Get the most out of your questions with IDNLearn.com's extensive resources. Our experts are available to provide in-depth and trustworthy answers to any questions you may have.
Sagot :
To solve this problem, we need to determine the average velocity for each leg of Adam's bus trip. The average velocity is found by dividing the distance traveled by the time taken for each leg of the trip. The formula for average velocity is:
[tex]\[ \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} \][/tex]
Let's calculate the average velocity for each leg (A, B, C, D, E):
Leg A:
- Distance = 18 km
- Time = 9 min
[tex]\[ \text{Average Velocity}_{\text{A}} = \frac{18 \text{ km}}{9 \text{ min}} = 2 \text{ km/min} \][/tex]
Leg B:
- Distance = 25 km
- Time = 15 min
[tex]\[ \text{Average Velocity}_{\text{B}} = \frac{25 \text{ km}}{15 \text{ min}} \approx 1.67 \text{ km/min} \][/tex]
Leg C:
- Distance = 24 km
- Time = 8 min
[tex]\[ \text{Average Velocity}_{\text{C}} = \frac{24 \text{ km}}{8 \text{ min}} = 3 \text{ km/min} \][/tex]
Leg D:
- Distance = 48 km
- Time = 12 min
[tex]\[ \text{Average Velocity}_{\text{D}} = \frac{48 \text{ km}}{12 \text{ min}} = 4 \text{ km/min} \][/tex]
Leg E:
- Distance = 15 km
- Time = 7 min
[tex]\[ \text{Average Velocity}_{\text{E}} = \frac{15 \text{ km}}{7 \text{ min}} \approx 2.14 \text{ km/min} \][/tex]
Next, we will arrange the legs from highest average velocity to lowest average velocity:
1. Leg D: [tex]\(4 \text{ km/min}\)[/tex]
2. Leg C: [tex]\(3 \text{ km/min}\)[/tex]
3. Leg E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
4. Leg A: [tex]\(2 \text{ km/min}\)[/tex]
5. Leg B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
Therefore, the legs of the trip arranged from highest velocity to lowest are:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
So, the average velocities and the ordered legs of the trip are:
Average Velocities:
- A: [tex]\(2 \text{ km/min}\)[/tex]
- B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
- C: [tex]\(3 \text{ km/min}\)[/tex]
- D: [tex]\(4 \text{ km/min}\)[/tex]
- E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
Order from highest to lowest velocity:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
[tex]\[ \text{Average Velocity} = \frac{\text{Distance}}{\text{Time}} \][/tex]
Let's calculate the average velocity for each leg (A, B, C, D, E):
Leg A:
- Distance = 18 km
- Time = 9 min
[tex]\[ \text{Average Velocity}_{\text{A}} = \frac{18 \text{ km}}{9 \text{ min}} = 2 \text{ km/min} \][/tex]
Leg B:
- Distance = 25 km
- Time = 15 min
[tex]\[ \text{Average Velocity}_{\text{B}} = \frac{25 \text{ km}}{15 \text{ min}} \approx 1.67 \text{ km/min} \][/tex]
Leg C:
- Distance = 24 km
- Time = 8 min
[tex]\[ \text{Average Velocity}_{\text{C}} = \frac{24 \text{ km}}{8 \text{ min}} = 3 \text{ km/min} \][/tex]
Leg D:
- Distance = 48 km
- Time = 12 min
[tex]\[ \text{Average Velocity}_{\text{D}} = \frac{48 \text{ km}}{12 \text{ min}} = 4 \text{ km/min} \][/tex]
Leg E:
- Distance = 15 km
- Time = 7 min
[tex]\[ \text{Average Velocity}_{\text{E}} = \frac{15 \text{ km}}{7 \text{ min}} \approx 2.14 \text{ km/min} \][/tex]
Next, we will arrange the legs from highest average velocity to lowest average velocity:
1. Leg D: [tex]\(4 \text{ km/min}\)[/tex]
2. Leg C: [tex]\(3 \text{ km/min}\)[/tex]
3. Leg E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
4. Leg A: [tex]\(2 \text{ km/min}\)[/tex]
5. Leg B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
Therefore, the legs of the trip arranged from highest velocity to lowest are:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
So, the average velocities and the ordered legs of the trip are:
Average Velocities:
- A: [tex]\(2 \text{ km/min}\)[/tex]
- B: [tex]\(\approx 1.67 \text{ km/min}\)[/tex]
- C: [tex]\(3 \text{ km/min}\)[/tex]
- D: [tex]\(4 \text{ km/min}\)[/tex]
- E: [tex]\(\approx 2.14 \text{ km/min}\)[/tex]
Order from highest to lowest velocity:
[tex]\[ \text{D}, \text{C}, \text{E}, \text{A}, \text{B} \][/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Accurate answers are just a click away at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.