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3 years ago

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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 apar

t. How many seconds (s) do the pilots have to take evasive action?

1 answer:

Naya [18.7K]3 years ago

5 0

Answer:

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.

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