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Old Oct 1st 2009, 04:39 PM   #1
VkL
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elevator problem

An elevator moves upward at 5.5 m/s. What is the minimum stopping time if the passengers are to remain on the floor?

what do i do with only the velocity ????
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Old Oct 1st 2009, 11:06 PM   #2
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If you travel in a car and it brakes suddenly, you are thrown forward. Like wise here, when the elevator brakes you have a tendency to be thrown up. This will happen when the elevator deccelerates with a value = g.

You have the initial vel, the final = 0. You know g . Find the time.
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Old Oct 3rd 2009, 04:59 AM   #3
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Physicsquest,

I would like to have an attempt at the question myself.

u=initial velocity
v=final velocity
t=time
a=acceleration
s=displacement

u=5.5 m/s

v^2 = 0 = u^2 - 2as

= 30.25 -30.25
s= 30.25/ (2a) = 1.543 m

t^2 = s/((1/2)a) = 0.3149 sq-rt = answer

time =
Spoiler:
0.561 secs


Is this correct?

Last edited by Paul46; Oct 3rd 2009 at 05:10 AM.
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Old Oct 3rd 2009, 10:56 AM   #4
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elevator problem

VkL,

Your answer looks right to me.
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Old Oct 4th 2009, 01:16 AM   #5
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Paul, you have the right answer. However it would be simpler to use

0 = 5.5 - g t. Accln = rate of change of vel = g = (5.5 - 0) / t.

We needn't find s as it hasn't been asked.
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