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Part 1: Calculate the distance traveled at \(20 \, \text{m/s}\) for 5 minutes
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( t_1 = 5 \, \text{minutes} = 300 \, \text{seconds} \) | Convert the time from minutes to seconds since the speed is given in meters per second. |
| 2 | \( v_1 = 20 \, \text{m/s} \) | The initial speed of the car is \(20 \, \text{m/s}\). |
| 3 | \( \Delta x_1 = v_1 \cdot t_1 \) | Use the formula for distance: \( \Delta x = v \cdot t \). |
| 4 | \( \Delta x_1 = 20 \, \text{m/s} \cdot 300 \, \text{s} = 6000 \, \text{m} \) | Calculate the distance traveled at the initial speed. |
Part 2: Calculate the distance traveled at \(40 \, \text{m/s}\) for an additional 2 km
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \Delta x_2 = 2 \, \text{km} = 2000 \, \text{m} \) | Convert the distance from kilometers to meters. |
| 2 | \( v_2 = 40 \, \text{m/s} \) | The speed of the car in the second part of the trip is \(40 \, \text{m/s}\). |
| 3 | \( t_2 = \frac{\Delta x_2}{v_2} \) | Use the formula for time: \( t = \frac{\Delta x}{v} \). |
| 4 | \( t_2 = \frac{2000 \, \text{m}}{40 \, \text{m/s}} = 50 \, \text{s} \) | Calculate the time taken to travel the additional distance. |
Part 3: Calculate the total distance and total time of travel
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( \Delta x_{\text{total}} = \Delta x_1 + \Delta x_2 \) | The total distance is the sum of the distances traveled in each part of the trip found in part 1 and 2. |
| 2 | \( \Delta x_{\text{total}} = 6000 \, \text{m} + 2000 \, \text{m} = 8000 \, \text{m} \) | Calculate the total distance traveled. |
| 3 | \( t_{\text{total}} = t_1 + t_2 \) | The total time is the sum of the times taken in each part of the trip. |
| 4 | \( t_{\text{total}} = 300 \, \text{s} + 50 \, \text{s} = 350 \, \text{s} \) | Calculate the total time of travel. |
Answer:
| Total Distance | Total Time |
|---|---|
| \( \boxed{8000 \, \text{m}} \) | \( \boxed{350 \, \text{s}} \) |
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There are two cables that lift an elevator, each with a force of \(10{,}000 \, \text{N}\). The \(1{,}000 \, \text{kg}\) elevator is lifted from the first floor and accelerates over \(10 \, \text{m}\) until it reaches its top speed of \(6 \, \text{m/s}\). What is the mass of the people in the elevator?
A rock is thrown vertically upward with a velocity of \( 20 \, \text{m/s} \) from the edge of a bridge \( 42 \, \text{m} \) above a river.
Two students are on a balcony 19.6 m above the street. One student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.
An object of unknown mass is acted upon by multiple forces:
The coefficients of friction are \(\mu_s = 0.6\) and \(\mu_k = 0.2\). Starting from rest, the object travels \(10 \, \text{m}\) in \(4.5 \, \text{s}\). What is the mass of the unknown object?
Ball A is dropped from the top of a cliff. At the same time, Ball B is thrown straight upward from the ground at \( 30 \) \( \text{m/s} \). The two balls pass each other after \( 2.0 \) \( \text{s} \).
A rocket, initially at rest, is fired vertically upward with an acceleration of \( 12.0 \, \text{m/s}^2 \). At an altitude of \( 1.00 \, \text{km} \), the rocket engine cuts off. Drag is negligible.
A blue sphere and a red sphere with the same diameter are released from rest at the top of a ramp. The red sphere takes a longer time to reach the bottom of the ramp. The spheres are then rolled off a horizontal table at the same time with the same speed and fall freely to the floor. Which sphere reaches the floor first?
A red car, initially at rest, travels east with an acceleration of \( 3.5 \, \text{m/s}^2 \). At the same time as the red car starts to move, a blue car is traveling west at \( 15 \, \text{m/s} \) and accelerating at \( 1.2 \, \text{m/s}^2 \). If they are \( 600 \, \text{m} \) apart the moment the red car starts to move and they are traveling towards each other, where and when will they meet?
A cart begins to move from rest on a horizontal track. Which of the following correctly indicates the magnitude of the average velocity of the cart during the interval shown and provides a valid explanation?
Hint: when solving this, its consider that the area of the acceleration vs time graph tells you the change in velocity.
A driver is driving at \( 40 \, \text{m/s} \) when the light turns red in front of her. It takes the driver \( 0.9 \, \text{s} \) to react and hit the brakes. After this, the car slows with an acceleration of \( 3.5 \, \text{m/s}^2 \). What is the total distance traveled by the car?
Total distance traveled: \( 8,000 \, \text{m} \)
Total time of travel: \( 350 \, \text{s} \)
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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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