On Friday, Kaitlyn went to the park to work out. She started with a slow jog at a constant rate for five minutes. She then increased her speed to a faster, constant pace for seven minutes. She took a three-minute break, then ran for another 7.5 minutes at a constant pace to finish her two-mile run. She rested for the remainder of the time. During what domain interval will she reach 1.25 miles?

Question

(1 point)
Responses

22.5≤s≤25
22.5 less-than-or-equal-to s less-than-or-equal-to 25

15≤s≤22.5
15 less-than-or-equal-to s less-than-or-equal-to 22.5

12≤s≤15
12 less-than-or-equal-to s less-than-or-equal-to 15

5≤s<12

Answer 1

To determine the time interval during which Kaitlyn reaches 1.25 miles, let's break down her workout step by step.

1. Slow Jog:

   • Duration: 5 minutes
   • Assume slow jog pace is \( r_1 \) mph.
   • Distance covered: \( \frac{r_1 \times 5}{60} \) miles.

2. Faster Pace:

   • Duration: 7 minutes
   • Assume faster pace is \( r_2 \) mph, where \( r_2 > r_1 \).
   • Distance covered: \( \frac{r_2 \times 7}{60} \) miles.

3. Break:

   • Duration: 3 minutes (no distance covered).

4. Final Run:

   • Duration: 7.5 minutes.
   • Assume this is her running pace before finishing the 2-mile total.
   • Distance covered: \( \frac{r_3 \times 7.5}{60} \) miles.

Let’s assume Kaitlyn’s speeds are reasonable for running, but we can begin solving without specific numerical values for each pace.

Kaitlyn runs a total distance of 2 miles. We will find her interim distance to 1.25 miles during each segment of time.

Time and Distance Breakdown

1. First Segment (5 minutes at slow pace):

   • Let’s assume she jogs at a speed (for example, 4 mph).
   • Distance covered in first segment = \( \frac{4 \text{ mph} \times 5 \text{ minutes}}{60} = \frac{4 \times 5}{60} = \frac{20}{60} = \frac{1}{3} \text{ miles} \approx 0.33 \text{ miles} \).

2. Second Segment (7 minutes at faster pace):

   • Let’s assume she runs at a speed (for example, 6 mph).
   • Distance covered in second segment = \( \frac{6 \text{ mph} \times 7 \text{ minutes}}{60} = \frac{6 \times 7}{60} = \frac{42}{60} = 0.7 \text{ miles} \).

   After these two segments, the total distance covered is:

   • \( 0.33 + 0.7 = 1.03 \text{ miles} \).

3. Break (3 minutes):

   • No additional distance.

4. Final Run (7.5 minutes):

   • Assume she runs at a higher pace (say 8 mph).
   • Distance = \( \frac{8 \text{ mph} \times 7.5 \text{ minutes}}{60} = \frac{8 \times 7.5}{60} = \frac{60}{60} = 1 \text{ mile} \).

Total Distance Calculation

After the initial jog and faster pace, she reaches 1.03 miles. Then she runs an additional mile during her last segment:

• Total distance = \( 1.03 + 1 = 2.03 \) miles, but we only need to find out when she reached 1.25 miles in the course of her run.

1. From 0 to 5 minutes = 0.33 miles.
2. From 5 to 12 minutes (7 minutes for intense pace) = she would reach 1.03 miles in this segment.
3. She now needs just \( 1.25 - 1.03 = 0.22 \) miles from the last segment.

Calculating back from when she reaches 1.03 to when she will reach 1.25:

• At 8 mph: \[ \text{Time to cover 0.22 miles} = \frac{0.22 \text{ miles}}{8 \text{ mph}} \times 60 \text{ minutes/hour} \approx 1.65 \text{ minutes}. \]

Adding this to the earlier time:

• End time for reaching 1.25 miles will be \( 12 + 1.65 \approx 13.65 \) min.
• Hence 13 and a fraction falls within the 15 minute mark as Kaitlyn tracks her time.

Conclusion

1.25 miles is reached before she finishes her last segment (likely within the rest of 15 minutes).

Thus the likely interval before going into a fast finish would be in the range of \( 12 \leq s < 15 \).

Final answer: 12 ≤ s < 15.