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| Albert... |
 Posted: Sat Oct 31, 2009 9:15 pm |
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Guest
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Hello,
I've attempted to antidifferentiate the function given but I can't
locate a pair of values to substitute in for finding the constant.
The problem:
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
Please help me calculate for / find something to substitute in.
Thanks In Advance,
Albert |
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| Arturo Magidin... |
| Posted: Sat Oct 31, 2009 9:15 pm |
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Guest
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On Oct 31, 9:15 pm, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]I've attempted to antidifferentiate the function given but I can't
locate a pair of values to substitute in for finding the constant.
The problem:
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
[/quote]
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
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Arturo Magidin |
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| Arturo Magidin... |
| Posted: Sun Nov 01, 2009 1:04 pm |
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On Nov 1, 4:48 pm, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]Arturo Magidin wrote:
Albert wrote:
snip
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
snip
Could you please point out where I have gone wrong below?
-8 = 0 - 0 + c
c = -8
v = t^2 / 18 - 5t / 9 - 8
0 = t^2 / 18 - 5t / 9 - 8
0 = t^2 - 10t - 144
0 = (t - 1 (t +
t = 18, -8 (rejected)
[/quote]
So, the elevator will have a velocity of 0 18 seconds after it passes
the 50th floor.
[quote] x = t^3 / 54 - 5t^2 / 18 - 8t
[/quote]
[quote] t = 18: x = a negative value.
[/quote]
Of course. because the way you have set things up, negative means "up"
and positive means "down". Will the elevator stop at a floor lower
than the 50th (which you have impllicitly decided lies at position 0),
or above the 50th floor? Since positive means down and negative means
up, should the answer be positive or negative?
--
Arturo Magidin |
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| Albert... |
| Posted: Sun Nov 01, 2009 5:48 pm |
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Guest
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Arturo Magidin wrote:
[quote]Albert wrote:
snip
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
snip
[/quote]
Could you please point out where I have gone wrong below?
-8 = 0 - 0 + c
c = -8
v = t^2 / 18 - 5t / 9 - 8
0 = t^2 / 18 - 5t / 9 - 8
0 = t^2 - 10t - 144
0 = (t - 1(t +
t = 18, -8 (rejected)
x = t^3 / 54 - 5t^2 / 18 - 8t
t = 18: x = a negative value. |
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| The Qurqirish Dragon... |
| Posted: Mon Nov 02, 2009 4:23 am |
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Guest
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On Nov 1, 5:48 pm, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]Arturo Magidin wrote:
Albert wrote:
snip
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
snip
Could you please point out where I have gone wrong below?
-8 = 0 - 0 + c
c = -8
v = t^2 / 18 - 5t / 9 - 8
0 = t^2 / 18 - 5t / 9 - 8
0 = t^2 - 10t - 144
0 = (t - 1(t +
t = 18, -8 (rejected)
x = t^3 / 54 - 5t^2 / 18 - 8t
t = 18: x = a negative value.
[/quote]
Nowhere. Since the velocity is negative for the time considered, you
should expect a negative displacement. In theory, your answer of -126m
gives you 21 floors in the direction the elevator was originally
moving. If the problem doesn't state otherwise, you are technically
allowed to assume either direction, but in "normal" situations (i.e.
the way most people think, and most textbooks leave as an unwritten
condition- or written once in the forward of the text), negative
refers to down and not up. This is also the way a "standard frame of
reference" usually defines it.
Thus, you should assume that the elevator stops 21 floors below the
given point, or the 29th floor. If this was a homework problem, and
you gave me or an answer of "the 71st floor", I would require an
explanation of your qoordinate system for full marks. |
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| Albert... |
| Posted: Mon Nov 02, 2009 5:10 am |
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Guest
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Arturo Magidin wrote:
[quote]Albert wrote:
Arturo Magidin wrote:
Albert wrote:
snip
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
snip
Could you please point out where I have gone wrong below?
-8 = 0 - 0 + c
c = -8
v = t^2 / 18 - 5t / 9 - 8
0 = t^2 / 18 - 5t / 9 - 8
0 = t^2 - 10t - 144
0 = (t - 1(t +
t = 18, -8 (rejected)
So, the elevator will have a velocity of 0 18 seconds after it passes
the 50th floor.
x = t^3 / 54 - 5t^2 / 18 - 8t
t = 18: x = a negative value.
Of course. because the way you have set things up, negative means "up"
and positive means "down". Will the elevator stop at a floor lower
than the 50th (which you have impllicitly decided lies at position 0),
or above the 50th floor?
[/quote]
Lower than the 50th.
Continuing from the above calculation, x = -126 m. Therefore, the lift
will stop 126 m below the 50th floor, or 21 floors below the 50th floor.
The answer is therefore "at the 29th floor".
[quote]Since positive means down and negative means up, should the answer be
positive or negative?
[/quote]
From what information (the question alone or the entire discussion) am
I supposed to come up with the right answer?
> <snip> |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 6:37 am |
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Guest
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On Nov 2, 4:10 am, Albert <albert.xtheunkno... at (no spam) gmail.com> wrote:
[quote]ArturoMagidinwrote:
Albert wrote:
ArturoMagidinwrote:
Albert wrote:
snip
The lift in a tall building passes the 50th floor with a velocity of -8
m/s and an acceleration of (1 / 9) * (t - 5) m/s^2.
If each floor spans a height of 6 metres, find at which floor the lift
will stop.
So far, I've written down that v = t^2 / 18 - 5t / 9 + c.
And you know the velocity when it passes the 50th floor, which you may
as well call t=0. So...
snip
Could you please point out where I have gone wrong below?
-8 = 0 - 0 + c
c = -8
v = t^2 / 18 - 5t / 9 - 8
0 = t^2 / 18 - 5t / 9 - 8
0 = t^2 - 10t - 144
0 = (t - 1(t +
t = 18, -8 (rejected)
So, the elevator will have a velocity of 0 18 seconds after it passes
the 50th floor.
x = t^3 / 54 - 5t^2 / 18 - 8t
t = 18: x = a negative value.
Of course. because the way you have set things up, negative means "up"
and positive means "down". Will the elevator stop at a floor lower
than the 50th (which you have impllicitly decided lies at position 0),
or above the 50th floor?
Lower than the 50th.
[/quote]
We were imagining things differently. When you said "lift", I
inadvertedly assumed the elevator was moving "up", and so the negative
velocity meant it was slowing down. Your frame of reference is the
more standard one, in which negative is down and positive is up.
[quote]Continuing from the above calculation, x = -126 m. Therefore, the lift
will stop 126 m below the 50th floor, or 21 floors below the 50th floor.
The answer is therefore "at the 29th floor".
From what information (the question alone or the entire discussion) am
I supposed to come up with the right answer?
[/quote]
I'm not sure what you mean. The question provided you with more than
sufficient information to solve the problem; however, you were unable
to do so with just the information provided by the question, so you
required the "entire discussion". That, I'm sorry to tell you, was not
because the *question* was lacking in necessary information....
--
Arturo Magidin |
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