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 please

Refer to the diagram shown below.

In 2.4 hours, the distance traveled by the first airplane heading a 51.3° at 750 mph is
a = 750*2.4 = 1800 miles.

The second airplane travels
b = 620*2.4 = 1488 mile

The angle between the two airplanes is
163° - 51.3° = 111.7°

Let c = the distance between the two airplanes after 2.4 hours.
From the Law of Cosines, obtain
c² = a² + b² - 2ab cos(111.7°)
= 3.24 x 10⁶ + 2.2141 x 10⁶
c = 2335.41 miles

Answer: 2335.4 miles

Edufirst Edufirst · Answer 2 · 2018-08-03T00:22:58+02:00

Answer: 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