Very high difficulty?

Question

The hands of clock A point to 6 o clock. At the moment past 6 the minute hand traveling downwards will have a increase of speed of x% every minute due to gravity. At 6:15 the speed of the minute minute hand will be of constant speed until 6:30. Since the hour hand at 6 is traveling upwards against gravity, its motion will be slowed by y% every hour until 9:00. After 9 the hand will have a constant speed until midnight. If clock B is place in a weightless enviroment, the minute and hour hand will perfectly match each other some time after 6 o clock. The hands of clock A will do the same, although at a different time. In terms of x and y, the absolute difference in times when this will happen is what?

Answer 1

the absolute difference is
(360/11)-(180*((2-x%)/(11-y%)))
when there is not gravity, the needle meet at 360/11 min from six.
you add or substract the difference(absolute value) when there is gravity.

Answer 2

Hi. Still need to know if the clock has analog motion or digital motion. This is necessary because the answer is either accurate but in an increment (digital type movement) or involves a calculus equation (analog type movement). My first answer follows.

"There have been similar questions and the same problem applies. The second hand can either sweep (analog style movement) or move in one second intervals (digital style movement). They still both have minute and hour hands, but if the movement is not perfectly smooth then the answer will be different.

Assuming analog type, the problem boils down to minute hand acceleration held back by hour hand deceleration. Both occur during the first 15 minutes after 6:00. The solution will contain these ratios."