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## Homework Statement

A building have 198 meters, an elevator take 40 seconds to go all the way up, knowing that the elevator takes 6 secods to speed up and another 6 seconds to speed down and the aceleration is constant during this period. Determine the top speed of the elevator.

## Homework Equations

(1) R = Ro + Vot +at²/2

(2) V² = Vo² + 2aΔt

## The Attempt at a Solution

I tryed to R1 + R2 + R3 = 198

Now using (1) in R1 in this approach Ro is always zero, and in R1 Vo is also zero

R1 = 18a

For R2 we have aceleration zero but we have initial speed (the final speed of R1) so we have

R2 = 28V1

and R3 = R1 since same aceleration only in oposite direction.

R3 = 18a

for R2 I elaborated V1 using (2)

V1² = Vo² + 2aR1

V1² = 0 + 36a²

V1 = 6a

use that on the R2 formula:

R2 = 28xV1

R2 = 168a

Now going to the final formula:

I tryed to R1 + R2 + R3 = 198

18a + 168a + 18a = 198

a = 0.9705

now using the a on the first 6 secs of the aceleration.

V = at

V = 0.9705*6

V = 5,823529 m/s

The list is telling me the correct answer should be 6... where did I go wrong? The list answer may be wrong? Do you guys have another approach for this exercise? I would like to see it.

Thank you...

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