To solve the problem of ranking the legs of the trip from lowest velocity to highest, follow these steps:
1. Convert the time for each leg from minutes to hours:
- Leg A: \( 10 \) minutes \( = \frac{10}{60} \) hours \( = 0.1667 \) hours (approximately)
- Leg B: \( 15 \) minutes \( = \frac{15}{60} \) hours \( = 0.25 \) hours
- Leg C: \( 12 \) minutes \( = \frac{12}{60} \) hours \( = 0.2 \) hours
- Leg D: \( 9 \) minutes \( = \frac{9}{60} \) hours \( = 0.15 \) hours
- Leg E: \( 14 \) minutes \( = \frac{14}{60} \) hours \( = 0.2333 \) hours (approximately)
This step is crucial because velocities need to be calculated with the same units for both distance and time.
2. Calculate the velocities for each leg (using the formula velocity = distance/time):
- Leg A: \( \frac{15 \, \text{km}}{0.1667 \, \text{hours}} \approx 90.0 \, \text{km/h} \)
- Leg B: \( \frac{20 \, \text{km}}{0.25 \, \text{hours}} = 80.0 \, \text{km/h} \)
- Leg C: \( \frac{24 \, \text{km}}{0.2 \, \text{hours}} = 120.0 \, \text{km/h} \)
- Leg D: \( \frac{36 \, \text{km}}{0.15 \, \text{hours}} = 240.0 \, \text{km/h} \)
- Leg E: \( \frac{14 \, \text{km}}{0.2333 \, \text{hours}} \approx 60.0 \, \text{km/h} \)
Now, we have the velocities for each leg:
- Leg A: \( 90.0 \, \text{km/h} \)
- Leg B: \( 80.0 \, \text{km/h} \)
- Leg C: \( 120.0 \, \text{km/h} \)
- Leg D: \( 240.0 \, \text{km/h} \)
- Leg E: \( 60.0 \, \text{km/h} \)
3. Sort the legs by their velocities from lowest to highest:
- Leg E: \( 60.0 \, \text{km/h} \)
- Leg B: \( 80.0 \, \text{km/h} \)
- Leg A: \( 90.0 \, \text{km/h} \)
- Leg C: \( 120.0 \, \text{km/h} \)
- Leg D: \( 240.0 \, \text{km/h} \)
Therefore, the legs of the trip, sorted from lowest velocity to highest velocity, are:
1. Leg E
2. Leg B
3. Leg A
4. Leg C
5. Leg D
