How Does Wind Affect the Maximum Roundtrip Distance of a Jetliner?

[pippintook]

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of 240 m/s. The plane is to make a roundtrip 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?

I've already calculated that the velocity of the plane on the way there is 182.2 m/s and on the way back is 297.8 m/s. I'm just not sure how to determine how to divide the distance. I tried using 3 hours there, 3 hours back, but the distances were uneven. Help! Is there a formula (or formulas) that I can use?

 

[LowlyPion]

A jetliner can fly 6.00 hours on a full load of fuel. Without any wind it flies at a speed of 240 m/s. The plane is to make a roundtrip 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?

I've already calculated that the velocity of the plane on the way there is 182.2 m/s and on the way back is 297.8 m/s. I'm just not sure how to determine how to divide the distance. I tried using 3 hours there, 3 hours back, but the distances were uneven. Help! Is there a formula (or formulas) that I can use?

You have 297.8 * x hours = 182.2 * y hours and you know that x + y = 6.

2 Equations, 2 unknowns

 

[thomate1]

I would approach this problem by first understanding the concept of relative velocities. In this case, the plane's velocity relative to the ground is affected by the wind, resulting in different velocities on the way there and back. The key to solving this problem is to find the point where the total distance traveled by the plane is equal to the total distance traveled by the wind, as this would mean the plane has reached its maximum distance and can still return home.

To find this point, we can use the formula d = vt, where d is the distance, v is the velocity, and t is the time. We know that the total time for the roundtrip is 6 hours, and we can set up the following equations:

Distance traveled by the plane on the way there: d1 = 182.2 m/s * 3 hours = 546.6 km
Distance traveled by the plane on the way back: d2 = 297.8 m/s * 3 hours = 893.4 km
Distance traveled by the wind: d3 = 57.8 m/s * 6 hours = 346.8 km

Now, we can set up the equation d1 + d2 = d3, as the total distance traveled by the plane (d1 + d2) should be equal to the total distance traveled by the wind (d3). Substituting the values from above, we get:

546.6 km + 893.4 km = 346.8 km
1440 km = 346.8 km

This equation is obviously not true, so we know that our initial assumption of 3 hours for each leg of the trip is incorrect. We need to adjust the time in order for the equation to be balanced. Since the plane is traveling with the wind on the way back, it will cover more distance in the same amount of time. This means that the time for the return trip needs to be shorter than 3 hours, while the time for the trip there needs to be longer than 3 hours.

We can approach this problem by trial and error, or we can use the formula t = d/v to calculate the time for each leg of the trip. Using this formula, we can rearrange it to t = d/v and solve for d, which gives us:

Distance traveled by the plane on the way there: d1 = vt = 182.2 m/s * t