Two planes approach each other head on. Each has a speed of 721km/h and they spot each other when they are initially 14.7km apart. How many seconds (s) do the pilots have to take evasive action?

The time the pilots have to take evasive action is approximately 36.699 seconds

Explanation:

The direction of the planes = Towards each other

The speed of each plane, v₁, v₂ = 721 km/h

The initial distance between the two planes, d = 14.7 km

The relative velocity of the two planes, v = v₁ + v₂

v = 721 km/h + 721 km/h = 1,442 km/h

The time it will take the planes to meet, t = d/v

∴ t = 14.7 km/(1,442 km/h) = (21/2,060) h × 60 min/hr × 60 s/min ≈ 36.699 s

The time it will take the planes to meet = The time the pilots have to take evasive action, t ≈ 36.699 seconds.

the answer would be a

the answer is 12 as i may remember

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