Two airplanes leave an airport at the same time. the velocity of the first airplane is 750 m/h at a heading of 51.3°. the velocity of the second is 620 m/h at a heading of 163°. how far apart are they after 2.4 h? answer in units of m. can someone explain

2727m

Explanation:

1) Position of first airplane after 2.4 h

Constant velocity ⇒ V = d / t ⇒ d = Vt =

d = 750m/h × 2.4h = 1,800 m; 51.3°

2) Position of the second airplaine after 2.4 h

Constant velocity ⇒d = 620m/h × 2.4 h = 1,488 m; 163°

3) Angle between the airplains, α:

α = 163° - 51.3° = 111.7°

4) Cosine theorem

a² = b² + c² - 2bc cosine (α)

b = 1,800 m (calcualted in part 1)

c = 1,488m (calculated in part 2)

α = 111.7°

⇒ a² = (1,800m)² + (1,488m)² - 2(1,800m)(1,488m) cos(111.7°) = 7,434,803 m²

⇒ a = √(7,434,803 m²) = 2,726.7 m ≈ 2,727m

a is the distance that separate the two airplains after 2.4 h

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