Respuesta :
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
[tex]f(x)=7000-50x[/tex] ...(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
[tex]g(x)=1000+70x[/tex] ...(ii)
Equate (i) and (ii), to find the time after which both planes are on the same height.
[tex]7000-50x=1000+70x[/tex]
[tex]7000-1000=50x+70x[/tex]
[tex]6000=120x[/tex]
Divide both sides by 120.
[tex]50=x[/tex]
Therefore, both planes are at the same height after 50 seconds.
