An elevator went from the bottom to the top of a 240 m tower, remained there for 12 seconds, and returned to the bottom in an elapsed time of 2 minutes. If the elevator traveled 1 m/s faster on the way down, find its speed going up.

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- Oct 14th 2012, 10:15 AMCFYRate time distance
An elevator went from the bottom to the top of a 240 m tower, remained there for 12 seconds, and returned to the bottom in an elapsed time of 2 minutes. If the elevator traveled 1 m/s faster on the way down, find its speed going up.

- Oct 14th 2012, 10:48 AMMaxJasperRe: Rate time distance
240/sup+12+240/(sup+1) = 2*60(Itwasntme)

- Oct 14th 2012, 11:54 AMCFYRe: Rate time distance
thanks for your help. I tweaked your formaula a bit and found my answer; I was just unclear about what you meant by using "sup.:

- Oct 14th 2012, 12:09 PMSorobanRe: Rate time distance
Hello, CFY!

Quote:

An elevator went from the bottom to the top of a 240-m tower, remained there for 12 seconds,

and returned to the bottom in an elapsed time of 2 minutes.

If the elevator traveled 1 m/s faster on the way down, find its speed going up.

Let $\displaystyle S$ = speed going up (meters per second).

Then $\displaystyle S\!+\!1$ = speed going down.

The elevator went up 240 meters at $\displaystyle S$ m/sec.

. . This took: $\displaystyle \frac{240}{S}$ seconds.

Then it stopped for 12 seconds.

Then it went down 240 meters at $\displaystyle S\!+\!1$ m/sec.

. . This took: $\displaystyle \frac{240}{S+1}$ seconds.

The total time is 2 minutes (120 seconds).

The equation is: .$\displaystyle \frac{240}{S} + 12 + \frac{240}{S+1} \:=\:120$

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Nice "tweaking", CFY!

You saw that Max meant: .$\displaystyle \frac{240}{s_{up}} + 12 + \frac{240}{s_{up} + 1} \;=\;2\cdot60$