Question

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? (1 point) Responses 12≤s≤15 12 less-than-or-equal-to s less-than-or-equal-to 15 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 5≤s<12

Answer 1

To determine the time interval during which Kaitlyn reaches 1.25 miles during her two-mile run, we need to analyze her different paces:

1. Slow jog for 5 minutes: Assuming an average jogging speed of \( x \) miles per minute, in 5 minutes, Kaitlyn covers: \[ 5x \text{ miles} \]

2. Faster pace for 7 minutes: Assuming her faster jogging speed is \( y \) miles per minute, in 7 minutes, she covers: \[ 7y \text{ miles} \]

3. Break for 3 minutes: During this time, she covers no distance.

4. Running for 7.5 minutes: Again at her faster speed \( y \) miles per minute, in 7.5 minutes she covers: \[ 7.5y \text{ miles} \]

Now, let's analyze the total distance covered. The equation for her total run will look like this: \[ 5x + 7y + 7.5y = 2 \text{ miles} \] which simplifies to: \[ 5x + 14.5y = 2 \] To find when she reaches 1.25 miles, we break the run into parts:

1. After her slow jog of 5 minutes: \[ \text{Distance covered} = 5x \] We need to check when \( 5x \) is equal to or exceeds 1.25 miles. So: \[ x \geq \frac{1.25}{5} \Rightarrow x \geq 0.25 \text{ miles per minute} \]

2. Then, she transitions to her faster pace for 7 minutes. The distance covered during this phase is: \[ \text{Total distance after the jog} = 5x + 7y \] We need: \[ 5x + 7y \geq 1.25 \] If \( 5x = 1.25 \), all distance during the jog contributes sufficiently. If not, we see how far into the faster run she needs to go to reach 1.25 miles.

Next, we assume a reasonable rate for \( y \). If \( x \) is \( 0.25 \), and substituting into our total distance formula at the point of her faster run yields: \[ 5(0.25) + 7y = 2 \] which gives: \[ 1.25 + 7y = 2 \Rightarrow 7y = 0.75 \Rightarrow y \approx 0.107 \text{ miles per minute} \] If we're trying to keep values realistic, we can use rates \( x=0.25 \) and select \( y=0.5 \) for faster running.

Calculating now:

• After 5 min of slow jogging, she covers 1.25 miles.
• Faster pace must be analyzed:
• To get to 1.25, she can reach this either sooner in time if her speed increases or in the time allotted if left constant.

Calculating reasonable time intervals for the various speeds indicates:

Given options to determine the interval for when she reaches 1.25 miles precisely fits in the faster run segment, looking at time progression with \( s = time in minutes \):

1. Standout suggestion is 15 ≤ s ≤ 22.5 minutes at sequences transitioning making sense with respect to mileage accumulation.

Thus, the correct domain interval during which Kaitlyn reaches 1.25 miles would be: \[ \text{Answer: } 15 \leq s \leq 22.5 \]