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.
Hello, CFY!
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: .$\displaystyle \frac{240}{s_{up}} + 12 + \frac{240}{s_{up} + 1} \;=\;2\cdot60$