| Topic: |
Science > Physics |
| User: |
"agent-s" |
| Date: |
14 Feb 2007 08:30:15 PM |
| Object: |
given vi, vf, and d -> solve for a |
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
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| User: "Greg Neill" |
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| Title: Re: given vi, vf, and d -> solve for a |
14 Feb 2007 09:38:34 PM |
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"agent-s" <shanekwon@gmail.com> wrote in message
news:1171506615.692033.284600@s48g2000cws.googlegroups.com...
I'm doing my freaking tycho and it won't accept my answer.
Meaningless terminology to the world at large.
Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
What's d? What are the units?
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
No units, so meaningless answer. Try entering the
answer with proper units.
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| User: "agent-s" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 12:17:06 AM |
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standard SI units
vi & vf: m/s
d: m
a: m/s
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| User: "Dirk Van de moortel" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 07:09:08 AM |
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"agent-s" <shanekwon@gmail.com> wrote in message news:1171506615.692033.284600@s48g2000cws.googlegroups.com...
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
If you have constant deceleration, then start from
vf = vi + a d
so
a = ( vf - vi )/d
so you get
a = -1.11
By the way, please use spaces when you multiply quantities.
Eg, don't write
dma = - (mvi^2)/2
but
d m a = - (m vi^2)/2
Dirk Vdm
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| User: "Greg Neill" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 08:04:04 AM |
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"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message
news:UXYAh.48$sY2.7@news.cpqcorp.net...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
The problem (as given) didn't mention breaking.
For all we can tell, the deceleration is due
to an applied force and no friction is involved.
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| User: "Dirk Van de moortel" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 10:10:21 AM |
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"Greg Neill" <gneillREM@VEsympatico.ca> wrote in message news:45d46660$0$7449$9a6e19ea@news.newshosting.com...
"Dirk Van de moortel" <dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote in message
news:UXYAh.48$sY2.7@news.cpqcorp.net...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
The problem (as given) didn't mention breaking.
For all we can tell, the deceleration is due
to an applied force and no friction is involved.
Yep.
Due to my silly mistaking d for t, I wrongly de-rationalized
the OP's line of thought.
Dirk Vdm
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| User: "Robert S" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 08:57:33 AM |
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On Feb 15, 1:09 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"agent-s" <shanek...@gmail.com> wrote in messagenews:1171506615.692033.284600@s48g2000cws.googlegroups.com...
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
If you have constant deceleration, then start from
vf = vi + a d
so
a = ( vf - vi )/d
so you get
a = -1.11
Oh dear. Time to add yourself to that fumble list.
Hint: ms^-1 and m^2s^-2 aren't the same thing.
By the way, please use spaces when you multiply quantities.
Eg, don't write
dma = - (mvi^2)/2
but
d m a = - (m vi^2)/2
The OP was right, you were wrong. That's more important, equation-
wise.
.
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| User: "Dirk Van de moortel" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 09:55:34 AM |
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"Robert S" <roberts218e@gmail.com> wrote in message news:1171551453.445380.168550@v45g2000cwv.googlegroups.com...
On Feb 15, 1:09 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"agent-s" <shanek...@gmail.com> wrote in messagenews:1171506615.692033.284600@s48g2000cws.googlegroups.com...
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
If you have constant deceleration, then start from
vf = vi + a d
so
a = ( vf - vi )/d
so you get
a = -1.11
Oh dear. Time to add yourself to that fumble list.
Oh dear - you're right - silly me. I took d for time :-)
So,
vf = vi + a t
giving
a = ( vf - vi )/t = -vi / t
and
d = vi t + 1/2 a t^2
giving
t = [ sqrt( vi^2 + 2 a d ) - vi ] / a
and therefore like the OP had:
a = -vi / [ [ sqrt( vi^2 + 2 a d ) - vi ] / a ]
= - vi^2 / (2 d)
Hint: ms^-1 and m^2s^-2 aren't the same thing.
By the way, please use spaces when you multiply quantities.
Eg, don't write
dma = - (mvi^2)/2
but
d m a = - (m vi^2)/2
The OP was right, you were wrong. That's more important, equation-
wise.
Yes, sorry to the OP :-)
Dirk Vdm
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| User: "John C. Polasek" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 11:38:19 AM |
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On Thu, 15 Feb 2007 15:55:34 GMT, "Dirk Van de moortel"
<dirkvandemoortel@ThankS-NO-SperM.hotmail.com> wrote:
"Robert S" <roberts218e@gmail.com> wrote in message news:1171551453.445380.168550@v45g2000cwv.googlegroups.com...
On Feb 15, 1:09 pm, "Dirk Van de moortel" <dirkvandemoor...@ThankS-NO-
SperM.hotmail.com> wrote:
"agent-s" <shanek...@gmail.com> wrote in messagenews:1171506615.692033.284600@s48g2000cws.googlegroups.com...
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
So you say that the Total Energy Difference equals
the Kinetic Energy Difference. This is clearly wrong.
When braking, you heat up the wheels. So you can't
use conservation of energy here. You could use it if
you wanted to find out how much heat is generated
by braking.
If you have constant deceleration, then start from
vf = vi + a d
so
a = ( vf - vi )/d
so you get
a = -1.11
Oh dear. Time to add yourself to that fumble list.
Oh dear - you're right - silly me. I took d for time :-)
So,
vf = vi + a t
giving
a = ( vf - vi )/t = -vi / t
and
d = vi t + 1/2 a t^2
giving
t = [ sqrt( vi^2 + 2 a d ) - vi ] / a
and therefore like the OP had:
a = -vi / [ [ sqrt( vi^2 + 2 a d ) - vi ] / a ]
= - vi^2 / (2 d)
Hint: ms^-1 and m^2s^-2 aren't the same thing.
By the way, please use spaces when you multiply quantities.
Eg, don't write
dma = - (mvi^2)/2
but
d m a = - (m vi^2)/2
The OP was right, you were wrong. That's more important, equation-
wise.
Yes, sorry to the OP :-)
Dirk Vdm
the simplest way:
m*a*d = m*vf^2/2 force x dist = kinetic energy
a = vf^2/2d
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| User: "Puppet_Sock" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 10:38:05 AM |
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On Feb 14, 9:30 pm, "agent-s" <shanek...@gmail.com> wrote:
I'm doing my freaking tycho and it won't accept my answer. Please take
a look at my work and correct. I'm trying to find acceleration using
conservation of energy, distance d, vi and vf are known. Basically
something is going vi and stops in d meters.
vf = final velocity = 0
vi = initial velocity = 16.6
W = KEf - KEi
(W = Fd)(F = ma)(K = (mv^2)/2)
dma = (mvf^2)/2 - (mvi^2)/2
since vf is 0
dma = - (mvi^2)/2
da = -(vi^2)/2
a = -(vi^2)/2d
with numbers:
a = -(16.6^2)/(2*15)
a = -9.1853...
I don't know what a "freaking tycho" is, nor why it won't accept
your answer. Maybe it, whatever it is, is expecting a postive
value. That is, maybe it is expecting only the magnitute.
Or possibly it wants some such thing as "in a backwards
direction" or some such thing.
Also, maybe it is expecting you to enter the units.
Socks
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| User: "agent-s" |
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| Title: Re: given vi, vf, and d -> solve for a |
15 Feb 2007 12:40:30 PM |
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Thanks for all the replies guys. Tycho is an online homework system.
It is expecting magnitude, I guess I should have clarified, but I
already took that into account when submitting my answer. Also, we
don't have to submit units. I ended up emailing my prof. Hopefully
he'll fix it for me.
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