Question

Two aircraft leave an airfield at the same time. One travels due south at an average velocity of 85 m s 1

and the second travels due east at an average velocity of 71 ms −1
. Calculate the distance between the planes after 6 hours. Give your answer in km, to the nearest km. km

Answer 1

Answer: 2392 kilometers.

Step-by-step explanation:

To find the distance between the two aircraft after 6 hours, we can use the Pythagorean theorem because they are moving at right angles to each other (one due south and the other due east). Here's how to calculate it:

First, calculate the distance traveled by each aircraft in 6 hours.

1. Aircraft traveling south:

Distance = Velocity × Time

Distance = 85 m/s × (6 hours × 3600 seconds/hour)

Distance = 85 m/s × 21600 s

Distance = 1836000 meters

2. Aircraft traveling east:

Distance = Velocity × Time

Distance = 71 m/s × (6 hours × 3600 seconds/hour)

Distance = 71 m/s × 21600 s

Distance = 1533600 meters

Now, we have the two sides of a right triangle:

- The southbound distance is 1836000 meters.

- The eastbound distance is 1533600 meters.

We can use the Pythagorean theorem to find the hypotenuse (distance between the aircraft):

Distance = √(southbound distance² + eastbound distance²)

Distance = √((1836000 m)² + (1533600 m)²)

Distance ≈ √(3376416000000 m² + 2352089600000 m²)

Distance ≈ √(5728505600000 m²)

Distance ≈ 2391760.78 meters

Now, let's convert the distance to kilometers:

Distance ≈ 2391.76 kilometers

Rounding to the nearest kilometer, the distance between the two aircraft after 6 hours is approximately 2392 kilometers.