ATSA Speed and Distance Calculations 2

Question 1

Aircraft A (ahead) is traveling at 300 knots. Aircraft B is 30 nm behind on the same route at 360 knots. How many minutes before B catches A?

Accepted Answer

30 minutes

Answer

Closure rate = 360 – 300 = 60 knots; 30 nm ÷ 60 knots × 60 = 30 minutes.

Question 2

Two aircraft are converging with a closure rate of 300 knots and are currently 20 nm apart. Minimum radar separation is 5 nm. How many minutes remain before minimum separation is reached?

Accepted Answer

3 minutes

Answer

The aircraft must close 15 nm (20 – 5 minimum) at 300 knots: 15 ÷ 300 × 60 = 3 minutes.

Question 3

Aircraft A is 100 nm from a fix flying at 500 knots. Aircraft B must arrive at the same fix exactly 3 minutes after A. B is 90 nm from the fix. What speed must B maintain?

Accepted Answer

360 knots

Answer

A arrives in 12 min; B must arrive in 15 min = 0.25 hr; 90 nm ÷ 0.25 hr = 360 knots.

Question 4

An aircraft at FL360 must cross a fix at FL180. The fix is 45 nm away and the aircraft is flying at 300 knots. What descent rate is required?

Accepted Answer

2,000 ft/min

Answer

Time to fix: 45 ÷ 300 × 60 = 9 min; altitude change: 36,000 – 18,000 = 18,000 ft; 18,000 ÷ 9 = 2,000 ft/min.

Question 5

Aircraft A is 120 nm from a fix at 400 knots. Aircraft B is 80 nm from the same fix at 500 knots. Which arrives first, and by approximately how many minutes?

Accepted Answer

B arrives ~8 minutes earlier

Answer

A: 120 ÷ 400 × 60 = 18 min; B: 80 ÷ 500 × 60 = 9.6 min; B arrives approximately 8.4 minutes sooner.

ATSA - Air Traffic Skills Assessment Practice Test

ATSA - Air Traffic Skills Assessment Practice Test

ATSA Speed and Distance Calculations 2