An aircraft is flying from 'A' to 'B' (distance 220 NM) at an average ground speed (GS) of 120 kt. It departs 'A' at 1200 UTC. After 70 NM along the course from 'A', the aircraft is 5 min ahead of the planned schedule. Using the actual GS, what is the revised estimated time of arrival (ETA) at B?
For this question, firstly determine the planned flight time and, through the actual Groundspeed, calculate the actual flight time:
1. According to the planned data, the aircraft would cover 70 NM in: Distance / Planned GS = 70 NM / 120 kt = 0.58 hr or (0.58x60) 35 min.
2. The aircraft is 5 min ahead of the planned schedule, so the 70 NM was actually covered in: 35 min - 5 min = 30 min or 0.5 hr.
3. Thus, the actual GS is: Distance / Actual Time = 70 NM / 0.5 hr = 140 kt.
4. So, the total distance from A to B will be actually covered in: Distance / Actual GS = 220 NM / 140 kt = 1.57 hr or 1 hr and (0.57x60) 35 min approximately.
5. Therefore, the revised estimated time of arrival (ETA) will be: 12:00 UTC + 01:35 = 13:35 UTC.
Your Notes (not visible to others)
This question has appeared on the real examination, you can find the related countries below.
-
Austro Control1
Use Keyboard Shortcuts
