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?

Question

(1 point)
Christian traveled 1.5 miles from
≤t≤
.

Answer 1

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

1. Assumptions about pace:

   • Let's assume Christian jogged at a pace of 6 miles per hour during the first 5 minutes.
   • For the second segment, let's assume his faster pace was 9 miles per hour for 7 minutes.

2. Calculating distance:

   • The distance covered while jogging for 5 minutes: \[ \text{Distance} = \text{Speed} \times \text{Time} = 6 , \text{miles/hour} \times \left(\frac{5}{60} , \text{hours}\right) = 0.5 , \text{miles} \]

   • The distance covered at the faster pace for 7 minutes: \[ \text{Distance} = 9 , \text{miles/hour} \times \left(\frac{7}{60} , \text{hours}\right) = 1.05 , \text{miles} \]

3. Finding total distance:

   • By the end of the jog and the faster pace: \[ \text{Total Distance} = 0.5 , \text{miles} + 1.05 , \text{miles} = 1.55 , \text{miles} \]

4. Determining the time intervals:

   • From 0 to 5 minutes: Christian travels 0.5 miles.
   • From 5 to 12 minutes (5 + 7): He travels an additional 1.05 miles, totaling 1.55 miles by the end of 12 minutes.

5. Finding the time interval for 1.5 miles:

   • He covers 1.5 miles between:
     • At 5 minutes he has traveled 0.5 miles
     • As he continues running, we can set up a proportion.

   The additional distance he needs to reach 1.5 miles is: \[ 1.5 , \text{miles} - 0.5 , \text{miles} = 1 , \text{mile} \]

   At his faster pace, it takes him: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{1 , \text{mile}}{9 , \text{miles/hour}} = \frac{1}{9} , \text{hours} \approx 6.67 , \text{minutes} \]

6. Calculating the interval:

   • Therefore, he travels 1.5 miles at approximately: \[ t \approx 5 , \text{minutes} + 6.67 , \text{minutes} = 11.67 , \text{minutes} \approx 11 , \text{minutes} , 40 , \text{seconds} \]

So, to summarize:

Christian traveled 1.5 miles from \(0 \leq t \leq 11.67\) minutes.

Final answer: Christian traveled 1.5 miles from \(5 , \text{minutes} \leq t \leq 11.67 , \text{minutes}\).