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Other models of drag due to air resistance: When some objects move at high speeds, air resistance has a more pronounced effect. For such objects, retardation due to air resistance is often modeled as being proportional to the square of velocity or to even higher powers of velocity. a. Arifle bullet fired downward has an initial velocity of 2100 feet per second. If we use the model that gives air resistance as the square of velocity, then the equation of change for the downward velocity \(V\), in feet per second, of a rifle bullet is $$ \frac{d V}{d t}=32-r V^{2} . $$ If the terminal velocity for a bullet is 1600 feet per second, find the drag coefficient \(r\). b. A meteor may enter the Earth's atmosphere at a velocity as high as 90,000 feet per second. If the downward velocity \(V\), in feet per second, of a meteor in the Earth's atmosphere is governed by the equation of change $$ \frac{d V}{d t}=32-2 \times 10^{-18} V^{4}, $$ how fast will it be traveling when it strikes the ground, assuming that it has reached terminal velocity?

Step by step solution

01

Understand Terminal Velocity Condition for Part (a)

For terminal velocity, the acceleration becomes zero. This occurs when the drag force equals the gravitational force. Given the terminal velocity is 1600 ft/s, substituting into the equation \( \frac{dV}{dt} = 32 - rV^2 \) gives us the equation \( 0 = 32 - r(1600)^2 \).

02

Solve for Drag Coefficient r in Part (a)

Rearrange the equation from Step 1 to solve for \( r \): \( r(1600)^2 = 32 \). Solve for \( r \) by dividing both sides by \( 1600^2 \): \( r = \frac{32}{1600^2} \).

03

Evaluate Drag Coefficient r

Calculate \( r \) using the values from Step 2: \( r = \frac{32}{1600^2} = \frac{32}{2560000} = 0.0000125 \).

04

Understand Terminal Velocity Condition for Part (b)

For terminal velocity, \( \frac{dV}{dt} \) becomes zero. Substitute \( V \) as the terminal velocity into \( \frac{dV}{dt} = 32 - 2 \times 10^{-18}V^4 \) and set it to zero: \( 0 = 32 - 2 \times 10^{-18}V^4 \).

05

Solve for Terminal Velocity in Part (b)

Rearrange the equation from Step 4 to solve for \( V \): \( 2 \times 10^{-18}V^4 = 32 \). Thus, \( V^4 = \frac{32}{2 \times 10^{-18}} = 1.6 \times 10^{19} \). Now, take the fourth root of both sides to find \( V \).

06

Calculate Terminal Velocity for the Meteor

Calculate \( V \) from \( V^4 = 1.6 \times 10^{19} \): \( V = (1.6 \times 10^{19})^{0.25} \). Performing the calculations yields \( V \approx 15811 \text{ ft/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Terminal Velocity

Terminal velocity is a crucial concept in understanding air resistance modeling. When an object falls through a fluid, like air, it experiences drag—a force opposing its motion. Eventually, the object reaches a speed where the drag force equals the gravitational force pulling it down.
This speed is called the terminal velocity. Once an object hits terminal velocity, its acceleration is zero, and it continues to move at a constant speed. In our problem, the bullet achieves a terminal velocity of 1600 feet per second, while the meteor, with a much greater potential terminal velocity, deduces how far drag can slow it down despite its huge initial speed.
Terminal velocity is especially important when modeling how objects behave at high speeds, like bullets or meteors, to ensure we can accurately predict their motion and impacts.

Drag Coefficient

The drag coefficient, usually denoted by the letter r, is a dimensionless number that characterizes the resistance an object encounters as it moves through a fluid, such as air.
In mathematical modeling, this coefficient is vital for quantifying the drag force experienced by the object. For the bullet example, calculating the drag coefficient involves using the terminal velocity condition—where the drag force and gravitational force balance out.

• The expression for drag used in the problem was proportional to the square of the velocity, leading to the equation \(r = \frac{32}{1600^2} \) —essentially determining the exact drag coefficient necessary for a bullet's terminal velocity computation.

The lower the drag coefficient, the less resistance the object experiences, which could be beneficial for understanding trajectories and speeds of projectiles or vehicles.

Differential Equations

Differential equations describe the relationship between functions and their rates of change. They are fundamental in physics for modeling dynamic systems, particularly motion against resistance forces.
In the given exercises, the change in velocity with respect to time, \( \frac{dV}{dt} \), is described using differential equations. This allows us to see how velocity evolves under specific conditions,

• For the bullet, the equation was \( \frac{dV}{dt} = 32 - rV^2 \)

, showing that the rate of velocity change depends on a balance of gravitational pull and drag force (which scales with the square of velocity).
In solving the problem, setting \( \frac{dV}{dt} \) to zero helped find consistent velocity scenarios like terminal velocity conditions. These insights highlight why differential equations are key tools in velocity modeling and predicting motion under various conditions.

Velocity Modeling

Velocity modeling helps us predict how the speed of an object changes over time. It accounts for forces like gravity and air resistance acting on it. In cases of high-speed objects, such as bullets and meteors, these models are especially critical.
Using velocity equations lets us estimate how quickly an object will reach the ground and at what speed. For instance, in the problem:

• The bullet's velocity is modeled by an equation factoring in gravity's constant addition and the squared velocity's drag.
• Meteors have a different drag model, incorporating the velocity to the fourth power.

These models can help determine trajectories, impacts, and necessary precautions. By adjusting the variable parameters, we can understand how changes in the environment or object shape affect the final speed and impact forces, aiding engineers and scientists in designing safer and more efficient projects.