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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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