Question

Two airplanes leave an airport at the same time. the velocity of the first airplane is 730 m/h at the heading of 66.4, the velocity of the second is 550 m/h at the heading of 86

How far apart are they after 1.9 h?
answer in units of m.

Answer 1

Answer: The two airplanes are 590.9 meters apart after 1.9 hours.

Step-by-step explanation:

Step 1: Calculate the distance traveled by each airplane.

To find the distance traveled by each airplane, we can use the formula:

Distance = Velocity × Time

For the first airplane:

Distance₁ = 730 m/h × 1.9 h = 1387 m

For the second airplane:

Distance₂ = 550 m/h × 1.9 h = 1045 m

Step 2: Calculate the horizontal and vertical components of each airplane's distance.

For the first airplane:

x₁ = Distance₁ × cos(θ₁) = 1387m × cos(66.4°) ≈ 586.8 m

y₁ = Distance₁ × sin(θ₁) = 1387m × sin(66.4°) ≈ 1241.6 m

For the second airplane:

x₂ = Distance₂ × cos(θ₂) = 1045 × cos(86°) ≈ 30.6 m

y₂ = Distance₂ × sin(θ₂) = 1045 × sin(86°) ≈ 1043.4 m

Step 3: Calculate the difference in horizontal and vertical components.

Δx = x₁ - x₂ = 586.8 - 30.6 = 556.2 m

Δy = y₁ - y₂ = 1241.6 - 1043.4 = 198.2 m

Step 4: Calculate the distance between the two airplanes.

We can use the Pythagorean theorem to find the distance between the two airplanes:

Distance = √(Δx² + Δy²) = √(556.2² + 198.2²) ≈ 590.9 m