SOLUTION: Two track team members ran a 200 m relay in 29 s. Each person ran 100 m. One sprinter ran 0.55 m/s faster than the other. what was the speed of each runner? thanks.. 😁

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Question 981386: Two track team members ran a 200 m relay in 29 s. Each person ran 100 m. One sprinter ran 0.55 m/s faster than the other. what was the speed of each runner?
thanks.. 😁

Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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S=speed of slower runner; S+0.55m/s=speed of faster runner
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(100m/S)+(100m/S+0.55m/s)=29 seconds Multiply by (S)(S+0.55m/s)
(100m)(S+0.55m/s)+(100m)(S)=29s(S)(S+0.55m/s)
100S+55m/s+100S=29s(S^2+0.55S)
200S+55m/s=29S^2+15.95S
0=29S^2-184.05S-55
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aS%5E2%2BbS%2Bc=0 (in our case 29S%5E2%2B-184.05S%2B-55+=+0) has the following solutons:

S%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=%28-184.05%29%5E2-4%2A29%2A-55=40254.4025.

Discriminant d=40254.4025 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--184.05%2B-sqrt%28+40254.4025+%29%29%2F2%5Ca.

S%5B1%5D+=+%28-%28-184.05%29%2Bsqrt%28+40254.4025+%29%29%2F2%5C29+=+6.63249996884891
S%5B2%5D+=+%28-%28-184.05%29-sqrt%28+40254.4025+%29%29%2F2%5C29+=+-0.285948244710973

Quadratic expression 29S%5E2%2B-184.05S%2B-55 can be factored:
29S%5E2%2B-184.05S%2B-55+=+29%28S-6.63249996884891%29%2A%28S--0.285948244710973%29
Again, the answer is: 6.63249996884891, -0.285948244710973. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+29%2Ax%5E2%2B-184.05%2Ax%2B-55+%29

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So S=6.63 m/s
The slower runner ran 6.63 meters per second.
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S+0.55m/s=6.63m/s+0.55m/s=7.18 m/s
The faster runner ran 7.18 meters per second
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CHECK:
(100m/6.63m/s)+(100m/7.18m/s)=29 seconds
15.08 seconds+13.92 seconds=29seconds
29 seconds=29 seconds