Question

an air traffic controller is tracking two planes. to start, plane a was at an altitude of 432 meters, and plane b was just taking off. plane a is gaining altitude at 16 meters per second, and plane b is gaining altitude at 25 meters per second. let x be the number of seconds that have passed. (a) for each plane, write an expression for the altitude of the plane after x seconds. altitude of plane a (in meters) = \boxed{} altitude of plane b (in meters) = \boxed{} (b) write an equation to show when the two planes would be at the same altitude. \boxed{}

Explanation:

Step1: Determine Plane A's altitude expression

Plane A starts at 432 meters and gains 16 meters per second. So after \( x \) seconds, the altitude is initial altitude plus gain: \( 432 + 16x \).

Step2: Determine Plane B's altitude expression

Plane B starts at 0 meters (just taking off) and gains 25 meters per second. So after \( x \) seconds, the altitude is \( 25x \).

Step3: Set altitudes equal for same altitude

To find when they are at the same altitude, set the two expressions equal: \( 432 + 16x = 25x \).

Answer:

(a) Altitude of Plane A (in meters) = \( 432 + 16x \)
Altitude of Plane B (in meters) = \( 25x \)
(b) Equation: \( 432 + 16x = 25x \)