Problem 65 - Calculus Single Variable

Question

At a given moment in time, two race cars are abreast and traveling at the speed of \(90 \mathrm{mi} / \mathrm{hr}\). Car A then begins to accelerate at a constant rate of \(6 \mathrm{mi} / \mathrm{min}^{2}\) and Car \(\mathrm{B}\) begins to accelerate at a constant rate of \(9 \mathrm{mi} / \mathrm{min}^{2}\). The cars are driving on a track of circumference 2 miles. How long will it take Car \(\mathrm{B}\) to lap Car A three times?

Step-by-Step Solution

Answer

It takes 2 minutes for Car B to lap Car A three times.

Step 1: Convert Units of Acceleration

First, convert the acceleration from miles per minute squared to miles per hour squared, since the speed is given in miles per hour. The conversion factor is that there are 60 minutes in an hour, so:For Car A: \[6 \text{ mi/min}^2 \times 60^2 = 21600 \text{ mi/hr}^2\]For Car B:\[9 \text{ mi/min}^2 \times 60^2 = 32400 \text{ mi/hr}^2\]

Step 2: Set Differential Equations

Because we are dealing with constant accelerations, we can use the formula for distance: \[ s = ut + \frac{1}{2}at^2 \]where:- \(s\) is the distance, - \(u\) is the initial speed (90 mi/hr), - \(a\) is the acceleration,- \(t\) is the time in hours.For Car A:\[ s_A = 90t + \frac{1}{2} \times 21600t^2 \]For Car B:\[ s_B = 90t + \frac{1}{2} \times 32400t^2 \]

Step 3: Set the Relative Distance Equation

To find when Car B has lapped Car A three times, Car B's total distance traveled must be 6 miles more than that of Car A (since it completes exactly 3 laps more on a 2-mile track). Thus:\[ s_B = s_A + 6 \]Plug in the equations:\[ 90t + \frac{1}{2} \times 32400t^2 = 90t + \frac{1}{2} \times 21600t^2 + 6 \]

Step 4: Simplify and Solve for t

First, cancel out \(90t\) from both sides:\[ \frac{1}{2} \times 32400t^2 = \frac{1}{2} \times 21600t^2 + 6 \]Simplify:\[ 32400t^2 - 21600t^2 = 12 \]\[ 10800t^2 = 12 \]Divide both sides by 10800:\[ t^2 = \frac{12}{10800} \]Simplify the fraction:\[ t^2 = \frac{1}{900} \]Take the square root of both sides:\[ t = \frac{1}{30} \text{ hours} \]

Step 5: Convert Time to Minutes

Convert the time from hours to minutes:\[ t = \frac{1}{30} \times 60 = 2 \text{ minutes} \]

Key Concepts

Differential Equations

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

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

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.

Distance Formula

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.