Problem 44 - College Algebra with Corequisite Support

Question

For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question. What is the total distance that two people travel in \(3 \mathrm{~h}\) if one of them is riding a bike at \(15 \mathrm{mi} / \mathrm{h}\) and the other is walking at \(3 \mathrm{mi} /\) h?

Step-by-Step Solution

Answer

The total distance traveled is 54 miles.

Step 1: Identify the known variables

We know the speeds of the two travelers. The biker travels at 15 mi/h and the walker at 3 mi/h. The time given for both travelers is 3 hours.

Step 2: Use the distance formula

The distance formula is \( d = rt \), where \(d\) is distance, \(r\) is the rate of speed, and \(t\) is time. We will use this formula to solve for the distances traveled by each person.

Step 3: Calculate the biker's distance

For the biker, \( r = 15 \) mi/h and \( t = 3 \) hours. Apply the formula: \( d = 15 \times 3 = 45 \) miles.

Step 4: Calculate the walker's distance

For the walker, \( r = 3 \) mi/h and \( t = 3 \) hours. Apply the formula: \( d = 3 \times 3 = 9 \) miles.

Step 5: Sum the distances

Add the distances traveled by both the biker and the walker: \( 45 + 9 = 54 \) miles.

Key Concepts

distance formula

The distance formula is a crucial tool in problems involving movement, and it's expressed simply as \( d = rt \). This formula breaks down how distance \(d\) relates to the rate of speed \(r\) and the time \(t\) spent traveling. The formula is straightforward and operates under the assumption of constant speed. Whenever you know any two of these three variables, you can solve for the third. For example, if you know that a runner maintains a consistent speed and travels for a specific duration, you can easily find out how far she runs using this formula.

• If you need to find the speed: Rearrange the formula to \( r = \frac{d}{t} \).
• If you need to find the time: Rearrange to \( t = \frac{d}{r} \).

Learning to manipulate and apply this formula in various scenarios is key to mastering distance-related problems.

rate of speed

Rate of speed refers to how fast an object is moving and is usually measured in units such as miles per hour (mi/h) or kilometers per hour (km/h). Understanding rate of speed is essential because it helps us compare how quickly different objects or people move. In distance problems, the rate of speed \(r\) allows us to determine how far someone or something will travel over a given period. For instance, in our exercise, we see two different rates of speed: the biker's 15 mi/h and the walker's 3 mi/h.

Different speeds fundamentally change the distances covered, even if the time is the same. Distinguishing between these rates helps to predict how both the biker and walker can cover different distances in the same duration.

distance calculation

Distance calculation involves using the distance formula to determine how far an object will travel based on its speed and duration of travel. For the biker traveling at 15 mi/h for 3 hours, we calculate the distance as follows:
\[ d = 15 \times 3 = 45 \text{ miles} \]
Similarly, for the walker moving at 3 mi/h for the same time span, we have:
\[ d = 3 \times 3 = 9 \text{ miles} \]

This step underscores the importance of inputting accurate values into the formula - it's crucial for achieving correct results. Once you've calculated individual distances, summing them up, as done in the problem, gives the total distance traveled by both individuals.

algebraic problem solving

Algebraic problem solving is all about breaking down problems into manageable parts and using equations to find solutions. In problems like these, you follow several steps:

• Identify the known variables: speed, distance, and time.
• Formulate the equation using known variables ─ here, using \( d = rt \).
• Insert the values into the formula for each traveler.
• Calculate the distance for each using proper multiplication.
• Finally, sum the resulting distances if needed.

Approaching each step methodically helps sort through even complex situations without feeling overwhelmed. Practicing this step-by-step approach not only makes algebraic problem solving less daunting but also builds confidence when facing similar real-world problems.