Question

An airplane has begun its descent for a landing. When the airplane is 150 miles west of its destination, its altitude is 25,000 feet. When the airplane is 90 miles west of its destination, its altitude is 19,000 feet. If the airplane's descent is modeled by a linear function, where will the airplane be in relation to the runway when it hits ground level?

Answer 1

airplane will over shoot the runway by 100 miles (for USA Test Prep people)

Explanation:

slope =

25,000 - 19,000

150 - 90

= 100

y = mx + b

25,000 = 100(150) + b

25,000 - 15,000 = b

b = 10,000

thus,

y = mx + b

0 = 100x + 10,000

-10,000 = 100x

x = -100

When the plane is at ground level (y = 0), it will be 100 miles past the end of the runway.

Answer 2

The airplane has descended (25,000 - 19,000) = 6,000 feet
while flying (150 - 90) = 60 miles.

If the descent is modeled by a linear function, then the slope
of the function is

(-6000 ft) / (60 miles) = - 100 ft/mile .

Since it still has 19,000 ft left to descend, at the rate of 100 ft/mi,
it still needs to fly

(19,000 ft) / (100 ft/mile) = 190 miles

to reach the ground.

It's located 90 miles west of the runway now. So if it continues
on the same slope, it'll be 100 miles past the runway (east of it)
when it touches down.

I sure hope there's another airport there.