Problem 36
Question
Tom and Joel are good runners, both able to run at a constant speed of 10 miles per hour. Their amazing dog Trot can do even better; he runs at 20 miles per hour. Starting from towns 60 miles apart, Tom and Joel run toward each other while Trot runs back and forth between them. How far does Trot run by the time the boys meet? Assume that Trot started with Tom running toward Joel and that he is able to make instant turnarounds. Solve the problem two ways. (a) Use a geometric series. (b) Find a shorter way to do the problem.
Step-by-Step Solution
Verified Answer
Trot runs 60 miles before the boys meet.
1Step 1: Calculate Meeting Time
First, determine how long it takes for Tom and Joel to meet. Tom and Joel are running towards each other, each at a speed of 10 miles per hour. Therefore, the distance between them closes at a rate of 20 miles per hour (10 miles per hour each). Since they start 60 miles apart, it will take them \(\frac{60}{20} = 3\) hours to meet.
2Step 2: Geometric Series Setup for Trot's Distance
While Tom and Joel are running towards each other, Trot runs back and forth between them. Let's calculate all the distances he travels using a geometric series. Trot runs for 3 hours at a constant speed of 20 miles per hour. Analyzing the scenario as a harmonic series allows us to conclude that: Trot runs a geometric sequence of distances each half time he makes a lap. Given the infinite back-and-forth, the sum of the series for a period he covers has a common ratio of \(\frac{1}{2}\).
3Step 3: Sum of Geometric Series
The distance he runs can be repeated for every half go-and-return as half of the previous loops. An infinite geometric sum for each infinite half-second covers 30 miles \(\text{Go}=20 \, \text{miles/hour}, \frac{20}{2}, \frac{20}{4}, \frac{20}{8} \), ... and so on. Calculate the sum by using the formula \(S = \frac{a}{1 - r}\) where \(a = 20\) miles/hour and \(r = \frac{1}{2}\): \[ S = \frac{20}{1 - \frac{1}{2}} = 40 \text{ miles} \].
4Step 4: Alternative Method to Find Trot's Distance
There's a simpler way to calculate the distance Trot runs. Since Trot runs constantly at 20 miles per hour for 3 hours (the time Tom and Joel spend running until they meet), the distance he covers is simply his speed multiplied by the time. Thus, the distance Trot runs is \(20 \times 3 = 60\) miles.
Key Concepts
Constant SpeedDistance CalculationProblem Solving Strategies
Constant Speed
Understanding constant speed is crucial for solving many distance-related problems like the one involving Tom, Joel, and Trot the dog. Constant speed is when an object covers equal distances in equal amounts of time. In this problem, Tom and Joel both run at a constant speed of 10 miles per hour. This means every hour, each runner covers exactly 10 miles, no more, no less.
Trot, their dog, runs at a constant speed of 20 miles per hour. This indicates that for every hour Trot runs, he travels 20 miles.
When objects move at constant speeds, predicting and calculating distances over time becomes straightforward, as seen with Trot running back and forth.
Trot, their dog, runs at a constant speed of 20 miles per hour. This indicates that for every hour Trot runs, he travels 20 miles.
When objects move at constant speeds, predicting and calculating distances over time becomes straightforward, as seen with Trot running back and forth.
- Tom's and Joel's speed: 10 miles/hour
- Trot's speed: 20 miles/hour
Distance Calculation
Distance calculation is one of the fundamental processes in understanding motion problems. To calculate the distance, you usually multiply speed by time. In the given problem, Tom and Joel start 60 miles apart and move towards each other.
Since both are running at 10 miles per hour, together they reduce the distance between them by 20 miles every hour. Thus, it takes them 3 hours to meet since \(\frac{60 \text{ miles}}{20 \text{ miles/hour}} = 3 \text{ hours}\).
Since both are running at 10 miles per hour, together they reduce the distance between them by 20 miles every hour. Thus, it takes them 3 hours to meet since \(\frac{60 \text{ miles}}{20 \text{ miles/hour}} = 3 \text{ hours}\).
- Time to meet: 3 hours
- Distance Trot runs: Speed × Time = 20 miles/hour × 3 hours = 60 miles
Problem Solving Strategies
Problem-solving strategies play a vital role in efficiently resolving challenges in exercises like this running scenario. Here, two different approaches were highlighted: using a geometric series, and a simpler, direct calculation.
Using a geometric series can accurately solve sub-problems involving repetitive or split distances, like Trot running to and fro. We analyze each segment of his run as a part of a decreasing series where each lap is half the distance of the previous one. Ultimately, we sum this infinite geometric series to get 40 miles.
However, the direct calculation method is less complex. By simply recognizing that Trot runs continuously at 20 miles per hour for the entire 3 hours Tom and Joel take to meet, you can calculate 20 miles/hour × 3 hours = 60 miles.
This solution shows how arithmetic can sometimes act as a powerful counterpart to more complex algebraic methods, employing strategies such as:
Using a geometric series can accurately solve sub-problems involving repetitive or split distances, like Trot running to and fro. We analyze each segment of his run as a part of a decreasing series where each lap is half the distance of the previous one. Ultimately, we sum this infinite geometric series to get 40 miles.
However, the direct calculation method is less complex. By simply recognizing that Trot runs continuously at 20 miles per hour for the entire 3 hours Tom and Joel take to meet, you can calculate 20 miles/hour × 3 hours = 60 miles.
This solution shows how arithmetic can sometimes act as a powerful counterpart to more complex algebraic methods, employing strategies such as:
- Recognizing constant speeds
- Breaking down complex tasks
- Choosing an appropriate approach based on simplicity and context
Other exercises in this chapter
Problem 35
In one version of Zeno's paradox, Achilles can run ten times as fast as the tortoise, but the tortoise has a 100-yard headstart. Achilles cannot catch the torto
View solution Problem 36
$$ \left|\frac{c^{2}-c}{\cos c}\right| ;\left[0, \frac{\pi}{4}\right] $$
View solution Problem 37
Assuming that \(u_{1}=\sqrt{3}\) and \(u_{n+1}=\sqrt{3+u_{n}}\) determine a convergent sequence, find \(\lim _{n \rightarrow \infty} u_{n}\) to four decimal pla
View solution Problem 37
Prove that if \(a_{n} \geq 0, b_{n}>0\), \(\lim _{n \rightarrow \infty} a_{n} / b_{n}=0\), and \(\Sigma b_{n}\) converges then \(\Sigma a_{n}\) converges.
View solution