# Mechanics Question 

## 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 aproach 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 aproach for this exercise? I would like to see it.

Thank you...

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vela
Staff Emeritus
Homework Helper
You work looks fine, [strike]but your calculated value for a is off. I got a=0.99 m/s2[/strike]. Keep in mind sig figs as well.

EDIT: Never mind. I can't add.

Last edited:
I got the same answer too.

34v=198
v=5.82352

I got the same answer too.

34v=198
v=5.82352
How you got to 34v = 198??

It can be easily shown if you sketch the velocity vs time graph.

vt=s
vx6x1/2 +vx6x1/2 + 28xv= 198
34v=198