Problem 23 - Algebra for College Students

Question

The time required for a car to travel a certain distance varies inversely as the rate at which it travels. If it takes 3 hours to travel the distance at 50 miles per hour, how long will it take at 30 miles per hour? 5 hours

Step-by-Step Solution

Answer

It takes 5 hours to travel the distance at 30 miles per hour.

Step 1: Understanding Inverse Variation

The time required to travel a certain distance varies inversely as the speed. This means that as the speed increases, the time taken decreases and vice versa. Mathematically, this relationship can be written as \( t = \frac{k}{r} \), where \( t \) is the time, \( r \) is the rate of travel in miles per hour, and \( k \) is a constant.

Step 2: Finding the Constant of Variation

We are given that it takes 3 hours to travel the distance at 50 miles per hour. Using the formula \( t = \frac{k}{r} \), substitute \( t = 3 \) and \( r = 50 \) to find the constant \( k \). This gives us \( 3 = \frac{k}{50} \). Solving for \( k \), we get \( k = 3 \times 50 = 150 \).

Step 3: Calculating the Time for 30 Miles Per Hour

Now that we know \( k = 150 \), we can find the time it will take to travel the same distance at 30 miles per hour. Using the equation \( t = \frac{k}{r} \) again, substitute \( k = 150 \) and \( r = 30 \), giving us \( t = \frac{150}{30} = 5 \).

Key Concepts

Mathematical Relationships

Inverse variation is a type of mathematical relationship where one quantity decreases as another quantity increases. In the context of rate and time, as speed increases, the time required to cover a fixed distance decreases, and vice versa. This relationship can be expressed mathematically through the formula:

• \( t = \frac{k}{r} \)

where:

• \( t \) is the time taken,
• \( r \) is the rate or speed, and
• \( k \) is a constant.

This kind of relationship helps us predict or calculate one variable when another changes proportionally in an opposite manner. In real-life applications, understanding inverse relationships is crucial as it lays the foundation for more complex problem-solving scenarios.

Rate and Speed Calculations

Calculating rates and speeds is essential in solving problems involving movement over time. In our exercise, we saw how changing speed affected the time it took to travel a fixed distance. These calculations involve determining how fast an object moves over a certain distance, in a given time frame.
For example, a car traveling at 50 miles per hour covers more ground in an hour compared to a car going 30 miles per hour. However, slower speed means longer time required for the same journey.
Understanding rate and speed calculations is key in various fields such as physics, engineering, and logistics. By mastering these concepts, we improve our ability to make decisions based on how altering speed affects time.

Algebraic Equations

Algebraic equations allow us to express relationships and solve problems through mathematical symbols and operations. In our exercise, the inverse relationship was expressed using the equation \( t = \frac{k}{r} \).

• This means that time \( t \) is proportional to a constant \( k \), divided by rate \( r \).

To find values such as time or speed, we rearrange and solve the equation by substituting known values and solving for the unknown.
In solving the problem, we first found the constant of variation, \( k = 150 \), and then used it to determine the time at a different speed (30 miles per hour).
These calculations show how algebra forms the basis of turning real-world scenarios into solvable mathematical problems, providing clear, logical insights into the nature of the relationship between the variables involved.