IDNLearn.com provides a reliable platform for finding accurate and timely answers. Discover in-depth and reliable answers to all your questions from our knowledgeable community members who are always ready to assist.
Sagot :
To solve the problem of determining how long Xóchitl needs to pedal to catch up to Cowessess, let's break it down step-by-step.
1. Understanding the Initial Distance Covered by Cowessess:
- Cowessess starts pedaling 12 minutes before Xóchitl and maintains a constant rate of [tex]\(10 \frac{ km }{ h }\)[/tex].
- First, convert the 12 minutes into hours:
[tex]\[ \text{Time}_\text{Cowessess} = \frac{12 \text{ minutes}}{60 \text{ minutes/hour}} = 0.2 \text{ hours} \][/tex]
- Determine how far Cowessess has pedaled during these 12 minutes using the formula for distance [tex]\((\text{Distance} = \text{Rate} \times \text{Time})\)[/tex]:
[tex]\[ \text{Distance}_\text{Cowessess} = 10 \frac{ km }{ h } \times 0.2 \text{ hours} = 2 \text{ km} \][/tex]
2. Setting Up the Equation for Xóchitl to Catch Up:
- Xóchitl starts pedaling at a rate of [tex]\(18 \frac{ km }{ h }\)[/tex].
- Both will travel the same distance to meet, so let’s denote the additional time Xóchitl needs to pedal as [tex]\( \text{time}_x \)[/tex].
- During this time, Xóchitl covers:
[tex]\[ \text{Distance}_\text{Xóchitl} = 18 \frac{ km }{ h } \times \text{time}_x \][/tex]
- During the same time, Cowessess continues to pedal at her constant rate, covering:
[tex]\[ \text{Distance}_\text{Cowessess}_\text{additional} = 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
- The total distance Cowessess would have traveled when Xóchitl catches up is the initial distance plus the additional distance:
[tex]\[ \text{Total Distance}_\text{Cowessess} = 2 \text{ km} + 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
3. Equating the Distances:
- To find when Xóchitl catches up, set the distance covered by Xóchitl equal to the total distance covered by Cowessess:
[tex]\[ 18 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} + 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
4. Solving for Time:
- Isolate [tex]\(\text{time}_x\)[/tex] by bringing the terms involving [tex]\(\text{time}_x\)[/tex] to one side:
[tex]\[ 18 \frac{ km }{ h } \times \text{time}_x - 10 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} \][/tex]
- Simplify:
[tex]\[ (18 \frac{ km }{ h } - 10 \frac{ km }{ h }) \times \text{time}_x = 2 \text{ km} \][/tex]
[tex]\[ 8 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} \][/tex]
- Divide both sides by [tex]\(8 \frac{ km }{ h }\)[/tex]:
[tex]\[ \text{time}_x = \frac{2 \text{ km}}{8 \frac{ km }{ h }} = 0.25 \text{ hours} \][/tex]
Therefore, Xóchitl would have to pedal for 0.25 hours (or 15 minutes) to catch up to Cowessess's distance.
1. Understanding the Initial Distance Covered by Cowessess:
- Cowessess starts pedaling 12 minutes before Xóchitl and maintains a constant rate of [tex]\(10 \frac{ km }{ h }\)[/tex].
- First, convert the 12 minutes into hours:
[tex]\[ \text{Time}_\text{Cowessess} = \frac{12 \text{ minutes}}{60 \text{ minutes/hour}} = 0.2 \text{ hours} \][/tex]
- Determine how far Cowessess has pedaled during these 12 minutes using the formula for distance [tex]\((\text{Distance} = \text{Rate} \times \text{Time})\)[/tex]:
[tex]\[ \text{Distance}_\text{Cowessess} = 10 \frac{ km }{ h } \times 0.2 \text{ hours} = 2 \text{ km} \][/tex]
2. Setting Up the Equation for Xóchitl to Catch Up:
- Xóchitl starts pedaling at a rate of [tex]\(18 \frac{ km }{ h }\)[/tex].
- Both will travel the same distance to meet, so let’s denote the additional time Xóchitl needs to pedal as [tex]\( \text{time}_x \)[/tex].
- During this time, Xóchitl covers:
[tex]\[ \text{Distance}_\text{Xóchitl} = 18 \frac{ km }{ h } \times \text{time}_x \][/tex]
- During the same time, Cowessess continues to pedal at her constant rate, covering:
[tex]\[ \text{Distance}_\text{Cowessess}_\text{additional} = 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
- The total distance Cowessess would have traveled when Xóchitl catches up is the initial distance plus the additional distance:
[tex]\[ \text{Total Distance}_\text{Cowessess} = 2 \text{ km} + 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
3. Equating the Distances:
- To find when Xóchitl catches up, set the distance covered by Xóchitl equal to the total distance covered by Cowessess:
[tex]\[ 18 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} + 10 \frac{ km }{ h } \times \text{time}_x \][/tex]
4. Solving for Time:
- Isolate [tex]\(\text{time}_x\)[/tex] by bringing the terms involving [tex]\(\text{time}_x\)[/tex] to one side:
[tex]\[ 18 \frac{ km }{ h } \times \text{time}_x - 10 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} \][/tex]
- Simplify:
[tex]\[ (18 \frac{ km }{ h } - 10 \frac{ km }{ h }) \times \text{time}_x = 2 \text{ km} \][/tex]
[tex]\[ 8 \frac{ km }{ h } \times \text{time}_x = 2 \text{ km} \][/tex]
- Divide both sides by [tex]\(8 \frac{ km }{ h }\)[/tex]:
[tex]\[ \text{time}_x = \frac{2 \text{ km}}{8 \frac{ km }{ h }} = 0.25 \text{ hours} \][/tex]
Therefore, Xóchitl would have to pedal for 0.25 hours (or 15 minutes) to catch up to Cowessess's distance.
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.