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Chapter 8: Problem 474

In the following exercises, solve. Joseph is traveling on a road trip. The distance, \(d\), he travels before stopping for lunch varies directly with the speed, \(v,\) he travels. He can travel 120 miles at a speed of \(60 \mathrm{mph}\). (a) Write the equation that relates \(d\) and \(v\). (b) How far would he travel before stopping for lunch at a rate of \(65 \mathrm{mph}\) ?

Short Answer

Expert verified
The equation is \(d = 2v\). He would travel 130 miles at 65 mph.

Step by step solution

01

- Understand Direct Variation

The problem states that the distance, \(d\), varies directly with the speed, \(v\). This means we can write the relationship as \(d = kv\), where \(k\) is a constant of proportionality.
02

- Find the Constant of Proportionality

Joseph can travel 120 miles at a speed of 60 mph. Substitute \(d = 120\) miles and \(v = 60\) mph into the equation \(d = kv\): \[ 120 = k \cdot 60 \] Solve for \(k\): \[ k = \frac{120}{60} = 2 \]
03

- Write the Equation

Now that the constant of proportionality \(k\) is known, substitute it back into the direct variation equation: \[ d = 2v \]
04

- Calculate the Distance for a Different Speed

Using the equation found, calculate the distance Joseph would travel before stopping for lunch at a speed of 65 mph: \[ d = 2 \cdot 65 \ d = 130 \text{ miles} \]

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Key Concepts

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

Proportionality Constant
In direct variation problems, a key term is the 'proportionality constant'. This constant, often represented by the symbol \( k \), links two variables that vary directly with each other. To better understand, let's break down the process:
  • We start with the formula, which often looks like this: \( d = kv \), where \( d \) is distance and \( v \) is speed.
  • The 'proportionality constant' \( k \) tells us how much one variable changes with the other.
In our example, Joseph's distance traveled before lunch changes directly with his speed. By using the given values, we find \( k \):
  • Joseph travels 120 miles at 60 mph. Substitute these values: \( 120 = k \times 60 \).
  • Solve for \( k \): \( k = \frac{120}{60} = 2 \). This means for every 1 mph increase in speed, Joseph travels 2 more miles before lunch.
Understanding the 'proportionality constant' is essential because it forms the backbone of the relationship between the variables in direct variation problems.
Distance-Speed Relationship
In the context of our problem, the distance-speed relationship is crucial. Knowing this relationship can help solve many real-world problems, including how far Joseph can travel before lunch.
  • The general formula for this relationship in direct variation is \( d = kv \).
  • After finding \( k \), we determine the specific relationship: \( d = 2v \).
This tells us Joseph's distance traveled is twice his speed. Let's use the formula to solve different scenarios:
  • If Joseph drives at 65 mph, substitute into the formula: \( d = 2 \times 65 = 130 \) miles.
  • This means if he maintains a speed of 65 mph, he could travel 130 miles before lunch.
This formula simplifies calculating distances for any given speed, making planning and estimating travel times straightforward.
Solving Algebraic Equations
Solving algebraic equations involves isolating the variable of interest by using basic arithmetic operations. Here's a step-by-step guide based on Joseph's problem:
  • We start with the direct variation formula: \( d = kv \). Given \( d = 120 \) and \( v = 60 \), substitute these into the formula to find \( k \).
  • This results in: \( 120 = k \times 60 \). To isolate \( k \), divide both sides by 60: \( k = \frac{120}{60} = 2 \).
After finding our constant, the problem asks for how far Joseph would travel at 65 mph:
  • Use the specific formula with \( k \) substituted: \( d = 2v \).
  • Substitute \( v = 65 \): \( d = 2 \times 65 = 130 \) miles.
This step-by-step approach is effective for solving similar algebraic equations, making direct variation problems easier to handle.

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Most popular questions from this chapter

Ken and Joe leave their apartment to go to a football game 45 miles away. Ken drives his car 30 mph faster Joe can ride his bike. If it takes Joe 2 hours longer than Ken to get to the game, what is Joe's speed?

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