Question

ActivityPlane A is descending toward the local airport, and plane B is ascending from the same airport. Plane A is descending at a rate of 2,500 feet perminute. Plane B is ascending at a rate of 4,000 feet per minute. If plane A is currently at an altitude of 14,000 feet and plane B is at an altitude of1,000 feet, how long will it take them to be at the same altitude? Represent time in minutes as the x-variable and altitude in thousands of feet asthe y-variable.

Answer 1

Let:

y1 = altitude of the plane A

y2 = altitude of the plane B

Let's find the equation for plane A:

$\begin{gathered} m1=-2500 \\ y1=-2500x+b \\ for \\ 14000=-2500(0)+b \\ b=14000 \\ y1=-2500x+14000 \end{gathered}$

And for plane B:

$\begin{gathered} m2=4000 \\ y2=4000x+b \\ for \\ 1000=4000(0)+b \\ b=1000 \\ y2=4000x+1000 \end{gathered}$

So:

$\begin{gathered} y1=y2 \\ -2500x+14000=4000x+1000 \\ solve_{\text{ }}for_{\text{ }}x\colon_{} \\ 6500x=13000 \\ x=\frac{13000}{6500} \\ x=2 \end{gathered}$

Answer:

2 minutes