Question

One airplane is at 7,000 feet and is descending 50 feet every second. Another plane is at

1,000 feet and is climbing at a rate of 70 feet per second, How long will it take for the
planes to be at the same heigh

Answer 1

Given:

One airplane is at 7,000 feet and is descending 50 feet every second.

Another plane is at 1,000 feet and is climbing at a rate of 70 feet per second.

To find:

The time taken by planes to reach at the same height.

Solution:

Let x be the number of seconds after which the planes be at the same height.

One airplane is at 7,000 feet and is descending 50 feet every second.

Change in 1 second = -50 feet (Here negative sign means decrease)

Change in x second = -50x feet

Height of first airplane after x seconds is

`f(x)=7000-50x` ...(i)

Another plane is at 1,000 feet and is climbing at a rate of 70 feet per second.

Change in 1 second = 70 feet

Change in x second = 70x feet

Height of first airplane after x seconds is

`g(x)=1000+70x` ...(ii)

Equate (i) and (ii), to find the time after which both planes are on the same height.

`7000-50x=1000+70x`

`7000-1000=50x+70x`

`6000=120x`

Divide both sides by 120.

`50=x`

Therefore, both planes are at the same height after 50 seconds.