Question

At 5:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 38520 ft and is descending at the rate of 750 ft/min. Airplane #2 has an elevation of 9400 ft and is climbing at the rate of 550 ft/min. (1) Let t represent the time in minutes since 5:00 am, and let E represent the elevation in feet. Write an equation for the elevation of each plane in terms of t . plane #1: E ( t ) = plane #2: E ( t ) =

Answer 1

Answer:

Plane 1

$y_{1} = 38250\,ft - \left(750\,\frac{ft}{min}\right)\cdot t$

Plane 2

$y_{2} = 9400\,ft + \left(550\,\frac{ft}{min}\right)\cdot t$

Step-by-step explanation:

As each plane is travelling at constant speed, each equation can be modelled after this expression:

$y = y_{o} + \dot y \cdot t$

Where:

$y_{o}$ - Intial elevation of the airplane, measured in feet.

$\dot y$ - Climbing/Descending rate of the airplane, measured in feet per minute. (Positive - Climbing, Negative - Descending)

$t$ - Time, measured in minutes.

Equations are described below:

Plane 1

$y_{1} = 38250\,ft - \left(750\,\frac{ft}{min}\right)\cdot t$

Plane 2

$y_{2} = 9400\,ft + \left(550\,\frac{ft}{min}\right)\cdot t$