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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?

(1 point)
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1 answer

To determine the time interval during which Christian traveled 1.5 miles, we need to find his jogging speed and the speed during his faster pace.

Let's assume:

  • Christian's jogging speed is \( v_1 \) (in miles per minute).
  • His faster pace speed is \( v_2 \) (in miles per minute).

1. Determine Distances

  1. Jogging for 5 minutes:
    Distance covered during jogging = \( 5 \times v_1 \) miles.

  2. Faster pace for 7 minutes:
    Distance covered during faster pace = \( 7 \times v_2 \) miles.

2. Total Distance After the First Interval

  • Total distance after jogging + faster pace =
    \[ \text{Total Distance} = 5v_1 + 7v_2 \text{ miles} \]

3. Analyzing the Time Intervals

  • He jogs for 5 minutes, then runs at a faster pace for 7 minutes, followed by a 4-minute rest.
  • Total time before rest = 5 minutes + 7 minutes = 12 minutes.

4. Calculate the Time Interval for 1.5 Miles

  1. If \( 5v_1 + 7v_2 < 1.5 \) miles:

    • Christian has not yet traveled 1.5 miles after 12 minutes.

  2. If \( 5v_1 + 7v_2 \geq 1.5 \) miles:

    • Christian has traveled 1.5 miles within the initial 12 minutes.

5. Determining Specific Speeds

Let's say Christian jogs at a pace of 6 minutes per mile while running faster at 3 minutes per mile.

  • Jogging speed:
    \( v_1 = \frac{1 \text{ mile}}{6 \text{ minutes}} = \frac{1}{6} \) miles per minute.

  • Faster pace speed:
    \( v_2 = \frac{1 \text{ mile}}{3 \text{ minutes}} = \frac{1}{3} \) miles per minute.

6. Calculate Distances for Speeds

  1. Jogging for 5 minutes:
    Distance = \( 5 \times \frac{1}{6} = \frac{5}{6} \) miles.

  2. Faster pace for 7 minutes:
    Distance = \( 7 \times \frac{1}{3} = \frac{7}{3} \) miles.

Total Distance After Both Activities

  • Total Distance =
    \[ \frac{5}{6} + \frac{7}{3} = \frac{5}{6} + \frac{14}{6} = \frac{19}{6} \text{ miles} \approx 3.17 \text{ miles} \]

Conclusion

Since he travels at least 1.5 miles within the first 12 minutes (in fact he covers more), the time interval during which Christian travels 1.5 miles is in the first 12 minutes.

Thus, the answer is: Christian traveled 1.5 miles from \( 0 \leq t \leq 12 \).