Ask a New Question
Menu

Use the image to answer the question.

A piecewise linear function consisting of five parts is plotted in quadrant 1 of a coordinate plane.

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

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
5 less-than-or-equal-to s less than 12

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

1 answer

To determine when Kaitlyn reaches 1.25 miles during her workout, let's break down her activity:

  1. Jogging at a slow rate for 5 minutes: Let's assume she is jogging at a slower constant pace. Typically, a slow jog might be around 4 miles per hour. In 5 minutes (which is 1/12 of an hour), she would cover: \[ \text{Distance} = \text{speed} \times \text{time} = 4 , \text{mph} \times \frac{1}{12} , \text{hour} \approx 0.33 \text{ miles} \]

  2. Faster pace for 7 minutes: Now let's assume her faster pace is about 6 miles per hour. In 7 minutes (which is 7/60 hours): \[ \text{Distance} = 6 , \text{mph} \times \frac{7}{60} , \text{hour} \approx 0.7 \text{ miles} \]

    After the jog and the faster run: \[ 0.33 , \text{miles} + 0.7 , \text{miles} = 1.03 , \text{miles} \]

  3. Taking a 3-minute break: This does not contribute to distance.

  4. Running for another 7.5 minutes: Suppose she maintains a pace of 6 miles per hour. In 7.5 minutes (7.5/60 hours): \[ \text{Distance} = 6 , \text{mph} \times \frac{7.5}{60} , \text{hour} = 0.75 \text{ miles} \]

Adding this to the previous total: \[ 1.03 , \text{miles} + 0.75 , \text{miles} = 1.78 , \text{miles} \]

To find when she reaches 1.25 miles:

  • In the 5 minutes of jogging, she reaches 0.33 miles.
  • In the next 7 minutes of faster running, she reaches 1.03 miles.
  • During the last 7.5 minutes of running, she must reach 1.25 miles somewhere between 1.03 miles and 1.78 miles.

To find out specifically when between the finish of her faster pace and the end of the last 7.5 minutes:

  • Starting distance when she begins the last 7.5 minutes is 1.03 miles.
  • She reaches 1.25 miles: \[ 1.25 - 1.03 = 0.22 \text{ miles to go} \]

At a speed of 6 mph: \[ \text{Time to cover 0.22 miles} = \frac{0.22}{6} \approx 0.0367 \text{ hours} \approx 2.2 \text{ minutes} \]

She starts this segment after \(5 + 7 + 3 = 15\) minutes (end of the break) and reaches 1.25 miles approximately 2.2 minutes later: \[ 15 \text{ minutes} + 2.2 \text{ minutes} = 17.2 \text{ minutes} \]

So, she reaches 1.25 miles between 15 minutes and 22.5 minutes (assuming the last segment of time adds to a total of 25 min for reaching 2 miles).

Thus, the answer is: 15 ≤ s ≤ 22.5.