Question

a jetliner can fly 4.5 hours on a full load of fuel. without any wind it flies at a speed of 3.04 x 10^2 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 59.0 - m/s wind from the jet stream, which blows from west to east. what is the maximum distance (in kilometers) that the plane can travel due west and just be able to return home?

Explanation:

Step1: Find the effective speeds

The speed of the plane going west (against the wind) is $v_{west}=3.04\times 10^{2}- 59.0=245$ m/s. The speed of the plane going east (with the wind) is $v_{east}=3.04\times 10^{2}+59.0 = 363$ m/s.

Step2: Let the distance traveled west be $d$.

The time taken to travel west is $t_{west}=\frac{d}{v_{west}}$ and the time taken to travel east is $t_{east}=\frac{d}{v_{east}}$. The total time $t = t_{west}+t_{east}$, and $t = 4.5\times3600$ s (since $4.5$ hours = $4.5\times3600$ s). So, $\frac{d}{245}+\frac{d}{363}=4.5\times3600$.

Step3: Solve for $d$.

First, find a common - denominator: $\frac{363d + 245d}{245\times363}=4.5\times3600$. Then, $\frac{608d}{88935}=4.5\times3600$. Cross - multiply: $608d=4.5\times3600\times88935$. So, $d=\frac{4.5\times3600\times88935}{608}\approx2.37\times 10^{6}$ m.

Step4: Convert to kilometers.

Since $1$ km = $1000$ m, $d = 2370$ km.

Answer:

$2370$ km