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macaw
2010-Mar-15, 05:31 AM
pzkpfw: split from a Q&A thread where it began (with me also at fault) to dominate.



Considering yourself to be at rest just means looking at things from your own perspective. This can be seen with just Galilean/Newtonian physics. If you are driving your car down a road at 30 miles an hour and another car passes you at 40 miles an hour, then from your perspective, you can consider yourself stationary and say the other car is passing you at 70 miles an hour .

...umm, you mean 10 miles/hr, right?

pzkpfw
2010-Mar-15, 06:25 AM
...umm, you mean 10 miles/hr, right?

We've been taking that as 30 and 40 mph in different directions. So when one was considered zero, the other was 70 mph.

(i.e. not taking "passing" as meaning "overtaking".)

macaw
2010-Mar-15, 02:29 PM
We've been taking that as 30 and 40 mph in different directions. So when one was considered zero, the other was 70 mph.

(i.e. not taking "passing" as meaning "overtaking".)

Passing means overtaking.

Jeff Root
2010-Mar-15, 02:51 PM
Passing means not failing, except when you are passing away.
Then passing means failing, as death overtakes you.

-- Jeff, in Minneapolis
__________________

korjik
2010-Mar-15, 03:40 PM
Passing means overtaking.

No, it dosent

macaw
2010-Mar-15, 06:08 PM
No, it dosent

In American English it does, grav is American.

NEOWatcher
2010-Mar-15, 06:17 PM
Passing means overtaking.No, it dosent
That might explain all those slow drivers in the passing lane.

macaw
2010-Mar-15, 06:20 PM
That might explain all those slow drivers in the passing lane.

Yes, it might explain :-)

pzkpfw
2010-Mar-15, 07:30 PM
In American English it does, grav is American.

You are nit-picking rather than admit you were wrong. Everyone in this thread, American or otherwise, was using a not-overtaking meaning for "passing". That's why everyone, not just grav, was happy to accept 70 mph as the relative speed between the two vehicles. e.g. see the diagram by WayneFrancis in post #42.

Even then, a google search for the very specific phrase ["passing a vehicle going the other way"] gave some hits, some of which do appear to be American in origin.

(More amusingly, the specific search for ["passing a roadside obstacle"] gets some hits whereas there are no hits for the specific phrase ["overtaking a roadside obstacle"] (there are some, for pages with that combination of words).)


What this shows, of course, is that language can be imprecise and open to misinterpretation, if one is not careful.

NEOWatcher
2010-Mar-15, 08:08 PM
Even then, a google search for the very specific phrase ["passing a vehicle going the other way"] gave some hits, some of which do appear to be American in origin.
"Passing each other" produces 1.8 million.


What this shows, of course, is that language can be imprecise and open to misinterpretation, if one is not careful.
Absolutely.
It's one of my peeves when people say that a word is bad. It's usage is usally what makes it bad. (with the exception of swear words and a few others)

Passing too. Look at how many entries are in Dictionary.com (http://dictionary.reference.com/browse/passing)

korjik
2010-Mar-15, 09:36 PM
In American English it does, grav is American.

Passing means 'in the act of going past.' It is direction independent.

TheHalcyonYear
2010-Mar-16, 12:59 AM
When in doubt, as here, ask the op what the relative velocity between the two vehicles is. The answer should straighten out the semantic uncertainty left by the grammatical ambiguity.

macaw
2010-Mar-16, 06:00 AM
Passing means 'in the act of going past.' It is direction independent.

The resulting speed is obviously not. You get "closing speed" of 70mph in one case vs. "separation speed" of 10mph in the other.

macaw
2010-Mar-16, 06:07 AM
You are nit-picking rather than admit you were wrong. Everyone in this thread, American or otherwise, was using a not-overtaking meaning for "passing". That's why everyone, not just grav, was happy to accept 70 mph as the relative speed between the two vehicles. e.g. see the diagram by WayneFrancis in post #42.

Yes, I missed the diagram. It is very clear. My mistake.
I went by grav's statement which was ambigous.




What this shows, of course, is that language can be imprecise and open to misinterpretation, if one is not careful.

Yes, language tends to be imprecise. This is why I insist on mathematical formulations, whenever possible.

TheHalcyonYear
2010-Mar-16, 06:13 AM
The resulting speed is obviously not. You get "closing speed" of 70mph in one case vs. "separation speed" of 10mph in the other.
This is a foolish argument since with relative frames of reference the two are identical. The "separation speed" of 10 mph could be a closing speed of 10 mph and the "closing speed" of 70 mph could be a "separation speed" of 70mph depending on one's frame of reference. Hence the conclusion: passing is direction independent.

macaw
2010-Mar-16, 06:33 AM
This is a foolish argument since with relative frames of reference the two are identical. The "separation speed" of 10 mph could be a closing speed of 10 mph and the "closing speed" of 70 mph could be a "separation speed" of 70mph depending on one's frame of reference. Hence the conclusion: passing is direction independent.

Err, no. "Separation" and "closing" speeds are always measured from the perspective of the same frame, the road.
In the example, if the cars move in the same sense ("direction" was not even correct, they both have the same direction, the direction of the road), then to 40mph car "closes" onto the 30mph car at 40-30=10mph.
On the other head, if the cars move in opposite senses, either towards each other or away from each other, their "separation" speed will be 40+30=70mph.

Jens
2010-Mar-16, 06:35 AM
I think it's clear that the fact that the phrasing can be ambiguous is quite well illustrated by the fact that we're even discussing it. As TheHalcyonYear quite well stated, the proper thing to do is just to ask the OP for clarification, which has already been done. English (and any other language really) is full of that kind of ambiguity. If you say, "he is standing in front of the car," and the car is facing away from me, does that mean he is standing between me and the car or is the car between the two of us? It could be either.

macaw
2010-Mar-16, 06:40 AM
I think it's clear that the fact that the phrasing can be ambiguous is quite well illustrated by the fact that we're even discussing it. As TheHalcyonYear quite well stated, the proper thing to do is just to ask the OP for clarification, which has already been done. English (and any other language really) is full of that kind of ambiguity. If you say, "he is standing in front of the car," and the car is facing away from me, does that mean he is standing between me and the car or is the car between the two of us? It could be either.

This is why physics is best expressed in terms of math and not in terms of natural language(s).

TheHalcyonYear
2010-Mar-16, 07:00 AM
Err, no. "Separation" and "closing" speeds are always measured from the perspective of the same frame, the road.
In the example, if the cars move in the same sense ("direction" was not even correct, they both have the same direction, the direction of the road), then to 40mph car "closes" onto the 30mph car at 40-30=10mph.
On the other head, if the cars move in opposite senses, either towards each other or away from each other, their "separation" speed will be 40+30=70mph.
But that's the point of the math. The equations for all three references, car1, car2, the road, are all valid. There is no "favored" reference. In the first scenario, does a passenger in car1 know that the vehicle is traveling 30mph? How does a passenger in car1 know that car1 isn't motionless and the road is moving? How does a passenger in car1 know whether both the road and car1 aren't both moving relative to each other.

That's the beauty of the math, from all three references, the car2 passes car1 at 10mph.We don't need to know any more than that.


[INDENT]From the point of view of car1, is car2 approaching it at 10mph or is

grapes
2010-Mar-16, 11:36 AM
My daughter is taking a calculus course where the homework is checked by computer, and you can re-submit. One problem caused her a lot of consternation, she checked and re-checked her work, it was a more complicated version of something like: A car position x (meters) along a path is x(t) = 2x - x2 for t (seconds) >=0, what is its deceleration at time t=2? The accepted answer was 2 m/s/s

Strange
2010-Mar-16, 11:51 AM
When in doubt, as here, ask the op what the relative velocity between the two vehicles is. The answer should straighten out the semantic uncertainty left by the grammatical ambiguity.

Possibly not, in the case of this OP :)

macaw
2010-Mar-16, 03:16 PM
But that's the point of the math. The equations for all three references, car1, car2, the road, are all valid. There is no "favored" reference.

Did I say anywhere that there was a "favored" reference frame? I simply explained to notions of "closing" and "separation" speeds.

macaw
2010-Mar-16, 05:07 PM
A car position x (meters) along a path is x(t) = 2x - x2 for t (seconds) >=0,

You mean x(t)=2t-t2, correct? It cannot be : x(t) = 2x - x2 because that would be an equation in x.

Then:

dx/dt=2-2t

d^x/dt^2=-2

Calculus gives the solution instantaneously and painlessly.

TheHalcyonYear
2010-Mar-16, 05:48 PM
Did I say anywhere that there was a "favored" reference frame? I simply explained to notions of "closing" and "separation" speeds.
But that's the point. Since there isn't any favored reference frame, there is no notion of "closing" or "separation" because with out the concept of a "favored" reference frame, there are only relative velocities. Each observer may consider their relative velocity as 0 which means there is no passing.

macaw
2010-Mar-16, 05:54 PM
Since there isn't any favored reference frame, there is no notion of "closing" or separation" except from the point of view of the observer.

Incorrect. The "closing" speed can be viewed from the perspective of any inertial frame.
The ground is a more "convenient" frame because cars have speedometers that measure their respective speeds wrt it. When we say "car A travels at 30mph" we mean wrt the ground. That's all.

TheHalcyonYear
2010-Mar-16, 07:14 PM
Incorrect. The "closing" speed can be viewed from the perspective of any inertial frame.
The ground is a more "convenient" frame because cars have speedometers that measure their respective speeds wrt it. When we say "car A travels at 30mph" we mean wrt the ground. That's all.
I have come to the conclusion that you are now just arguing for the sake of arguing. Oh kiddo, you do it your way and the rest of us will do physics.

macaw
2010-Mar-16, 07:37 PM
I have come to the conclusion that you are now just arguing for the sake of arguing. Oh kiddo, you do it your way and the rest of us will do physics.

I tried to teach you, I give up.

pzkpfw
2010-Mar-16, 07:45 PM
... d^x/dt^2=-2

Calculus gives the solution instantaneously and painlessly.

I think the point might have been that the accepted answer was 2 not -2.

Taeolas
2010-Mar-16, 08:11 PM
-2 units per second (or whatever the units were) indicates it is slowing down; a negative acceleration. Since it is already asking for deceleration, you need to toggle the signs, so it is decelerating at 2 units per second.

cjl
2010-Mar-16, 08:31 PM
I have come to the conclusion that you are now just arguing for the sake of arguing. Oh kiddo, you do it your way and the rest of us will do physics.

The closing velocity between two objects should be observable and identical regardless of the frame from which the system is being observed. The velocity of each object may vary depending on frame, but the velocity of the first object with respect to the second (the closing velocity) should be the same.

TheHalcyonYear
2010-Mar-16, 09:00 PM
The closing velocity between two objects should be observable and identical regardless of the frame from which the system is being observed. The velocity of each object may vary depending on frame, but the velocity of the first object with respect to the second (the closing velocity) should be the same.
The poster is making a distinction between "closing velocity" and "separation velocity". Since there is no preferred frame of reference, there is no preferred velocity for the observer. If the observer is in the car, the observed velocity may be the road passing underneath the car or the car passing over the road; the speedometer will give the same reading in either case.

So, if the observer makes the assumption that their velocity is 0, then "closing velocity" is the "speed" of the object before it passes the observer and the "separation velocity" is the "speed" of the object after it passes the observer. However, since there is no "preferred" frame of reference, we cannot determine an absolute velocity of car1, car2, or the road. This may seem a bit trite with respect to cars on a highway, but it's most assuredly not when discussing astronomical bodies. However, it is true for both cases.

pzkpfw
2010-Mar-16, 10:05 PM
-2 units per second (or whatever the units were) indicates it is slowing down; a negative acceleration. Since it is already asking for deceleration, you need to toggle the signs, so it is decelerating at 2 units per second.

Exactly.

My point being that macaw, too, would have had trouble getting the computerised homework exercise to accept his answer.

grapes
2010-Mar-16, 11:25 PM
You mean x(t)=2t-t2, correct? Oops, yes, thanks.

I think the point might have been that the accepted answer was 2 not -2.Yes, that's part of the point I was making. I had her try it as soon as I read the problem (which was just a bit more complicated than the problem I gave above, so she struggled to find her error).

-2 units per second (or whatever the units were) indicates it is slowing down; a negative acceleration.
That's the other part. In that problem, it is definitely not slowing down. :)

macaw
2010-Mar-17, 05:18 AM
Exactly.

My point being that macaw, too, would have had trouble getting the computerised homework exercise to accept his answer.

Too bad for the computer, I would have felt sorry for it.
This is also why none of my three cars has any chip in them. :-)

TheHalcyonYear
2010-Mar-17, 06:10 AM
Exactly.

My point being that macaw, too, would have had trouble getting the computerised homework exercise to accept his answer.
Computerization is the least of macaw's problems. Without a basic understanding of the concept "frames of reference" they are going to have a lot of difficulty with physics whether it be on a computer, calculator, or (as in the case of my father's generation) an old-fashioned slide-rule. :)

macaw
2010-Mar-21, 03:44 AM
Computerization is the least of macaw's problems. Without a basic understanding of the concept "frames of reference" )

:lol:

Snake Charmer
2010-Mar-22, 04:09 PM
-2 units per second (or whatever the units were) indicates it is slowing down; a negative acceleration.

v(t)=2-2t

So at t=1.9, the velocity is -1.8, at t=2.0, it has slowed down to a velocity of -2, and at t=2.1, it has slowed down further still to a velocity of -2.2. At t=10, it will be going really slowly, with a velocity of -18.


The poster is making a distinction between "closing velocity" and "separation velocity". Since there is no preferred frame of reference, there is no preferred velocity for the observer. If the observer is in the car, the observed velocity may be the road passing underneath the car or the car passing over the road; the speedometer will give the same reading in either case.

So, if the observer makes the assumption that their velocity is 0, then "closing velocity" is the "speed" of the object before it passes the observer and the "separation velocity" is the "speed" of the object after it passes the observer. However, since there is no "preferred" frame of reference, we cannot determine an absolute velocity of car1, car2, or the road. This may seem a bit trite with respect to cars on a highway, but it's most assuredly not when discussing astronomical bodies. However, it is true for both cases.

It seems pretty clear to me that by "closing velocity", macaw means the speed of two objects relative to each other, not relative to some other observer. If you read it in this way, everything makes perfect sense. cjl has already pointed out that the closing velocity, interpreted in this way, between two objects does not depend on the motion of the person observing the two objects, even though the velocities of each of the two objects does. That's all true according to Newton, the only way the closing velocity is frame-dependent is if you want to start invoking time and distance dilation effects. If we assume the observer isn't streaking past the two cars at a substantial fraction of the speed of light, then as cjl has pointed out, macaw's "closing velocity" is frame independent.