SOLUTION: An elevator went from the bottom to the top at 240m tower, remained there for 12 sec, and returned to the bottom in an elapsed time of 2 minutes. I fthe elevator traveled 1 m/s fas

Question 21644: An elevator went from the bottom to the top at 240m tower, remained there for 12 sec, and returned to the bottom in an elapsed time of 2 minutes. I fthe elevator traveled 1 m/s faster on the way down, find its speed going up.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Time up + time down = 120 sec -12 sec = 108 sec
Up data: distance = 240 m; rate= x; time= 240/x
Down data: dist = 240 m; rate = x+1; time = 240/(x+1)
EQUATION:
Time up + Time down = 108 sec
240/x + 240/(x+1) = 108
1/x + 1/(x+1) = .45
2x +1= .45(x^2+x)
9x^2-31x - 20 =0

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=1681 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 0.555555555555556, -4. Here's your graph:

Looks like rate up is 0.55... meters/sec
Cheers,
Stan H.
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