Physics Help Forum elevator problem

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Oct 1st 2009, 04:39 PM #1 Junior Member   Join Date: Sep 2009 Posts: 3 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 ????
 Oct 1st 2009, 11:06 PM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 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.
 Oct 3rd 2009, 04:59 AM #3 Banned   Join Date: Aug 2009 Location: UK Posts: 240 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.
 Oct 3rd 2009, 10:56 AM #4 Senior Member   Join Date: Jul 2009 Location: Kathu Posts: 131 elevator problem VkL, Your answer looks right to me.
 Oct 4th 2009, 01:16 AM #5 Physics Team   Join Date: Feb 2009 Posts: 1,425 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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