Cities A, B, and C are located on equator, separated by equal distances.?

Question

Airplane departed from A at noon and arrived to B at noon.
Without delay airplane departed from B at noon and arrived to C at noon.
Without delay airplane departed from C at noon and arrived to A at noon of the next day.

Second identical airplane provided similar servce in opposite direction. Second airplane departed from A at noon and proceeded to C, and then to B.

What are arrival/departure times at cities C and B of the second airplane?

Answer 1

C: 4 am
B: 8 pm

The first plane flies west at one time zone per hour (fast plane!). It crosses the date line between C and A.

So it takes 8 hours to fly from A to B, & B to C & then C to A; circumnavigating the Earth in 24 hours (and at a distance of approximately 24,000 miles, a rate of 1000 miles per hour!)

At noon at A, it is 8pm previous day at C and 4am at B.

In 8 hours the second plane will land at C, so it will be 4am same day local time.

In 16 hours it will land at B at 4am + 16 = 8pm same day local time.

In 24 hours it will land back at A at noon the next day (of course, since our watch registered midnight a while ago, ) , and landing with the first plane.

Answer 2

In the frame of reference of the earth, the first plane moves with the sun. The second plane moves opposite to the sun, and takes 8 hours to get to C. At the start it was 16 before noon (8 pm previous day) at C. So it arrives at C @ 4am the same day.

When the plane is at A, it is 8 hours before noon (4am same day) at B. Then it takes 16 hours to get there, so it arrives at B at 8pm (on the same day).

Then it finally reaches back at A at noon (next day).

*EDIT*
The first 2 answerers assumed that the absolute speed of the plane in space is the same. Relative to the sun, the first plane is standing still while the earth rotates under it. But identical should mean relative to the earth, not the sun.

Answer 3

The first airplane traveled just fast enough to follow the apparent motion of the sun around the earth, making a complete trip around the earth in one day. If the airplane keeps traveling around the earth at the same speed, the local time for the plane will be perpetual noon or thereabouts.

There is insufficient information to determine the locations of cities B and C with respect to A so the problem cannot be answered as stated. It can be said, however, that the International Date Line lies between point C and point A and that the first plane travels in a generally westerly direction. It can also be said that the second plane generally travels east and also takes one day to make the trip around the earth. If points A and C are close enough together, it is possible that the second plane could land at point C almost by as much as a day before leaving point A.

Answer 4

noon; noon.

If it's equal distances between the cities, and airplane 1 arrived at them all at noon, why would it be any different for airplane 2?

Answer 5

sin regulation that shizz sin48.3/527= sinABC/319 sin^-a million= 319sin48.3/527 (im no longer doing this menial artwork for you) as quickly as u get ABC, do one hundred eighty- 40 8.3- ABC to locate the attitude of ACB (shall we call that perspective y) so which you get y, and placed it lower back into the sin regulation stuff sin48.3/527= sinACB/AB AB= 527sinACB/sin48.3 and ta- da you obtain AB!

Answer 6

at C noon of previous day
at B noon of the same day.