Speed and Distance Calculations Flashcards

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?

Answer: 30 minutes

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

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?

Answer: 3 minutes

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

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?

Answer: 360 knots

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

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?

Answer: 2,000 ft/min

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

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?

Answer: B arrives ~8 minutes earlier

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

An aircraft is 150 nm from an airport and must cross an intermediate fix that is 90 nm from the airport in 15 minutes. What speed must it maintain to the fix?

Answer: 240 knots

Distance to fix: 150 – 90 = 60 nm; 60 ÷ (15/60 hr) = 60 × 4 = 240 knots.

Two aircraft depart the same airport simultaneously. Aircraft A flies north at 400 knots and Aircraft B flies east at 300 knots. After 30 minutes, how far apart are the two aircraft?

Answer: 250 nm

A travels 200 nm north; B travels 150 nm east; separation = √(200² + 150²) = √62,500 = 250 nm.