| 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]. |
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A horizontal, uniform board of weight \( 125 \, \text{N} \) and length \( 4 \, \text{m} \) is supported by vertical chains at each end. A person weighing \( 500 \, \text{N} \) is hanging from the board. The tension in the right chain is \( 250 \, \text{N} \).
An object is projected vertically upward from ground level. It rises to a maximum height [katex] H [/katex]. If air resistance is negligible, which of the following must be true for the object when it is at a height [katex] H/2 [/katex] ?
A \( 0.20 \) \( \text{kg} \) object moves along a straight line. The net force acting on the object varies with the object’s displacement as shown in the graph above. The object starts from rest at displacement \( x = 0 \) and time \( t = 0 \) and is displaced a distance of \( 20 \) \( \text{m} \). Determine each of the following.
In the diagram shown, a \(20 \, \text{N}\) force is applied to block \(B\) (\(7 \, \text{kg}\)). Block \(A\) has a mass of \(3 \, \text{kg}\). Assume frictionless conditions.
A block sliding down an frictionless inclined plane is experiencing both gravitational and normal forces; which force’s magnitude changes when the angle of the incline is increased?
A 0.035 kg bullet moving horizontally at 350 m/s embeds itself into an initially stationary 0.55 kg block. Air resistance is negligible.
At time \( t = 0 \) an object is traveling to the right along the \( +x \) axis at a speed of \( 10.0 \) \( \text{m/s} \) with acceleration \( -2.0 \) \( \text{m/s}^2 \). Which statement is true?
A block is given a brief push so that it slides up a ramp. After the block reaches its highest point, it slides back down, but the magnitude of its acceleration is less on the descent than on the ascent. Why?
The moment of inertia of a uniform solid sphere (mass \( M \), radius \( R \)) about a diameter is \( \frac{2}{5}MR^2 \). The sphere is placed on an inclined plane (angle \( \theta \)) and released from rest.
A rocket is sent to shoot down an invading spacecraft that is hovering at an altitude of \( 1500 \, \text{m} \). The rocket is launched with an initial velocity of \( 180 \, \text{m/s} \). Find the following:
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] |
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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