An air traffic controller is tracking two planes. To start, Plane A is at an altitude of 3586 feet and Plane B is at an altitude of 5000 feet. Plane A is gaining altitude at 55.5 feet per second and Plane B is gaining altitude at 30.25 feet per second.

How many seconds will pass before the planes are at the same altitude?

What will their altitude be when theyre at the same altitude?

To answer this question, you need to determine the altitude difference and speed difference of plane B and plane A.

The initial altitude difference should be: 5000 feet - 3586 feet = 1414 feet.

The speed difference should be: 30.25 ft/s - 55.5 ft/s = - 25.25ft/s

After that, you can determine how long will pass before the plane in the same altitude. The calculation would be:

Final altitude difference = Initial altitude difference + speed difference*time

0 ft = 1414 ft + (-25.25ft/s * time)

25.25 time = 1414s

time = 56 second

To determine the altitude you just need to sample either plane A or plane B. Let's use plane B for easier initial altitude. The calculation would be:

Final altitude = initial altitude + speed*time = 5000ft + 30.25ft/s * 56s = 6694ft