13 \) \( \text{N} \) force is applied. How far does it stretch when a \( 26 \) \( \text{N} \) force is applied?
When a skier skis down a hill, the normal force exerted on the skier by the hill is
2.04 s
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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] |
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] |
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