Question

Christian went on a run. He jogged for five minutes, then increased his rate to a faster constant pace for seven minutes. He then took a rest for four minutes.
During what time interval has Christian traveled 1.5 miles?

Christian traveled 1.5 miles from \square \leq t \leq \square

Answer 1

To determine during what time interval Christian traveled 1.5 miles, we need to analyze his run.

1. **Jogging Phase (5 minutes):**
Assume Christian jogs at a speed of \( x \) miles per minute. In 5 minutes, the distance he covers is:
\[
\text{Distance} = x \times 5 \text{ minutes} = 5x \text{ miles}
\]

2. **Faster Pace Phase (7 minutes):**
Next, he increases his pace to \( y \) miles per minute for 7 minutes. The distance covered during this phase is:
\[
\text{Distance} = y \times 7 \text{ minutes} = 7y \text{ miles}
\]

3. **Total Distance Calculation:**
By the end of his run (before resting), the total distance he would have traveled is:
\[
\text{Total Distance} = 5x + 7y \text{ miles}
\]

4. **Total Time without Rest:**
The total running time before he rests is:
\[
\text{Total Time} = 5 \text{ minutes} + 7 \text{ minutes} = 12 \text{ minutes}
\]

5. **Finding Specific Interval:**
We want to find how long it takes him to reach a total of 1.5 miles.

- If we assume Christian jogs at a reasonable pace, let's set \( x = 0.1 \) miles per minute (about a 10-minute mile pace), which means:
\[
5 \text{ minutes} \Rightarrow 5 \times 0.1 = 0.5 \text{ miles}
\]

During this phase, he reaches 0.5 miles by the end of 5 minutes.

- After that, let's assume he speeds up to \( y = 0.15 \) miles per minute (about a 6.67-minute mile pace), leading to:
\[
7 \text{ minutes} \Rightarrow 7 \times 0.15 = 1.05 \text{ miles}
\]

Therefore, at the end of the first running segment (5 minutes), he has covered 0.5 miles.

By the end of the faster running segment (12 minutes total), he has run:
\[
0.5 + 1.05 = 1.55 \text{ miles}
\]

6. **Finding when he hits 1.5 miles:**
Since 1.5 miles is hit between the 5-minute mark and the 12-minute mark, we can calculate when during the faster pace period he reaches a total of 1.5 miles.

He needs to cover an additional:
\[
1.5 - 0.5 = 1 \text{ mile}
\]

At the speed during the faster pace, \( y = 0.15 \) miles per minute, the time \( t \) required to cover 1 mile is:
\[
t = \frac{1 \text{ mile}}{0.15 \text{ miles/min}} = \frac{1}{0.15} \approx 6.67 \text{ minutes}
\]


7. **Total time to reach 1.5 Miles:**
He would have reached 0.5 miles at 5 minutes and would take an extra 6.67 minutes to cover the remaining mile:

\[
5 + 6.67 \approx 11.67 \text{ minutes} \approx 11 \text{ minutes, } 40 \text{ seconds}
\]

Christian traveled 1.5 miles during the interval:
\[
\text{From } 5 \text{ minutes to approximately } 11.67 \text{ minutes.}
\]

Thus, expressing this interval:
\[
Christian traveled 1.5 miles from \textbf{5} \leq t \leq \textbf{11.67} \text{ (approximately 11 minutes, 40 seconds).}
\]