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

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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) About Me  (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 ax%5E2%2Bbx%2Bc=0 (in our case 9x%5E2%2B31x%2B-20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%2831%29%5E2-4%2A9%2A-20=1681.

Discriminant d=1681 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-31%2B-sqrt%28+1681+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2831%29%2Bsqrt%28+1681+%29%29%2F2%5C9+=+0.555555555555556
x%5B2%5D+=+%28-%2831%29-sqrt%28+1681+%29%29%2F2%5C9+=+-4

Quadratic expression 9x%5E2%2B31x%2B-20 can be factored:
9x%5E2%2B31x%2B-20+=+9%28x-0.555555555555556%29%2A%28x--4%29
Again, the answer is: 0.555555555555556, -4. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+9%2Ax%5E2%2B31%2Ax%2B-20+%29

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