Aero Course ATPL Mach questions

[CamdenW]

Can someone help me solve this question from the aero course ATPL workbook 6th edition? Mostly just wondering what the formula to be used here is?

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!

[GooseDriver]

Skip it, the question is incomplete. Relation between Mach and true airspeed varies with altitude. Without altitude information the question cannot be solved.

If they gave you an altitude you would solve simply by punching numbers into a flight computer or referencing some graphs. If you don’t have any, then the answer is C.

[triplebarrel]

I used an E6B app. I went to "plan mach no." Enter the 0.83 at -45C, this gives you a TAS of 488.53
mentally you know it's westbound and winds are a 40-knot headwind but you could toss that in the calculator as well. In short, it was 448.53 GS

Then you go to the leg time function for leg time or you could do it by a normal calc= 150nm / 448.53knots = (answer) x 60 = roughly 20:00

**BTW you had a typo in your post. question says 150 nm, not 115nm

[Conflicting Traffic]

Relation between Mach and true airspeed varies with altitude.

The relationship between Mach and true airspeed varies with temperature, which is given.

V_mach = 661*sqrt(T/T_ref), where T and T_ref are in Kelvin. T_ref is 288K (15C), and T in this case is 228K (-45C).

The answer I get is 15.4 min for 115 NM, or 20 min for 150 NM.

[CamdenW]

Thank you guys!

Sorry about the typo, yup it was 150!

[CodyJohnston]

Thank you for helping me out as well. I appreciate you guys.

[bekean]

subway surfers What an insightful piece!

[canadianfly]

M .83 is roughly 8nm/min, that gives 19 mins for 150 nm.