Calculating Distance Using Speed and Time: A Mathematical Dive into Real-World Applications

Every day, countless workers like Mr. Johnson face an interesting challenge as they commute from their homes to their factories. In this article, we will explore a real-world example where Mr. Johnson’s commute time is affected by his walking speed. Understanding these concepts not only helps in solving such problems but also in various real-world applications in fields such as logistics, transportation, and even personal travel planning.

Problem Statement

Mr. Johnson has a routine commute where, if his speed from his house to the factory is 5 km/hr, he is 8 minutes late for his shift. If he increases his speed to 10 km/hr, he arrives 7 minutes early. Let's work through the mathematics to find the distance between his house and the factory.

Step 1: Setting Up Equations

Let the distance from his house to the factory be (d) km.

When walking at 5 km/hr:

Time taken (frac{d}{5}) hours. He is 8 minutes late which is (frac{8}{60} frac{2}{15}) hours late. Let the scheduled time be (T). Therefore, the scheduled time can be expressed as:

(T frac{d}{5} - frac{2}{15})

When walking at 10 km/hr:

Time taken (frac{d}{10}) hours. He is 7 minutes early which is (frac{7}{60}) hours early. So, the scheduled time can also be expressed as:

(T frac{d}{10} frac{7}{60})

Step 2: Setting Equations Equal to Each Other

Since both expressions equal (T), we can set them equal to each other:

(frac{d}{5} - frac{2}{15} frac{d}{10} frac{7}{60})

To eliminate the fractions, we can multiply through by the least common multiple of the denominators which is 60:

(60 cdot frac{d}{5} - 60 cdot frac{2}{15} 60 cdot frac{d}{10} - 60 cdot frac{7}{60})

This simplifies to:

(12d - 8 6d 7)

Step 3: Solving for (d)

Rearranging the equation gives:

( 12d - 6d 7 8)

(6d 15)

(d frac{15}{6} 2.5)

Conclusion: The distance from Mr. Johnson's house to the factory is 2.5 km.

Alternative Approach

Another method to solve this problem involves expressing minutes in terms of hours:

8 minutes (frac{8}{60} frac{2}{15}) hours 7 minutes (frac{7}{60}) hours Let (T) be the on-time duration in hours. We know that (d) is related to speed (s) and time (t) as (d st)

In each scenario, (d) is fixed.

We are told that (frac{d}{5} T frac{2}{15}) (frac{d}{10} T - frac{7}{60})

Thus, we get:

(frac{d}{10} - frac{d}{5} - frac{2}{15} - frac{7}{60} )

This simplifies to:

(frac{-d}{10} - frac{8 7}{60} )

(frac{d}{10} frac{15}{60} frac{1}{4} )

(d 10 cdot frac{1}{4} frac{10}{4} 2.5)

The answer remains the same: the distance between the house and the factory is 2.5 km.

Conclusion

This problem demonstrates the practical application of distance, speed, and time relationships in solving real-world issues. Understanding these concepts is crucial in various fields, such as logistics, transportation, and personal travel planning. By breaking down the problem step-by-step, we can ensure a clear and comprehensive solution, enhancing our problem-solving skills and aiding in more efficient planning and decision-making.