Solving Real-World Problems with Speed, Distance, and Time: A Mathematical Journey
Understanding the relationship between speed, distance, and time is crucial for solving a variety of real-world problems. This article explores a mathematical journey where we apply these concepts to solve a common problem, providing you with a deeper insight into the practical applications of these fundamental principles.
In the first scenario, a car travels a distance of 715 km at a uniform speed. If the speed is increased by 10 km/hr, the time taken to travel the same distance is reduced by 2 hours. Let's delve into how we can determine the original speed of the car through a step-by-step mathematical process.
Original Speed Calculation
Let the original speed of the car be x km/h. The time taken to travel 715 km at this speed is given by:
Time Distance / Speed 715 / x
When the speed is increased by 10 km/h, the new speed is x - 10 km/h. The time taken at this new speed is:
Time Distance / New Speed 715 / (x - 10)
According to the problem, the time taken with the increased speed is 2 hours less than the time taken at the original speed. Therefore, we can set up the equation:
(715 / x) - (715 / (x - 10)) 2
To solve this equation, we first find a common denominator, which is x(x - 10):
(715(x - 10) - 715x) / [x(x - 10)] 2
This simplifies to:
(715x - 7150 - 715x) / [x(x - 10)] 2
Further simplification yields:
7150 / [x(x - 10)] 2
Multiplying both sides by x(x - 10):
7150 2x^2 - 2
Rearranging the equation gives:
2x^2 - 2 - 7150 0
Dividing the entire equation by 2 to simplify:
x^2 - 1 - 3575 0
Now we can use the quadratic formula:
x (-b ± √(b^2 - 4ac)) / 2a
where a 1, b -10, and c -3575:
Calculate the discriminant:
10^2 - 4 * 1 * (-3575) 100 14300 14400
Applying the quadratic formula:
x (-(-10) ± √14400) / (2 * 1) (10 ± 120) / 2
This gives us two potential solutions:
x (110) / 2 55 and x (-130) / 2 -65
Since speed cannot be negative, we take the positive solution:
x 55 km/h
Therefore, the original speed of the car is 55 km/h.
Second Scenario
In the second scenario, a car travels a distance of 840 km at a uniform speed. If the speed of the car is increased by 10 km/hr, it takes 2 hours less to cover the same distance. Let's apply the same principles to solve for the original speed.
Let S denote the required original speed in km/hr of the given car. From the data, we get the relation:
840 / S - 840 / (S 10) 2
Simplifying, we get:
S - 10 840 / 2 420
Rearranging the equation gives:
S^2 - 10S - 4200 0
Applying the quadratic formula again:
S (10 ± √(10^2 4 * 4200)) / 2 (10 ± 120) / 2
This gives us two potential solutions:
S (130) / 2 70 and S (-110) / 2 -55
Since speed cannot be negative, we take the positive solution:
S 70 km/hr
Thus, the original speed of the car is 70 km/hr.
Key Takeaways
1. **Speed, Distance, and Time Relationship**: Understanding the fundamental relationship between speed, distance, and time is essential for solving real-world problems.
2. **Mathematical Application**: By applying the principles of algebra and quadratic equations, we can find the original speed of a car given a set of conditions.
3. **Real-World Applications**: These mathematical concepts have numerous applications in fields such as physics, engineering, and transportation, making them highly valuable knowledge.
These problems not only test your mathematical skills but also help you understand the importance of these fundamental principles in solving practical scenarios. By practicing similar problems, you can enhance your problem-solving abilities and gain a deeper understanding of the relationships between speed, distance, and time.