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3 years ago

14

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 veloci

ty 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

2 answers:

Slav-nsk [51]3 years ago

6 0

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

abruzzese [7]3 years ago

4 0

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

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