Determining flight time with Mach number ATPL
 

Can someone help me solve this question from the aero course ATPL workbook 6th edition?

The following wind reports relate to a westbound flight of a small business jet aircraft:
274045
How long would it take for this aircraft to cover a distance of 115 NM if it cruised at Mach .83?

31 mins
25 mins
20 mins
17mins

Any help would be greatly appreciated!

You need to know the air temperature at the cruising altitude. From that you can convert Mach to TAS.

Then factor wind, and it should just drop out.

G

What Ghengis said for ATPL exams, or if you’re sat in the cockpit without a TAS readout…….Mach 0.83 equals roughly 8.3 miles per minute, covert to TAS, knock off the 45kts headwind and you’ve got a ball park figure. Useful for gross error checking.

17min

Impressive! Without cruising altitude / temperature, or aircraft heading, I have no idea how you worked that out. Clearly much better than I am. :8:D:O

G

Genghis (as always) has hit the nail on the head. You don't have enough information. But, having used Aerostudies myself, you seem to have left out a lot of the information that is provided in the book to help you answer the question. A similarly worded question in the 4th edition provides the cruise altitude and temperature in addition to the wind.

Let me help. generally a westbound flight is heading 270 ;)

Sat many hours looking at .83 mid Atlantic...at normal business jet altitude... Approx 480kts
-45= 7 'n a bit nm per min= 17ish mins. 😆
Not sure I could get through ATPL now!
Good luck with it.

Ah bother, missed that magic word.

Still need temperature at altitude mind you.

G

In the absence of any additional information, I think it would be reasonable to assume the small business jet is flying in an ISA temperature tropopause, which gives 17 minutes.