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Differential equations are crucial tools in mathematics, especially when studying motion and change over time. In the context of our exercise, differential equations relate to how the position of each car changes with respect to time. Since both cars start at the same speed and accelerate uniformly, we can use a differential equation to track their respective distances. The general form of the equation for position with constant acceleration is:

• \[ s = ut + \frac{1}{2}at^2 \]

Here, \( s \) represents the distance traveled, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) stands for time. By developing these equations for both cars, you can figure out the moment when one car laps the other multiple times. Differential equations translate these real-world motion problems into mathematically solvable scenarios.

Acceleration is a measure of how quickly an object's speed changes. In this problem, Car A and Car B start at the same initial speed but have different accelerations.

• Car A accelerates at \(6 \, \mathrm{mi/min}^2\).
• Car B accelerates at \(9 \, \mathrm{mi/min}^2\).

Constant acceleration means the speed of both cars increases at a steady rate, making it predictable and uniform. By knowing both the initial speed and acceleration of each car, you can calculate their speeds at any point in time using the formula: \[ v = u + at \] Where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time. Understanding how acceleration impacts the motion of the cars helps us determine their positions in relation to each other over time.

Unit conversion is an essential skill that helps bridge different measurement systems. In the given exercise, our speeds and accelerations need to be compatible to solve for the time correctly. We start with acceleration in \(\mathrm{mi/min}^2\), but our cars' speed is in \(\mathrm{mi/hr}\). We must convert these units so that they are consistent when used in equations. To convert acceleration from \(\mathrm{mi/min}^2\) to \(\mathrm{mi/hr}^2\), use the factor:

• 1 minute = 60 seconds or 1 hour = 60 minutes.

Thus, \[ 6 \times 60^2 = 21600 \text{ mi/hr}^2 \] for Car A and \[ 9 \times 60^2 = 32400 \text{ mi/hr}^2 \] for Car B. This conversion ensures that all calculations are in the same unit system, avoiding any mathematical inconsistencies.

The distance formula is a handy tool for understanding how far an object travels over time, especially under uniform acceleration. The specific formula used in our exercise is \[ s = ut + \frac{1}{2}at^2 \]. Here, \( s \) is the distance, \( u \) is the starting speed (initial velocity), \( a \) is the acceleration, and \( t \) is the time elapsed. By using this formula, you can calculate the distance each car travels during a certain period of time.For Car A and B, plugged-in values provide an expression of their journeys:

• For Car A: \[ s_A = 90t + \frac{1}{2} \times 21600t^2 \]
• For Car B: \[ s_B = 90t + \frac{1}{2} \times 32400t^2 \]

The challenge is to figure out how long it will take for Car B to cover a greater distance than Car A by 6 miles, which corresponds to completing 3 more laps around a circuit that's 2 miles long. This setup requires an understanding of the distance formula, which unveils significant insights when examining the cars' respective positions over time.