A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of `2.40xx10^(2)m//s`. The plane is to make a round-trip by heading due west for a certain distance, turning around, and then heading due east for the return trip. During the entire flight, however, the plane encounters a 57.8 m/s wind from the jet stream, which blows from west to east. What is the maximum distance that the plane can travel due west and just be able to return home?

Options

A. 1870 km
B. 2160 km
C. 2440 km
D. 1950 km

Answer

To solve the problem, we need to determine the maximum distance the jetliner can travel due west before turning around and returning home, considering the wind's effect on its speed.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Time of flight (T) = 6 hours
- Speed of the jetliner without wind (V) = \(2.40 \times 10^2 \, \text{m/s} = 240 \, \text{m/s}\)
- Speed of the wind (W) = 57.8 m/s (from west to east)

2. **Calculate the Effective Speeds:**
- When flying west (against the wind):
\[ V_{\text{west}} = V - W = 240 \, \text{m/s} - 57.8 \, \text{m/s} = 182.2 \, \text{m/s} \]
- When returning east (with the wind):
\[ V_{\text{east}} = V + W = 240 \, \text{m/s} + 57.8 \, \text{m/s} = 297.8 \, \text{m/s} \]

3. **Convert Time to Seconds:**
- Convert 6 hours into seconds:
\[ T = 6 \, \text{hours} \times 60 \, \text{minutes/hour} \times 60 \, \text{seconds/minute} = 21600 \, \text{seconds} \]

4. **Set Up the Equation for Total Time:**
- Let \(D\) be the distance traveled due west. The time taken to travel west and return east can be expressed as:
\[ \text{Total Time} = \frac{D}{V_{\text{west}}} + \frac{D}{V_{\text{east}}} \]
- Therefore:
\[ \frac{D}{182.2} + \frac{D}{297.8} = 21600 \]

5. **Combine the Fractions:**
- Find a common denominator and combine:
\[ D \left(\frac{1}{182.2} + \frac{1}{297.8}\right) = 21600 \]
- The common denominator is \(182.2 \times 297.8\):
\[ D \left(\frac{297.8 + 182.2}{182.2 \times 297.8}\right) = 21600 \]
- Simplifying gives:
\[ D \left(\frac{480}{182.2 \times 297.8}\right) = 21600 \]

6. **Solve for D:**
- Rearranging gives:
\[ D = 21600 \times \frac{182.2 \times 297.8}{480} \]
- Calculate \(D\):
\[ D = 21600 \times \frac{182.2 \times 297.8}{480} \approx 2440.62 \, \text{meters} \]

7. **Convert Distance to Kilometers:**
- Convert meters to kilometers:
\[ D \approx 2.44 \, \text{km} \]

### Final Answer:
The maximum distance that the plane can travel due west and just be able to return home is approximately **2.44 km**.