Question

am air traffic controller is tracking two planes. to start, plane A is at an altitude of 2667 feet and plane B is at an altitude of 4500 feet. plane A is getting altitude at 65.5 feet per second and plane B is getting altitude at 30.25 feet per second.

Answer 1

Hello!

Let's write some information:

Plane A:

• altitude: 2667 feet

,

• each second: +65.5 feet

Plane B:

• altitude: 4500 feet

,

• each second: +30.25 feet

Knowing it, we can write an equation for each airplane:

• A: 2667 + 65.5s

,

• B: 4500 + 30.25s

Now, let's solve the exercise:

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

To solve this exercise we just have to equal the two equations that I create above and solve it, look:

`\begin{gathered} 2667+65.5s=4500+30.25s \\ 65.5s-30.25s=4500-2667 \\ 35.25s=1833 \\ s=(1833)/(35.25) \\ \\ s=52\text{ seconds} \end{gathered}`

What will their altitude be when they're at the same altitude?

Now we will use the same equations again, but replace where's "s" by 52:

`\begin{gathered} 2667+65.5s=4500+30.25s \\ 2667+(65.5\cdot52)=4500+(30.25\cdot52) \\ 2667+3406=4500+1573 \\ 6073=6073 \end{gathered}`

The altitude will be 6073 feet.