## RE: [puzzles] Puzz

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• There are many problems in mathematics which can only be solved if u avoid points of singularity, which in this case is at s=0....hence for all practical
Message 1 of 25 , Jul 25, 2002
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There are many problems in mathematics which can only be solved if u avoid points of singularity, which in this case is at s=0....hence for all practical purposes let us say that he has an initial velocity lim -> 0....this way we can avoid this point and end up with the same solution.
-----Original Message-----
From: Prashant Kothari [mailto:pk7ul@...]
Sent: Thursday, July 25, 2002 3:46 PM
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

Unfortunately, no unique solution exists for this
problem...and I can proove that. Unless, of course, u
know the initial velocity..

--- Amir Esufally <amir@...> wrote:
> Here is the solution to the puzzle...
> By the problem statement
> s->distance travelled
> t-> time taken
> => s = v  at every instant of time
> => s =ds / dt
> =>dt = (1/s) ds
> integrating on both sides of the eq.
> t= ln(s) +c---------1
> for all practical purposes let us assume c=0;
> at the end point s=1000mts
> substituting in eq 1
> t=6.9 sec.

>
> -----Original Message-----
> From: Vijaykumar Nayak (RBIN/EDC1) *
> [mailto:Vijaykumar.Nayak@...]
> Sent: Friday, July 19, 2002 1:42 PM
> To: Puzzle (E-mail)
> Subject: [puzzles] Puzz
>
>

>
>
>

>
> A man starts with  some initial velocity...He has to
> travel a distance of 1km.
>
> He has decided to travel at  a the velocity  such
> that,
>
> at each point of time the distance he has travelled
> from the starting point
> is his velocity.
>
> how much time does he need to cover the disatnce?
>

>

>
>
>
>
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> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>Here
> is the solution to the puzzle...</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>By
> the
> problem statement</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff
> size=2>s-&gt;distance travelled </FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff
> size=2>t-&gt;
> time taken</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>=&gt;
> s = v&nbsp; at every instant of
> time</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>=&gt;
> s =ds / dt</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff
> size=2>=&gt;dt = (1/s) ds</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff
> size=2>integrating on both sides of the
> eq.</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>t=
> ln(s) +c---------1</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>for
> all practical purposes let us assume
> c=0;</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>at
> the
> end point s=1000mts</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff
> size=2>substituting in eq 1</FONT></SPAN></DIV>
> <DIV><SPAN class=756014211-22072002><FONT face=Arial
> color=#0000ff size=2>t=6.9
> sec.</FONT></SPAN></DIV>
> <DIV><SPAN
> class=756014211-22072002>&nbsp;</SPAN></DIV>
> <BLOCKQUOTE>
> align=left><FONT face=Tahoma
>   size=2>-----Original Message-----<BR><B>From:</B>
> Vijaykumar Nayak
> (RBIN/EDC1)
>   *
>
[mailto:Vijaykumar.Nayak@...]<BR><B>Sent:</B>
> Friday, July 19,
> 2002
>   1:42 PM<BR><B>To:</B> Puzzle
> (E-mail)<BR><B>Subject:</B> [puzzles]
>   Puzz<BR><BR></FONT></DIV>
>   <DIV><FONT face=Arial color=#000080
> size=2></FONT>&nbsp;</DIV>
>   <P></P>
>   <P><FONT face=Fixedsys color=#000080
> size=2></FONT>&nbsp;</P><FONT face=Arial
>   color=#000080 size=2>
>   <P><SPAN class=523290508-19072002>A man starts
> with&nbsp; some initial
>   velocity...He has to travel a distance of
> 1km.</SPAN></P>
>   <P><SPAN class=523290508-19072002>He has decided
> to travel at&nbsp; a the
>   velocity&nbsp; such that,</SPAN></P>
>   <P><SPAN class=523290508-19072002>at each point of
> time the distance he has
>   travelled from the starting point&nbsp;is his
> velocity.</SPAN></P>
>   <P><SPAN class=523290508-19072002>how much time
> does he need to cover the
>   disatnce?</SPAN></P>
>   <P><SPAN
> class=523290508-19072002></SPAN>&nbsp;</P></FONT>
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• But that s going away from the problem definition...whatever point of singularity means! and initial velocity of 0 isnt practical for all purposes! ...
Message 2 of 25 , Jul 25, 2002
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But that's going away from the problem
definition...whatever point of singularity means!

and initial velocity of 0 isnt practical for all
purposes!

--- "Vijaykumar Nayak (RBIN/EDC1) *"
<Vijaykumar.Nayak@...> wrote:
> There are many problems in mathematics which can
> only be solved if u avoid points of singularity,
> which in this case is at s=0....hence for all
> practical purposes let us say that he has an initial
> velocity lim -> 0....this way we can avoid this
> point and end up with the same solution.
>
> -----Original Message-----
> From: Prashant Kothari
> [mailto:pk7ul@...]
> Sent: Thursday, July 25, 2002 3:46 PM
> To: puzzles@yahoogroups.com
> Subject: RE: [puzzles] Puzz
>
>
> Unfortunately, no unique solution exists for this
> problem...and I can proove that. Unless, of course,
> u
> know the initial velocity..
>
>
> --- Amir Esufally <amir@...> wrote:
> > Here is the solution to the puzzle...
> > By the problem statement
> > s->distance travelled
> > t-> time taken
> > => s = v at every instant of time
> > => s =ds / dt
> > =>dt = (1/s) ds
> > integrating on both sides of the eq.
> > t= ln(s) +c---------1
> > for all practical purposes let us assume c=0;
> > at the end point s=1000mts
> > substituting in eq 1
> > t=6.9 sec.
> >
> >
> > -----Original Message-----
> > From: Vijaykumar Nayak (RBIN/EDC1) *
> > [mailto:Vijaykumar.Nayak@...]
> > Sent: Friday, July 19, 2002 1:42 PM
> > To: Puzzle (E-mail)
> > Subject: [puzzles] Puzz
> >
> >
> >
> >
> >
> >
> >
> >
> > A man starts with some initial velocity...He has
> to
> > travel a distance of 1km.
> >
> > He has decided to travel at a the velocity such
> > that,
> >
> > at each point of time the distance he has
> travelled
> > from the starting point
> > is his velocity.
> >
> > how much time does he need to cover the disatnce?
> >
> >
> >
> >
> >
> >
> >
> >
> <
>
http://rd.yahoo.com/M=228862.2128520.3581629.1829184/D=egroupweb/S=1705297
>
<http://rd.yahoo.com/M=228862.2128520.3581629.1829184/D=egroupweb/S=1705297>
>
> >
> 356:HM/A=1155067/R=0/*
>
> >
> >
> >
> > To unsubscribe from this list, go to the ONElist
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> >
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>Here
> > is the solution to the
> puzzle...</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>By
> > the
> > problem statement</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>s->distance travelled
> </FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>t->
> > time taken</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>=>
> > s = v  at every instant of
> > time</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>=>
> > s =ds / dt</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>=>dt = (1/s) ds</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>integrating on both sides of the
> > eq.</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>t=
> > ln(s) +c---------1</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>for
> > all practical purposes let us assume
> > c=0;</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>at
> > the
> > end point s=1000mts</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>substituting in eq 1</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>t=6.9
> > sec.</FONT></SPAN></DIV>
> > <DIV><SPAN
> > class=756014211-22072002> </SPAN></DIV>
> > <BLOCKQUOTE>
> > align=left><FONT face=Tahoma
> > size=2>-----Original
> Message-----<BR><B>From:</B>
> > Vijaykumar Nayak
> > (RBIN/EDC1)
> > *
> >
>
[mailto:Vijaykumar.Nayak@...]<BR><B>Sent:</B>
> > Friday, July 19,
> > 2002
> > 1:42 PM<BR><B>To:</B> Puzzle
> > (E-mail)<BR><B>Subject:</B> [puzzles]
> > Puzz<BR><BR></FONT></DIV>
> > <DIV><FONT face=Arial color=#000080
> > size=2></FONT> </DIV>
> > <P></P>
> > <P><FONT face=Fixedsys color=#000080
> > size=2></FONT> </P><FONT face=Arial
> > color=#000080 size=2>
> > <P><SPAN class=523290508-19072002>A man starts
> > with  some initial
> > velocity...He has to travel a distance of
> > 1km.</SPAN></P>
> > <P><SPAN class=523290508-19072002>He has decided
> > to travel at  a the
> > velocity  such that,</SPAN></P>
> > <P><SPAN class=523290508-19072002>at each point
> of
> > time the distance he has
> > travelled from the starting point is his
> > velocity.</SPAN></P>
> > <P><SPAN class=523290508-19072002>how much time
> > does he need to cover the
> > disatnce?</SPAN></P>
> > <P><SPAN
> > class=523290508-19072002></SPAN> </P></FONT>
> > <DIV> </DIV><BR><BR><TT>To
> > unsubscribe from this list, go to the ONElist
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• Singularity is a point where conventional laws of physics and mathematics breakdown!!!(Like a singularity that arises in complex analysis or black hole for
Message 3 of 25 , Jul 25, 2002
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Singularity is a point where conventional laws of physics and mathematics breakdown!!!(Like a singularity that arises in complex analysis or black hole for that matter!!)
Also i didn't say velocity is  "0" ....i said velocity tends to 0(lim ->0).
-----Original Message-----
From: Prashant Kothari [mailto:pk7ul@...]
Sent: Thursday, July 25, 2002 4:05 PM
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

But that's going away from the problem
definition...whatever point of singularity means!

and initial velocity of 0 isnt practical for all
purposes!

--- "Vijaykumar Nayak (RBIN/EDC1) *"
<Vijaykumar.Nayak@...> wrote:
> There are many problems in mathematics which can
> only be solved if u avoid points of singularity,
> which in this case is at s=0....hence for all
> practical purposes let us say that he has an initial
> velocity lim -> 0....this way we can avoid this
> point and end up with the same solution.
>
> -----Original Message-----
> From: Prashant Kothari
> [mailto:pk7ul@...]
> Sent: Thursday, July 25, 2002 3:46 PM
> To: puzzles@yahoogroups.com
> Subject: RE: [puzzles] Puzz
>
>
> Unfortunately, no unique solution exists for this
> problem...and I can proove that. Unless, of course,
> u
> know the initial velocity..
>
>
> --- Amir Esufally <amir@...> wrote:
> > Here is the solution to the puzzle...
> > By the problem statement
> > s->distance travelled
> > t-> time taken
> > => s = v  at every instant of time
> > => s =ds / dt
> > =>dt = (1/s) ds
> > integrating on both sides of the eq.
> > t= ln(s) +c---------1
> > for all practical purposes let us assume c=0;
> > at the end point s=1000mts
> > substituting in eq 1
> > t=6.9 sec.
> >
> >
> > -----Original Message-----
> > From: Vijaykumar Nayak (RBIN/EDC1) *
> > [mailto:Vijaykumar.Nayak@...]
> > Sent: Friday, July 19, 2002 1:42 PM
> > To: Puzzle (E-mail)
> > Subject: [puzzles] Puzz
> >
> >
> >
> >
> >
> >
> >
> >
> > A man starts with  some initial velocity...He has
> to
> > travel a distance of 1km.
> >
> > He has decided to travel at  a the velocity  such
> > that,
> >
> > at each point of time the distance he has
> travelled
> > from the starting point
> > is his velocity.
> >
> > how much time does he need to cover the disatnce?
> >
> >
> >
> >
> >
> >
> >
> >
> <
>
http://rd.yahoo.com/M=228862.2128520.3581629.1829184/D=egroupweb/S=1705297
>
<http://rd.yahoo.com/M=228862.2128520.3581629.1829184/D=egroupweb/S=1705297>
>
> >
> 356:HM/A=1155067/R=0/*
>
> >
> >
> >
> > To unsubscribe from this list, go to the ONElist
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> puzzle...</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>By
> > the
> > problem statement</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>s-&gt;distance travelled
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> > time taken</FONT></SPAN></DIV>
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> face=Arial
> > color=#0000ff size=2>=&gt;
> > s = v&nbsp; at every instant of
> > time</FONT></SPAN></DIV>
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> face=Arial
> > color=#0000ff size=2>=&gt;
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> face=Arial
> > color=#0000ff
> > size=2>=&gt;dt = (1/s) ds</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>integrating on both sides of the
> > eq.</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>t=
> > ln(s) +c---------1</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>for
> > all practical purposes let us assume
> > c=0;</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>at
> > the
> > end point s=1000mts</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff
> > size=2>substituting in eq 1</FONT></SPAN></DIV>
> > <DIV><SPAN class=756014211-22072002><FONT
> face=Arial
> > color=#0000ff size=2>t=6.9
> > sec.</FONT></SPAN></DIV>
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> > class=756014211-22072002>&nbsp;</SPAN></DIV>
> > <BLOCKQUOTE>
> > align=left><FONT face=Tahoma
> >   size=2>-----Original
> Message-----<BR><B>From:</B>
> > Vijaykumar Nayak
> > (RBIN/EDC1)
> >   *
> >
>
[mailto:Vijaykumar.Nayak@...]<BR><B>Sent:</B>
> > Friday, July 19,
> > 2002
> >   1:42 PM<BR><B>To:</B> Puzzle
> > (E-mail)<BR><B>Subject:</B> [puzzles]
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• Thanks a lot for this enlightening piece of singularity :)) and velocity tending to 0 seems as much impractical as it tending to 0! ...
Message 4 of 25 , Jul 25, 2002
• 0 Attachment
Thanks a lot for this enlightening piece of
"singularity" :))

and velocity tending to 0 seems as much impractical as
it tending to 0!

--- "Vijaykumar Nayak (RBIN/EDC1) *"
<Vijaykumar.Nayak@...> wrote:
> Singularity is a point where conventional laws of
> physics and mathematics
> breakdown!!!(Like a singularity that arises in
> complex analysis or black
> hole for that matter!!)
> Also i didn't say velocity is "0" ....i said
> velocity tends to 0(lim ->0).
>
> -----Original Message-----
> From: Prashant Kothari
> [mailto:pk7ul@...]
> Sent: Thursday, July 25, 2002 4:05 PM
> To: puzzles@yahoogroups.com
> Subject: RE: [puzzles] Puzz
>
>
> But that's going away from the problem
> definition...whatever point of singularity means!
>
> and initial velocity of 0 isnt practical for all
> purposes!
>
> --- "Vijaykumar Nayak (RBIN/EDC1) *"
> <Vijaykumar.Nayak@...> wrote:
> > There are many problems in mathematics which can
> > only be solved if u avoid points of singularity,
> > which in this case is at s=0....hence for all
> > practical purposes let us say that he has an
> initial
> > velocity lim -> 0....this way we can avoid this
> > point and end up with the same solution.
> >
> > -----Original Message-----
> > From: Prashant Kothari
> > [mailto:pk7ul@...]
> > Sent: Thursday, July 25, 2002 3:46 PM
> > To: puzzles@yahoogroups.com
> > Subject: RE: [puzzles] Puzz
> >
> >
> > Unfortunately, no unique solution exists for this
> > problem...and I can proove that. Unless, of
> course,
> > u
> > know the initial velocity..
> >
> >
> > --- Amir Esufally <amir@...> wrote:
> > > Here is the solution to the puzzle...
> > > By the problem statement
> > > s->distance travelled
> > > t-> time taken
> > > => s = v at every instant of time
> > > => s =ds / dt
> > > =>dt = (1/s) ds
> > > integrating on both sides of the eq.
> > > t= ln(s) +c---------1
> > > for all practical purposes let us assume c=0;
> > > at the end point s=1000mts
> > > substituting in eq 1
> > > t=6.9 sec.
> > >
> > >
> > > -----Original Message-----
> > > From: Vijaykumar Nayak (RBIN/EDC1) *
> > > [mailto:Vijaykumar.Nayak@...]
> > > Sent: Friday, July 19, 2002 1:42 PM
> > > To: Puzzle (E-mail)
> > > Subject: [puzzles] Puzz
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > A man starts with some initial velocity...He
> has
> > to
> > > travel a distance of 1km.
> > >
> > > He has decided to travel at a the velocity
> such
> > > that,
> > >
> > > at each point of time the distance he has
> > travelled
> > > from the starting point
> > > is his velocity.
> > >
> > > how much time does he need to cover the
> disatnce?
> > >
> > >
> > >
> > >
> > >
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> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>Here
> > > is the solution to the
> > puzzle...</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>By
> > > the
> > > problem statement</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>s->distance travelled
> > </FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>t->
> > > time taken</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>=>
> > > s = v  at every instant of
> > > time</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>=>
> > > s =ds / dt</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>=>dt = (1/s) ds</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>integrating on both sides of the
> > > eq.</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>t=
> > > ln(s) +c---------1</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>for
> > > all practical purposes let us assume
> > > c=0;</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>at
> > > the
> > > end point s=1000mts</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>substituting in eq 1</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>t=6.9
> > > sec.</FONT></SPAN></DIV>
> > > <DIV><SPAN
> > > class=756014211-22072002> </SPAN></DIV>
> > > <BLOCKQUOTE>
> > > <DIV class=OutlookMessageHeader dir=ltr
> > > align=left><FONT face=Tahoma
> > > size=2>-----Original
> > Message-----<BR><B>From:</B>
> > > Vijaykumar Nayak
> > > (RBIN/EDC1)
> > > *
> > >
> >
>
[mailto:Vijaykumar.Nayak@...]<BR><B>Sent:</B>
> > > Friday, July 19,
> > > 2002
> > > 1:42 PM<BR><B>To:</B> Puzzle
> > > (E-mail)<BR><B>Subject:</B> [puzzles]
> > > Puzz<BR><BR></FONT></DIV>
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> > > size=2></FONT> </P><FONT face=Arial
> > > color=#000080 size=2>
> > > <P><SPAN class=523290508-19072002>A man starts
> > > with  some initial
> > > velocity...He has to travel a distance of
> > > 1km.</SPAN></P>
> > > <P><SPAN class=523290508-19072002>He has
> decided
> > > to travel at  a the
> > > velocity  such that,</SPAN></P>
> > > <P><SPAN class=523290508-19072002>at each
> point
> > of
> > > time the distance he has
> > > travelled from the starting point is his
> > > velocity.</SPAN></P>
> > > <P><SPAN class=523290508-19072002>how much
> time
> > > does he need to cover the
> > > disatnce?</SPAN></P>
> > > <P><SPAN
> > >
> class=523290508-19072002></SPAN> </P></FONT>
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• You may not realise it; but tending to 0 is very different from tending to 0! as 0!
Message 5 of 25 , Jul 25, 2002
• 0 Attachment
<<and velocity tending to 0 seems as much impractical as
it tending to 0!>>

You may not realise it; but tending to 0 is very different from tending
to 0! as 0! = Factorial 0, which is 1.

LKG
• OK, mention not! velocity tending to 0 is not unpractical . Every object in motion has to have its velocity to tend to 0...in order to be actually 0(Rest).
Message 6 of 25 , Jul 25, 2002
• 0 Attachment
OK,
mention not!
velocity tending to 0 is not unpractical .
Every object in motion has to have  its velocity to tend to 0...in order to be actually 0(Rest).

-----Original Message-----
From: Prashant Kothari [mailto:pk7ul@...]
Sent: Thursday, July 25, 2002 10:36 PM
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

Thanks a lot for this enlightening piece of
"singularity" :))

and velocity tending to 0 seems as much impractical as
it tending to 0!

--- "Vijaykumar Nayak (RBIN/EDC1) *"
<Vijaykumar.Nayak@...> wrote:
> Singularity is a point where conventional laws of
> physics and mathematics
> breakdown!!!(Like a singularity that arises in
> complex analysis or black
> hole for that matter!!)
> Also i didn't say velocity is  "0" ....i said
> velocity tends to 0(lim ->0).
>
> -----Original Message-----
> From: Prashant Kothari
> [mailto:pk7ul@...]
> Sent: Thursday, July 25, 2002 4:05 PM
> To: puzzles@yahoogroups.com
> Subject: RE: [puzzles] Puzz
>
>
> But that's going away from the problem
> definition...whatever point of singularity means!
>
> and initial velocity of 0 isnt practical for all
> purposes!
>
> --- "Vijaykumar Nayak (RBIN/EDC1) *"
> <Vijaykumar.Nayak@...> wrote:
> > There are many problems in mathematics which can
> > only be solved if u avoid points of singularity,
> > which in this case is at s=0....hence for all
> > practical purposes let us say that he has an
> initial
> > velocity lim -> 0....this way we can avoid this
> > point and end up with the same solution.
> >
> > -----Original Message-----
> > From: Prashant Kothari
> > [mailto:pk7ul@...]
> > Sent: Thursday, July 25, 2002 3:46 PM
> > To: puzzles@yahoogroups.com
> > Subject: RE: [puzzles] Puzz
> >
> >
> > Unfortunately, no unique solution exists for this
> > problem...and I can proove that. Unless, of
> course,
> > u
> > know the initial velocity..
> >
> >
> > --- Amir Esufally <amir@...> wrote:
> > > Here is the solution to the puzzle...
> > > By the problem statement
> > > s->distance travelled
> > > t-> time taken
> > > => s = v  at every instant of time
> > > => s =ds / dt
> > > =>dt = (1/s) ds
> > > integrating on both sides of the eq.
> > > t= ln(s) +c---------1
> > > for all practical purposes let us assume c=0;
> > > at the end point s=1000mts
> > > substituting in eq 1
> > > t=6.9 sec.
> > >
> > >
> > > -----Original Message-----
> > > From: Vijaykumar Nayak (RBIN/EDC1) *
> > > [mailto:Vijaykumar.Nayak@...]
> > > Sent: Friday, July 19, 2002 1:42 PM
> > > To: Puzzle (E-mail)
> > > Subject: [puzzles] Puzz
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > >
> > > A man starts with  some initial velocity...He
> has
> > to
> > > travel a distance of 1km.
> > >
> > > He has decided to travel at  a the velocity
> such
> > > that,
> > >
> > > at each point of time the distance he has
> > travelled
> > > from the starting point
> > > is his velocity.
> > >
> > > how much time does he need to cover the
> disatnce?
> > >
> > >
> > >
> > >
> > >
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> > > Transitional//EN">
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> > CONTENT="text/html;
> > > charset=iso-8859-1">
> > >
> > >
> > > <META content="MSHTML 6.00.2600.0"
> > > <BODY>
> > >
> > >
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>Here
> > > is the solution to the
> > puzzle...</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>By
> > > the
> > > problem statement</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>s-&gt;distance travelled
> > </FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>t-&gt;
> > > time taken</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>=&gt;
> > > s = v&nbsp; at every instant of
> > > time</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>=&gt;
> > > s =ds / dt</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>=&gt;dt = (1/s) ds</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>integrating on both sides of the
> > > eq.</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>t=
> > > ln(s) +c---------1</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>for
> > > all practical purposes let us assume
> > > c=0;</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>at
> > > the
> > > end point s=1000mts</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff
> > > size=2>substituting in eq 1</FONT></SPAN></DIV>
> > > <DIV><SPAN class=756014211-22072002><FONT
> > face=Arial
> > > color=#0000ff size=2>t=6.9
> > > sec.</FONT></SPAN></DIV>
> > > <DIV><SPAN
> > > class=756014211-22072002>&nbsp;</SPAN></DIV>
> > > <BLOCKQUOTE>
> > >   <DIV class=OutlookMessageHeader dir=ltr
> > > align=left><FONT face=Tahoma
> > >   size=2>-----Original
> > Message-----<BR><B>From:</B>
> > > Vijaykumar Nayak
> > > (RBIN/EDC1)
> > >   *
> > >
> >
>
[mailto:Vijaykumar.Nayak@...]<BR><B>Sent:</B>
> > > Friday, July 19,
> > > 2002
> > >   1:42 PM<BR><B>To:</B> Puzzle
> > > (E-mail)<BR><B>Subject:</B> [puzzles]
> > >   Puzz<BR><BR></FONT></DIV>
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> > >   <P></P>
> > >   <P><FONT face=Fixedsys color=#000080
> > > size=2></FONT>&nbsp;</P><FONT face=Arial
> > >   color=#000080 size=2>
> > >   <P><SPAN class=523290508-19072002>A man starts
> > > with&nbsp; some initial
> > >   velocity...He has to travel a distance of
> > > 1km.</SPAN></P>
> > >   <P><SPAN class=523290508-19072002>He has
> decided
> > > to travel at&nbsp; a the
> > >   velocity&nbsp; such that,</SPAN></P>
> > >   <P><SPAN class=523290508-19072002>at each
> point
> > of
> > > time the distance he has
> > >   travelled from the starting point&nbsp;is his
> > > velocity.</SPAN></P>
> > >   <P><SPAN class=523290508-19072002>how much
> time
> > > does he need to cover the
> > >   disatnce?</SPAN></P>
> > >   <P><SPAN
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• First, let me remind myself of the puzzle s wording: ***A man starts with some initial velocity*** ...He has to travel a distance of 1km. ***He has decided
Message 7 of 25 , Jul 26, 2002
• 0 Attachment
First, let me remind myself of the puzzle's wording:

" ***A man starts with some initial velocity*** ...He has to travel
a distance of 1km.

***He has decided to travel at a the velocity such that,

at each point of time the distance he has travelled from the starting
point is his velocity.***

how much time does he need to cover the distance?"

Well,

#1 A Distance is NOT a Velocity!

Honestly now, that phrase about the velocity equaling the distance
travelled means nothing (or little). For example, let's say he
travelled one meter. Now, should he be travelling at one meter per
hour, one meter per second, or one meter per nanosecond?

If this problem would have more solutions, it's this ambiguity to
blame.

But of course, there's only one solution that makes sense ...

#2 Forever.

Yep, that's the time required to finish the trip. The reason is very
simple, and it has been said before. Bearing that odd decision about
the velocity in mind:

The journey would obviously start at the beginning, "0" (true 0, not
lim->0). (It starts at 0, because if any bit of distance were
travelled, there'd be a question of when that was travellled and thus
the beginning was in fact at an earlier time). So, 0 is your initial
distance, and hence, 0 is your initial speed. 0 meters per second,
nanosecond or hour is the same as standing still. Remaining at 0
distance would also imply remaining at 0 speed.

No calculus is needed. Sheer logic suffices. Btw, the problem is
worded inconsistently, in that it stipulates there is some initial
velocity, while at the same time making it possible to infer that the
initial velocity is null. Odd.

But true. Also, "zero distance" may be called a singularity point,
but it may not be ignored. It is part of the problem, and one cannot
just ignore bits of a problem to make it easier to solve.

In a nutshell, I agree with anyone who says that this journey,
carried out like this, will never be completed.
• Hi, sorry for chipping in so late, but I find a more fundamental problem with your solution. When you equate s = ds / dt you are ignoring the unit of time (is
Message 8 of 25 , Jul 30, 2002
• 0 Attachment
Hi,
sorry for chipping in so late, but I find a more

When you equate
s = ds / dt
you are ignoring the unit of time (is it seconds,
hours, or what?).

So the correct answer would be 6.9 units. It could be
6.9 secs or 6.9 hours!

To get a unique answer the problem should state under
what units the distance and velocity are equal (in
magnitude).

regards,

--- "Vijaykumar Nayak (RBIN/EDC1) *"
<Vijaykumar.Nayak@...> wrote:
> Here is the solution to the puzzle...
> By the problem statement
> s->distance travelled
> t-> time taken
> => s = v at every instant of time
> => s =ds / dt
> =>dt = (1/s) ds
> integrating on both sides of the eq.
> t= ln(s) +c---------1
> for all practical purposes let us assume c=0;
> at the end point s=1000mts
> substituting in eq 1
> t=6.9 sec.
>
>
> -----Original Message-----
> From: Vijaykumar Nayak (RBIN/EDC1) *
> [mailto:Vijaykumar.Nayak@...]
> Sent: Friday, July 19, 2002 1:42 PM
> To: Puzzle (E-mail)
> Subject: [puzzles] Puzz

>
> A man starts with some initial velocity...He has to
> travel a distance of 1km.
>
> He has decided to travel at a the velocity such
> that,
>
> at each point of time the distance he has travelled
> from the starting point is his velocity.
>
> how much time does he need to cover the disatnce?
>
>

=====
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at the same time. I think I've forgotten
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• Hi,I m not convinced about putting the integration constant to 0 without sufficient reason. We use boundary conditions (or some other known conditions) to
Message 9 of 25 , Jul 30, 2002
• 0 Attachment
Hi,

I'm not convinced about putting the integration constant to 0 without sufficient reason. We use boundary conditions (or some other known conditions) to derive the value of the constant.

Also, unit should not play any role unlike the case here. You use 1000 meter and substitute it in the place of s. But if I use unit as km, I've to put s = 1. This gives t = 0. i.e., distance traveled is 1 km at the begining only!!! Instead, if I use cm as the unit, t will become 11.51!!. The smaller the unit, larger will be the time! Completely absurd. Could you guys please correct me if I'm wrong?

Thanks and regards,
Gandhar

----- Original Message -----
Date: Tue, 30 Jul 2002 08:48:53 -0700 (PDT)
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

> Hi,
> sorry for chipping in so late, but I find a more
> fundamental problem with your solution.
>
> When you equate
> s = ds / dt
> you are ignoring the unit of time (is it seconds,
> hours, or what?).
>
> So the correct answer would be 6.9 units. It could be
> 6.9 secs or 6.9 hours!
>
> To get a unique answer the problem should state under
> what units the distance and velocity are equal (in
> magnitude).
>
> regards,
>
> --- "Vijaykumar Nayak (RBIN/EDC1) *"
> <Vijaykumar.Nayak@...> wrote:
> > Here is the solution to the puzzle...
> > By the problem statement
> > s->distance travelled
> > t-> time taken
> > => s = v at every instant of time
> > => s =ds / dt
> > =>dt = (1/s) ds
> > integrating on both sides of the eq.
> > t= ln(s) +c---------1
> > for all practical purposes let us assume c=0;
> > at the end point s=1000mts
> > substituting in eq 1
> > t=6.9 sec.
> >
> >
> > -----Original Message-----
> > From: Vijaykumar Nayak (RBIN/EDC1) *
> > [mailto:Vijaykumar.Nayak@...]
> > Sent: Friday, July 19, 2002 1:42 PM
> > To: Puzzle (E-mail)
> > Subject: [puzzles] Puzz
>
> >
> > A man starts with some initial velocity...He has to
> > travel a distance of 1km.
> >
> > He has decided to travel at a the velocity such
> > that,
> >
> > at each point of time the distance he has travelled
> > from the starting point is his velocity.
> >
> > how much time does he need to cover the disatnce?
> >
> >
>
>
> =====
> "Right now I'm having amnesia and deja vu
> at the same time. I think I've forgotten
> this before".
>
> __________________________________________________
> Do You Yahoo!?
> Yahoo! Health - Feel better, live better
> http://health.yahoo.com
>

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• Yes, The unit controversy can be solved if we only stick to (Only one) mks system of units. as i had given the distance in meters and time must be measured in
Message 10 of 25 , Jul 30, 2002
• 0 Attachment
Yes,
The unit controversy can be solved if we only stick to (Only one) mks system of units.
as i had given the distance in meters and time must be measured in seconds.
one can't say i will substitute 1km in place of S because, the unit followed in here is meters for distance ,and hence the result observed is in seconds.
There is good reason as to why a smaller unit gets us larger time, thats because even though we change the unit of distance from meters to cm we stick on to the unit of time as second.As  for mathematics replacing meters by centimeters will only result in changing 1000 units to 1 unit.
I agree that the constant  of integration can't be chosen 0 without any satisfactory initial condition ,so the problem has to be changed such that C=0, that can be done if we assume that he starts from 1 and travels up to 1000 meters.
now as
t=ln(s)+C
at t=0 s=1 =>
c=0;

-----Original Message-----
From: Gandhar Gokhale [mailto:gandhar@...]
Sent: Wednesday, July 31, 2002 9:11 AM
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

Hi,

I'm not convinced about putting the integration constant to 0 without sufficient reason. We use boundary conditions (or some other known conditions) to derive the value of the constant.

Also, unit should not play any role unlike the case here. You use 1000 meter and substitute it in the place of s. But if I use unit as km, I've to put s = 1. This gives t = 0. i.e., distance traveled is 1 km at the begining only!!! Instead, if I use cm as the unit, t will become 11.51!!. The smaller the unit, larger will be the time! Completely absurd. Could you guys please correct me if I'm wrong?

Thanks and regards,
Gandhar

----- Original Message -----
Date: Tue, 30 Jul 2002 08:48:53 -0700 (PDT)
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

> Hi,
> sorry for chipping in so late, but I find a more
> fundamental problem with your solution.
>
> When you equate
> s = ds / dt
> you are ignoring the unit of time (is it seconds,
> hours, or what?).
>
> So the correct answer would be 6.9 units. It could be
> 6.9 secs or 6.9 hours!
>
> To get a unique answer the problem should state under
> what units the distance and velocity are equal (in
> magnitude).
>
> regards,
>
> --- "Vijaykumar Nayak (RBIN/EDC1) *"
> <Vijaykumar.Nayak@...> wrote:
> > Here is the solution to the puzzle...
> > By the problem statement
> > s->distance travelled
> > t-> time taken
> > => s = v  at every instant of time
> > => s =ds / dt
> > =>dt = (1/s) ds
> > integrating on both sides of the eq.
> > t= ln(s) +c---------1
> > for all practical purposes let us assume c=0;
> > at the end point s=1000mts
> > substituting in eq 1
> > t=6.9 sec.
> >
> >
> > -----Original Message-----
> > From: Vijaykumar Nayak (RBIN/EDC1) *
> > [mailto:Vijaykumar.Nayak@...]
> > Sent: Friday, July 19, 2002 1:42 PM
> > To: Puzzle (E-mail)
> > Subject: [puzzles] Puzz
>
> >
> > A man starts with  some initial velocity...He has to
> > travel a distance of 1km.
> >
> > He has decided to travel at  a the velocity  such
> > that,
> >
> > at each point of time the distance he has travelled
> > from the starting point is his velocity.
> >
> > how much time does he need to cover the disatnce?
> >
> >
>
>
> =====
> "Right now I'm having amnesia and deja vu
> at the same time. I think I've forgotten
> this before".
>
> __________________________________________________
> Do You Yahoo!?
> Yahoo! Health - Feel better, live better
> http://health.yahoo.com
>

--
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• So, finally college algebra makes sense :) ... From: Vijaykumar Nayak (RBIN/EDC1) * To: puzzles@yahoogroups.com Sent: Wednesday, July 31, 2002 9:09 AM
Message 11 of 25 , Jul 30, 2002
• 0 Attachment
So, finally college algebra makes sense :)
----- Original Message -----
Sent: Wednesday, July 31, 2002 9:09 AM
Subject: RE: [puzzles] Puzz

Yes,
The unit controversy can be solved if we only stick to (Only one) mks system of units.
as i had given the distance in meters and time must be measured in seconds.
one can't say i will substitute 1km in place of S because, the unit followed in here is meters for distance ,and hence the result observed is in seconds.
There is good reason as to why a smaller unit gets us larger time, thats because even though we change the unit of distance from meters to cm we stick on to the unit of time as second.As  for mathematics replacing meters by centimeters will only result in changing 1000 units to 1 unit.
I agree that the constant  of integration can't be chosen 0 without any satisfactory initial condition ,so the problem has to be changed such that C=0, that can be done if we assume that he starts from 1 and travels up to 1000 meters.
now as
t=ln(s)+C
at t=0 s=1 =>
c=0;

-----Original Message-----
From: Gandhar Gokhale [mailto:gandhar@...]
Sent: Wednesday, July 31, 2002 9:11 AM
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

Hi,

I'm not convinced about putting the integration constant to 0 without sufficient reason. We use boundary conditions (or some other known conditions) to derive the value of the constant.

Also, unit should not play any role unlike the case here. You use 1000 meter and substitute it in the place of s. But if I use unit as km, I've to put s = 1. This gives t = 0. i.e., distance traveled is 1 km at the begining only!!! Instead, if I use cm as the unit, t will become 11.51!!. The smaller the unit, larger will be the time! Completely absurd. Could you guys please correct me if I'm wrong?

Thanks and regards,
Gandhar

----- Original Message -----
Date: Tue, 30 Jul 2002 08:48:53 -0700 (PDT)
To: puzzles@yahoogroups.com
Subject: RE: [puzzles] Puzz

> Hi,
> sorry for chipping in so late, but I find a more
> fundamental problem with your solution.
>
> When you equate
> s = ds / dt
> you are ignoring the unit of time (is it seconds,
> hours, or what?).
>
> So the correct answer would be 6.9 units. It could be
> 6.9 secs or 6.9 hours!
>
> To get a unique answer the problem should state under
> what units the distance and velocity are equal (in
> magnitude).
>
> regards,
>
> --- "Vijaykumar Nayak (RBIN/EDC1) *"
> <Vijaykumar.Nayak@...> wrote:
> > Here is the solution to the puzzle...
> > By the problem statement
> > s->distance travelled
> > t-> time taken
> > => s = v  at every instant of time
> > => s =ds / dt
> > =>dt = (1/s) ds
> > integrating on both sides of the eq.
> > t= ln(s) +c---------1
> > for all practical purposes let us assume c=0;
> > at the end point s=1000mts
> > substituting in eq 1
> > t=6.9 sec.
> >
> >
> > -----Original Message-----
> > From: Vijaykumar Nayak (RBIN/EDC1) *
> > [mailto:Vijaykumar.Nayak@...]
> > Sent: Friday, July 19, 2002 1:42 PM
> > To: Puzzle (E-mail)
> > Subject: [puzzles] Puzz
>
> >
> > A man starts with  some initial velocity...He has to
> > travel a distance of 1km.
> >
> > He has decided to travel at  a the velocity  such
> > that,
> >
> > at each point of time the distance he has travelled
> > from the starting point is his velocity.
> >
> > how much time does he need to cover the disatnce?
> >
> >
>
>
> =====