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Old Oct 7th 2010, 05:35 PM   #1
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[SOLVED] Determine position vs time

This is the question




Alright so i eliminated choices 1, 2 , 4 , 7 and 8.

Which leave numbers 3, 5, 6 being left...

Am i on the right track?
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Old Oct 7th 2010, 10:48 PM   #2
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Yes, good so far. I assume that you based this on the fact that the elevator is still accelerating up until 15 seconds.

One last thing that you can do, but which takes some time is finding the exact displacement x.

From W = mg, we know that the mass is 40 kg.

Hence, using F = ma, we know that the resultant acceleration is 12.5 m/s^2 by substituting F = 500 N and m = 40 kg. So, the acceleration of the elevator alone is = 12.5 - 10 = 2.5 m/s^2.

The distance covered can be obtained by s = ut + 0.5 at^2 and for 5 seconds, you get the displacement as 32.25 m.

Next part, we get resultant acceleration as 600/40 = 15 m/s^2, which means the elevator goes up at 5 m/s^2.

Now, we need to find the velocity after 10 seconds to determine the distance covered between 10 and 15 seconds.

v = u + at

We get the final velocity between 5 and 10 seconds as 12.5 m/s

Hence, the displacement for the second part is:

s = (12.5)(5) + 0.5(5)(5)^2 = 125 m

This is the additional distance covered.

I think you can finish this problem now

~~~~~~~~~~~~~~~~~~~~~~

If you don't like that, you can still make first an acceleration time graph by using F = ma.

When done, convert it into a velocity time graph by taking the area under the acceleration time graph. You get a straight line for each piece of 5 seconds.

Then, the area under the velocity time graph will give you the displacement time graph at any instant. I think it's easier, but more difficult to explain through a post.

I hope it helps!
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Old Oct 8th 2010, 09:34 AM   #3
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hmmm, I seemed to do that.

To calculate the velocity from 15 seconds to 20 seconds, would the result give me a negative value of -15.5 m/s?

And to get the displacement for the last 5 seconds is confusing me.

I do understand how to apply the formulas, I am just having trouble with the graphs
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Old Oct 8th 2010, 09:46 AM   #4
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I gave you the displacement completed at 10 seconds. This should show you the right answer right away.

But... I'm curious now...

How did you get -15.5 m/s?
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Old Oct 8th 2010, 09:50 AM   #5
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I did 32.5 times 2 =77.5 / 5 = 15.5.
I am trying to find the velocity and the displacement from 15 seconds to 20 seconds and that where I am stuck.
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Old Oct 8th 2010, 09:59 AM   #6
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The distance he covered in all from 0 to 15 seconds is (125 + 32.25) = 157.25 m

So, you have to take (157.25 x 2)/(5^2).

But you'll see that it's not necessary. As I previously said, the additional distance covered is 157.25 m. So, look in the graphs you have, for a graph where the distance at t = 15 s is 157.25 m.

Here, 6. is the answer.

As for the last part, from 15 to 20 seconds, the balance reads 0 N. That could mean the acceleration of the elevator to be anything smaller than -10 m/s^2. A balance won't record any value less than 0 N, so, you would get the same reading (0 N) whether the acceleration of the elevator is -10 m/s^2 or -100 m/s^2 (exaggeration on the acceleration here).

Last edited by Unknown008; Oct 8th 2010 at 11:12 AM.
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Old Oct 8th 2010, 11:07 AM   #7
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Thank you very much!

But my other questions were pertaining on how to get the velocity and displacement

1)What is the velocity of the elevator at the end of the fourth 5 s interval (at 20 s)?

and

2)What is the displacement of the elevator above the starting point at the end of the fourth 5 s interval (at 20 s)?

For number 1, I get velocity to be -10 m/s?
for number 2, the displacement would be 0? since the student is feeling weightlessness?
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Old Oct 8th 2010, 11:26 AM   #8
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Acceleration for the 4th 5 s interval cannot be determined for the reasons I told you in my previous post.

But if you want to, the highest acceleration possible is -10 m/s^2.

Between 10 and 15, the acceleration is 5 m/s^2, giving a final velocity of = 12.5 + (5)(5) = 37.5 m/s

(It started with 12.5 m/s from 10 s, hence an initial velocity of 12.5 m/s)

Now, from 15 to 20 s, the acceleration is -10 m/s^2, giving a final velocity of = 37.5 - (10)(5) = -12.5 m/s

(It started with 37.5 m/s from 15 s, hence an initial velocity of 37.5 m/s)

Now, to get the distance covered during the 4rd part, we can make use of: v^2 = u^2 + 2as since we now have the initial and final velocities for a shortcut.

-12.5^2 = 37.5^2 + 2(-10)s

s = 62.5 m

Hence, the total displacement covered by the elevator = 0 + 32.25 + 125 + 62.5 = 220 m

Well, it seems that none of those answers correspond to the last part of the acceleration being -10 m/s

I told you that acceleration cannot be determined
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Old Oct 8th 2010, 11:31 AM   #9
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Ah wow, you make it looks so easy.

I understand it completely now! Also, thank you for taking your time to help me comprehend sir!
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Old Oct 8th 2010, 11:37 AM   #10
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I'm not a sir, lol.

I'm still in High School and have my exams starting in 5 days. I'm glad to see so many interesting questions on the forum and helps me get sometimes into thinking differently about problems and even find some shortcuts within a big and long problem.

Thank you for your question!
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