Question

A jetliner leaves san francisco for new york, 4600 km away. with a strong tailwind, its speed is 1100 km/hr. at the same time, a second jet leaves new york for san francisco. flying into the wind, it makes only 700 km/hr. when and where do the two planes pass?

Answer 1

First, make an illustration of the problem. In reality, the direction traveled by the planes is not linear. Strictly speaking, it should curve out following the shape of the sphere. But I suppose, we could do away with that because we are given with the quantity for distances. I'm just showing a guide to see where the planes would meet.

So, we know that the total distance is 4,600 km. It would consequently mean that x and y would equal to 4,600. The distances x and y could be determined by multiplying their respective speeds to the time. Since they meet at the same time, let's denote it as t.

1,100t + 700t = 4,600
Solving for t,
1,800t = 4600
t = 4600/1800
t = 2.556 hours

So, we know that the two planes would meet after 2.56 hours from the time they leave the airports. Next, we have to know where they meet exactly.

1,100(2.556) = 2,812 km
700(2.556) = 1,788 km

Therefore, they meet at approximately 2,812 km from San Francisco airport, or 1,788 km from New York airport after 2.56 hours since they left their respective ports.