91Ó°ÊÓ

When analyzing problems involving vehicles, especially at high speeds, it’s crucial to express velocity in consistent units. This often involves converting speeds into feet per second (ft/s) or meters per second (m/s), which are standard in physical calculations.
In this exercise, the velocity of the space shuttle orbiter is initially given in miles per hour (mi/hr). To convert to feet per second, follow these simple steps:

• Understand the basic conversion factors: 1 mile is equal to 5280 feet, and 1 hour is equal to 3600 seconds.
• Use the formula: \( v_{ft/s} = v_{mi/hr} \times \frac{5280}{3600} \).

Applying this to our problem:

• For a speed of 200 mi/hr, the conversion is \( 200 \times \frac{5280}{3600} \approx 293.33 \text{ ft/s} \).
• For a speed of 35 mi/hr, the conversion results in \( 35 \times \frac{5280}{3600} \approx 51.33 \text{ ft/s} \).

This provides a consistent unit of measure, making equations involving motion more manageable.

Differential equations are equations that involve the rates of change of quantities. In the context of physics, they are often used to represent how velocities, accelerations, and other physical phenomena change over time or space.
In this exercise, the differential equation governing the shuttle’s motion is based on its deceleration, which is linked to the square of its velocity:\[ a(v) = \frac{dv}{dt} = -0.0003 v^2 \]Here, \( a(v) \) is the acceleration (or deceleration when negative), \( dv/dt \) is the rate of change of velocity over time, and \( v \) is velocity. This setup reflects a situation where the deceleration is directly proportional to \( v^2 \).
The goal is to rearrange this equation to find out more about the state of the system—for instance, by expressing acceleration in terms of distance instead of time, which leads us to the relationship:\[ v \frac{dv}{ds} = -0.0003 v^2 \]This can then be simplified to:\[ \frac{dv}{ds} = -0.0003v \]Understanding these transformations is key to solving the problem analytically and gaining insights into the shuttle's behavior during deceleration.

Integration is a core technique in solving differential equations. It allows us to move from a rate of change expressed in derivatives to an expression that describes a total change—a vital step in understanding different aspects of motion, such as displacement and velocity changes.
In the context of the problem, once the equation has been set up as:\[ \frac{dv}{ds} = -0.0003v \]The next step is to integrate both sides of the equation to find the total distance traveled:

• Rearrange the equation: \( \int \frac{1}{v} dv = \int -0.0003 ds \).
• Perform the integration. The left side uses a simple logarithmic integration, resulting in: \( \ln |v| = -0.0003s + C \).

The constant \( C \) is determined by applying initial conditions, which is addressed in the next concept. This technique bridges the gap between differential equation formulations and tangible physical quantities like distance.

Initial conditions are specific values at the beginning of a problem that allow us to solve for constants in our equations. Without these conditions, it would be impossible to find unique solutions to differential equations.
For the shuttle orbiter problem, the initial condition is the known velocity when the parachute deploys, \( v_1 = 293.33 \text{ ft/s} \), at the starting point \( s = 0 \).
Using these conditions:

• Substitute \( s = 0 \) and \( v = 293.33 \) into the equation \( \ln |v| = -0.0003s + C \).
• In this case, \( 293.33 = e^{C} \) leads to finding \( C \) as \( \ln 293.33 \).

Once \( C \) is known, it can be used in further calculations to find the distance traveled until the parachute is jettisoned at the velocity \( v_2 = 51.33 \text{ ft/s} \). This approach turns the mathematical expression into a solution with practical relevance.