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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."
The occupants of a car traveling at a speed of \( 30 \) \( \text{m/s} \) note that on a particular part of a road their apparent weight is \( 15\% \) higher than their weight when driving on a flat road.
Suppose an object is accelerated by a force of \( 100 \) \( \text{N} \). Suddenly a second force of \( 100 \) \( \text{N} \) in the opposite direction is exerted on the object, so that the forces cancel. The object
A vehicle is moving at a speed of 12.3 m/s on a decline when the brakes of all four wheels are fully applied, causing them to lock. The slope of the decline forms an angle of 18.0 degrees with the horizontal plane. Given that the coefficient of kinetic friction between the tires and the road surface is 0.650.
A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only \( 0.75 \) of her regular weight. Calculate the acceleration of the elevator, and find the direction of acceleration.
The steepest street in the world is Baldwin Street in Dunedin, New Zealand. It has an inclination angle of \( 38.0^\circ \) with respect to the horizontal. Suppose a wooden crate with a mass of \( 25.0 \) \( \text{kg} \) is placed on Baldwin Street. An additional force of \( 59 \) \( \text{N} \) must be applied to the crate perpendicular to the pavement in order to hold the crate in place. If the coefficient of static friction between the crate and the pavement is \( 0.599 \), what is the magnitude of the frictional force?
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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