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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]. |
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A girl, standing still, tosses a ball vertically upwards. One second later, she tosses up another ball at the same velocity. The balls collide \( 0.5 \) \( \text{s} \) after the second ball is tossed. With what velocity were they tossed? The acceleration due to gravity is \( 9.8 \) \( \text{m/s}^2 \).
An object can move upward while having a downward acceleration.
A mass moving with a constant speed \( u \) encounters a rough surface and comes to a stop. The mass takes a time \( t \) to stop after encountering the rough surface. The coefficient of dynamic friction between the rough surface and the mass is \( 0.40 \). Which of the following expressions gives the initial speed \( u \)?
The speed of a \(40 \, \text{N}\) hockey puck, sliding across a level ice surface, decreases at the rate of \(0.61 \, \text{m/s}^2\). The coefficient of kinetic friction between the puck and ice is
Which of these scenarios involve accelerated motion? (Select all that apply)
A “doomsday” asteroid with a mass of \( 1010 \, \text{kg} \) is hurtling through space. Unless the asteroid’s speed is changed by about \( 0.20 \, \text{cm/s} \), it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of \( 2.5 \, \text{N} \). For how long must this force act?
The magnitude of the gravitational field on the surface of a new planet is \(20 \, \text{N/kg}\). The planet’s mass is half the mass of Earth. The radius of Earth is \(6400 \, \text{km}\). What is the radius of the new planet?
A \( 4700 \, \text{kg} \) truck carrying a \( 900 \, \text{kg} \) crate is traveling at \( 25 \, \text{m/s} \) to the right along a straight, level highway, as shown above. The truck driver then applies the brakes, and as it slows down, the truck travels \( 55 \, \text{m} \) in the next \( 3.0 \, \text{s} \). The crate does not slide on the back of the truck.
The driver of a car makes an emergency stop by slamming on the car’s brakes and skidding to a stop. How far would the car have skidded if it had been traveling twice as fast?
Two objects (49.0 and 24.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley hangs from the ceiling. Find the acceleration of the objects and the tension in the string.
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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