A and B are 2 positions in coordinate system, particles p,q are at A
and B respectively. p moves perpendicular to AB, q is directed towards
p always, both have speed= v m/s, distance b/w them initially is d,
find the distance b/w them at time=infinity.

plz help me solve this problem, or giving various ideas
regards,
Divij

On Jul 1, 5:07 am, Sam Wormley wrote:
Divij Rao wrote:
A and B are 2 positions in coordinate system, particles p,q are at A
and B respectively. p moves perpendicular to AB, q is directed towards
p always, both have speed= v m/s, distance b/w them initially is d,
find the distance b/w them at time=infinity.

Infinity is not a point in time.
Pick some value for time and add one.

but the question stands that way. it means, at the end, when the
particles are almost collinear to A, the distance b/w them is _______.
one may imagine q is on A-p, position A and particle p.

one idea that i have is that from the frame of reference of p, q is
moving in a straight line towards itself....but even thn how to
proceed?

thanks for trying.

regards,
Divij

On 30 juin, 14:17, Divij Rao wrote:
A and B are 2 positions in coordinate system, particles p,q are at A
and B respectively. p moves perpendicular to AB, q is directed towards
p always, both have speed= v m/s, distance b/w them initially is d,
find the distance b/w them at time=infinity.

plz help me solve this problem, or giving various ideas
regards,
Divij

Very interesting problem.

One thing is certain, q will never catch up with p since they
have the same velocity.

First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

Hope this helps.

André Michaud

First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

i dont know so much...in simpler words...plz

Hope this helps.
didnot understand the meaning .
André Michaud

thank you,
Divij

On 8 juil, 07:50, Divij Rao wrote:
First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

i dont know so much...in simpler words...plz

Hope this helps.

didnot understand the meaning .

André Michaud

thank you,
Divij

Sorry for the delay, I just saw your post.

I requote your question here for convenience:

A and B are 2 positions in coordinate system, particles p,q are at A
and B respectively. p moves perpendicular to AB, q is directed towards
p always, both have speed= v m/s, distance b/w them initially is d,
find the distance b/w them at time=infinity.

I will try to make you see what I conclude.

Imagine position A and B as lying on a horizontal line (x axis ?)

you have p initially located at A and q initially located at B

the distance between p and q is d, and both p and q have the
same velocity, so imagine p and q tied together by a string,
which will force q to always move towards p as p moves away.

now if you set the particles in motion at velocity v, p is going
to move upwards vertically with q being pulled along, always
moving towards p.

Now, the string is mentioned only so that you can clearly
see how q can always point towards p.

At the beginning, q will move almost horizontally but will
progressively angle upwards since it is following p which
is moving vertically upwards.

After some time, q will end up moving vertically as it
trails behind p. since both have the same velocity,
q will be unable to catch up with p and will forever
follow behind.

You are dealing with uniform motion for both p and q

I did not do the actual math, so I don't know whether
q will catch up some with p. I expect so, but I really
wouldn't know without doing the actual math.

André Michaud

On Jul 10, 9:47 pm, wrote:
On 8 juil, 07:50, Divij Rao wrote:

First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

i dont know so much...in simpler words...plz

Hope this helps.

didnot understand the meaning .

André Michaud

thank you,
Divij

Sorry for the delay, I just saw your post.

I requote your question here for convenience:

A and B are 2 positions in coordinate system, particles p,q are at A
and B respectively. p moves perpendicular to AB, q is directed towards
p always, both have speed= v m/s, distance b/w them initially is d,
find the distance b/w them at time=infinity.

I will try to make you see what I conclude.

Imagine position A and B as lying on a horizontal line (x axis ?)

you have p initially located at A and q initially located at B

the distance between p and q is d, and both p and q have the
same velocity, so imagine p and q tied together by a string,
which will force q to always move towards p as p moves away.

now if you set the particles in motion at velocity v, p is going
to move upwards vertically with q being pulled along, always
moving towards p.

Now, the string is mentioned only so that you can clearly
see how q can always point towards p.

At the beginning, q will move almost horizontally but will
progressively angle upwards since it is following p which
is moving vertically upwards.

After some time, q will end up moving vertically as it
trails behind p. since both have the same velocity,
q will be unable to catch up with p and will forever
follow behind.

You are dealing with uniform motion for both p and q

I did not do the actual math, so I don't know whether
q will catch up some with p. I expect so, but I really
wouldn't know without doing the actual math.

André Michaud

excellent idea!
but i m still not sure... suppose p has moved a distance x, after some
time say t, by speed, we cant say that distance between p and q is the
same, as per the xample, the string MAY become loose. maybe u r
correct.
thanks a lot, all comments will be appreciated.
Regards,
Divij

On 15 juil, 01:43, Divij Rao wrote:
On Jul 10, 9:47 pm, wrote:



On 8 juil, 07:50, Divij Rao wrote:

First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

i dont know so much...in simpler words...plz

Hope this helps.

didnot understand the meaning .

AndréMichaud

thank you,
Divij

Sorry for the delay, I just saw your post.

I requote your question here for convenience:

A and B are 2 positions in coordinate system, particles p,q are
at A and B respectively. p moves perpendicular to AB, q is
directed towards p always, both have speed= v m/s, distance
b/w them initially is d, find the distance b/w them at time=infinity.

I will try to make you see what I conclude.

Imagine position A and B as lying on a horizontal line (x axis ?)

you have p initially located at A and q initially located at B

the distance between p and q is d, and both p and q have the
same velocity, so imagine p and q tied together by a string,
which will force q to always move towards p as p moves away.

now if you set the particles in motion at velocity v, p is going
to move upwards vertically with q being pulled along, always
moving towards p.

Now, the string is mentioned only so that you can clearly
see how q can always point towards p.

At the beginning, q will move almost horizontally but will
progressively angle upwards since it is following p which
is moving vertically upwards.

After some time, q will end up moving vertically as it
trails behind p. since both have the same velocity,
q will be unable to catch up with p and will forever
follow behind.

You are dealing with uniform motion for both p and q

I did not do the actual math, so I don't know whether
q will catch up some with p. I expect so, but I really
wouldn't know without doing the actual math.

AndréMichaud

excellent idea!

but i m still not sure... suppose p has moved a distance x,
after some time say t, by speed, we cant say that distance
between p and q is the same, as per the xample, the string
MAY become loose.

That was my point, precisely. You must do the calculation
to ascertain the final distance between p and q.

maybe u r correct.

What is uncertain in my mind, without doing the actual
calculation, is whether or not on the first leg of the motion
p will come any closer to p, but I am positive that once
q tends towards moving vertically (following p since it is always
pointing towards p) the distance will tend to become stable,
and remain stable, since both have the same velocity.

André Michaud

thanks a lot, all comments will be appreciated.
Regards,
Divij

On Jul 15, 11:03 pm, wrote:
On 15 juil, 01:43, Divij Rao wrote:





On Jul 10, 9:47 pm, wrote:

On 8 juil, 07:50, Divij Rao wrote:

First hunch is that you are dealing with an asymptote,
with q trailing behind p at AB - deltaAB

i dont know so much...in simpler words...plz

Hope this helps.

didnot understand the meaning .

AndréMichaud

thank you,
Divij

Sorry for the delay, I just saw your post.

I requote your question here for convenience:

A and B are 2 positions in coordinate system, particles p,q are
at A and B respectively. p moves perpendicular to AB, q is
directed towards p always, both have speed= v m/s, distance
b/w them initially is d, find the distance b/w them at time=infinity.

I will try to make you see what I conclude.

Imagine position A and B as lying on a horizontal line (x axis ?)

you have p initially located at A and q initially located at B

the distance between p and q is d, and both p and q have the
same velocity, so imagine p and q tied together by a string,
which will force q to always move towards p as p moves away.

now if you set the particles in motion at velocity v, p is going
to move upwards vertically with q being pulled along, always
moving towards p.

Now, the string is mentioned only so that you can clearly
see how q can always point towards p.

At the beginning, q will move almost horizontally but will
progressively angle upwards since it is following p which
is moving vertically upwards.

After some time, q will end up moving vertically as it
trails behind p. since both have the same velocity,
q will be unable to catch up with p and will forever
follow behind.

You are dealing with uniform motion for both p and q

I did not do the actual math, so I don't know whether
q will catch up some with p. I expect so, but I really
wouldn't know without doing the actual math.

AndréMichaud

excellent idea!

but i m still not sure... suppose p has moved a distance x,
after some time say t, by speed, we cant say that distance
between p and q is the same, as per the xample, the string
MAY become loose.

That was my point, precisely. You must do the calculation
to ascertain the final distance between p and q.

maybe u r correct.

What is uncertain in my mind, without doing the actual
calculation, is whether or not on the first leg of the motion
p will come any closer to p, but I am positive that once
q tends towards moving vertically (following p since it is always
pointing towards p) the distance will tend to become stable,
and remain stable, since both have the same velocity.

André Michaud



thanks a lot, all comments will be appreciated.
Regards,
Divij- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

yes, the distabnce bet them becomes constant after time infinity, but
how to calculate that?
can resolving the speeds of q along p be useful?

if so how?