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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

1 Answer

1 vote

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.

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