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
...(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
...(ii)
Equate (i) and (ii), to find the time after which both planes are on the same height.
![7000-50x=1000+70x](https://tex.z-dn.net/?f=7000-50x%3D1000%2B70x)
![7000-1000=50x+70x](https://tex.z-dn.net/?f=7000-1000%3D50x%2B70x)
![6000=120x](https://tex.z-dn.net/?f=6000%3D120x)
Divide both sides by 120.
![50=x](https://tex.z-dn.net/?f=50%3Dx)
Therefore, both planes are at the same height after 50 seconds.