Question

an air traffic controller is tracking two planes. to start, plane a is at an altitude of 5000 feet and plane b is at an altitude of 3866 feet. plane a is gaining altitude at 45.25 feet per second and plane b is gaining altitude at 65.5 feet per second. how many seconds will pass before the planes are at the same altitude? seconds what will their altitude be when theyre at the same altitude? feet

Explanation:

Step1: Define Variables and Equations

Let \( t \) be the time in seconds. The altitude of Plane A after \( t \) seconds is \( A(t) = 5000 + 45.25t \). The altitude of Plane B after \( t \) seconds is \( B(t) = 3866 + 65.5t \). We set them equal to find when they are at the same altitude: \( 5000 + 45.25t = 3866 + 65.5t \).

Step2: Solve for \( t \)

Subtract \( 45.25t \) and \( 3866 \) from both sides:
\( 5000 - 3866 = 65.5t - 45.25t \)
\( 1134 = 20.25t \)
Divide both sides by \( 20.25 \):
\( t = \frac{1134}{20.25} = 56 \) seconds.

Step3: Find the Altitude

Substitute \( t = 56 \) into Plane A’s altitude formula:
\( A(56) = 5000 + 45.25(56) \)
\( 45.25 \times 56 = 2534 \)
\( A(56) = 5000 + 2534 = 7534 \) feet. (We can also check with Plane B: \( 3866 + 65.5(56) = 3866 + 3668 = 7534 \) feet.)

Answer:

First box: 56
Second box: 7534