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

1 answer

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