Measuring The Future?

[Motor Daddy]

A train traveled down tracks that were measured and marked like a ruler. When the train contacted a specific mark on the tracks (starting line), a timer started. The train traveled down the tracks for a duration of time and the timer stopped. The distance the train traveled from the start of the timer to the stop of the timer was 200 meters. The timer indicated a total duration of the event of 10 seconds. The following distances and times were noted:

 

1 second-20 meter mark

2 seconds-40 meter mark

3 seconds-60 meter mark

4 seconds-80 meter mark

5 seconds-100 meter mark

6 seconds-120 meter mark

7 seconds-140 meter mark

8 seconds-160 meter mark

9 seconds-180 meter mark

10 seconds-200 meter mark

 

After the entire train passed the 200 meter mark on the tracks the train slowed down, pulled into the train station, and stopped.

 

During the event there was a guy standing on the back of the train facing the opposite direction of the train's travel. He decided to throw a ball off the back of the train, back towards the starting line. He released the ball at the 100 meter mark and the ball remained 100 meters away from the start line.

 

Q: What is the ball's velocity relative to the train? (negating friction and gravity for the example)

[DrP]

He threw the ball so that it came to a halt then right? Therefore it stops on the track and has no velocity wrt the track. So it should have a velocity of 20 m/s wrt to the train (as the train is going 20 m/s). (oh and zero if you mean after the train has continued on and come to a halt) To do this however hwe would have to throw the ball at 20m/s wrt the train (as the ball ON the train is already going 20 m/s in the direction of the train) backward, thus cancelling out the balls velocity and so it remains level with the track.

 

IF!!!! he threw it fowards at 20 m/s (ignoring friction gravity etc..) the ball would continue to travel at 20m/s away from the train but would actually be going 40 m/s wrt the track.

 

 

To answer your question (as it seems like a trick) I'll say 0m/s because the train comes to a halt at the station and the ball has stopped also. If you mean whilst the train is still moving (i.e. before it pulls into the station) then 20m/s

 

 

 

I am finding that I want to type the following sentance to end this post:

 

"Whats the hidden catch then? What crap are you going to come out with now?"

 

But I don't want to be rude.

 

 

 

 

PS - someone please correct me if I'm wrong here - it seems REALLY simple, but I may have missed something. I have a couple of questions myself about relative velocities as you approach C, but will will post them at a later date if thats OK. I thought I understood it but have got confused reading some of the recent threads regarding the matter.

Edited June 23, 2008 by DrP

[Sisyphus]

Yup, 20m/s.

 

Well, what's the catch?

[Edtharan]

As we are dealing with velocities far below the speed of light, we can simplify the whole thing by using classical mechanics. This of course only gives an approximation, but this approximation would have an error about the same size (or less) then we would get from the effects of quantum mechanics (uncertainty principle).

[/disclaimer]

 

Q: What is the ball's velocity relative to the train? (negating friction and gravity for the example)

Firstly we must consider vectors. A vector is a composite mathematical entity. That is it is made up of multiple components. A vector has a magnitude which is always expressed as a positive number, and it has a direction.

 

Any object that has a description that includes a direction and the magnitude of that direction can be said to have a vector.

 

It is possible to separate the components of a vector and treat them separately. However, in doing so you have to be careful how you do it and under what circumstances you do. Leaving out a particular component can lead to several errors that, if the full vector is used would produce a different answer.

 

The train is indicated as moving along the tracks in a specific direction. This direction is given as "Down the tracks". So we can use this as a reference. This means that the term "Up the tracks" will be a direction 180 degrees away from the direction "Down the tracks".

 

This Up or Down the tracks does not indicate a speed in any way, only a direction. We could be using compass bearings, but as none was given in the example, we will use these directions instead.

 

The train is also stated (well calculated) as having a specific speed. This is the magnitude of the direction the train is travelling in. So the train can be said to have a vector which we will call its velocity.

 

This defines several important terms:

Speed: The magnitude of a velocity vector.

Facing: The direction component of the velocity vector.

Velocity: A vector that has the components of Speed and Facing.

 

For instance a train that was not moving relative to the tracks could be said to have a speed of 0, and a direction down the tracks. It could also be said under that specific circumstance (a speed of 0) to also have a direction up the tracks, but as the train is indicated as moving down the tracks we can assume that a train will typically be moving in the direction it is facing and so use Down the tracks as the direction that the train is facing in. As common linguistic usage would then indicate, we will us ether default of the direction an object is facing as a common usage for its direction. Thus when referencing a train at rest relative to the tracks we can say that it is moving at 0m/s down the tracks, even though it would be perfectly valid to also state that it was moving at 0m/s up the tracks.

 

This only applies when the train is at rest relative to the tracks, as any movement would give a vector and that vector would then contain the direction of importance.

 

Now at various point within the scenario the train is behaving differently. At some points it is in constant motion relative to the tracks, at others it is accelerating (getting faster or slowing down) and it is also at rest relative to the tracks.

 

This gives us 3 frames of reference with which to examine the relative motions of the ball, train and tracks.

 

1) Constant velocity: The train is moving at a relative velocity of 20m/s (speed) down the tracks (facing). This frame exists while the train is between the start line and the end line.

 

2) Acceleration: Although it is safe to assume that the train accelerated prior to crossing the start line, this is not specified as being part of the scenario. So although we could perform calculations for the frames of reference before the train crosses the start line, we are not given any details with which to do so. And, as these frames of reference are placed out side the scenario in question, we can safely ignore them for the sake of this scenario and simplicity.

 

However, we are give details of the train after it crosses the finish line and these are considered as part of the scenario as they are included in the description. Thus this situation counts as another frame of reference.

 

This is a distinct frame of reference that is fundamentally different from a constant velocity frame as objects within an accelerating frame can perform experiments that indicate that their velocity is changing. Such simple ones as a plum-bob (a weight on the end of a string) will indicate a force is acting on the subjects within that frame of reference.

 

3) Constant Velocity: Finally, the scenario covers the train when it is at rest relative to the tracks. This is a constant velocity frame of reference as the velocity is not changing, even though it is at 0.

 

The scenario states that the train stops changing its velocity when it's speed relative to the tracks reaches 0. As no further information about the behaviour of the train past this point is given, we can therefore assume that the behaviour of any part of the scenario after that point is irrelevant to the scenario and not of interest. It can therefore safely be ignored.

 

Ok, now on to the answers:

For the first frame of reference (1), the train is stated to travel a distance of 20 metres of track in 1 second. Using the formula Speed = Distance / Time, we can therefore calculate that the speed of the train is 20m/s relative to the tracks

 

It is also stated that the direction of the train at this point is "Down the tracks".

 

So we can combine this information with the speed to give a velocity of: 20m/s relative to the tracks, down the tracks.

 

An event is stated that at the 100m mark of the track whereby the ball was released with a velocity of 0m/s relative to the tracks.

 

Prior to this event, the ball is assumed to have the same velocity (speed and facing) as the train as no details were given other wise and the ball is stated to be carried by someone on the train.

 

After this event the ball will be considered separate from the train as it is moving differently to the train. At which point we will consider it as its own entity.

 

As stated above, if the ball has a velocity of 0m/s relative to the tracks, then the direction of this vector can be any direction with equal validity. It could even be considered as perpendicular (at right angles) to the tracks and still be equally valid.

 

However, its velocity relative to the train is different. As its speed relative to the train is not 0m/s we must consider its facing component of it vector.

 

As this direction is not given, we have to calculate it. This involves vector addition.

 

We know the velocity (vector) of the ball relative to the track (Speed 0m/s direction any) and the velocity (vector) of the train relative to the track. We can then add the vectors of the train to the track and the track to the ball to calculate the vector of the ball relative to the train.

 

The vector of the train at this point in time (the frame of reference of constant velocity after the ball was released and before the train crosses the finish line) is 20m/s down the track relative to the track. This give the ball a vector of 20m/s up the track relative to the train.

 

Consequently, the train can also be said to have a velocity of 20m/s down the track, relative to the ball.

 

As the ball has a velocity of 0m/s relative to the tracks, then as a check the ball should have the same relative velocity compared to the train as the track does. And as we can see, it does.

 

In the second frame of reference (2), the train is undergoing acceleration. This means that the velocity (a vector) is changing. This change has both a magnitude (the rate of acceleration) and a direction (which direction this change is taking place in). Therefore acceleration is also a vector.

 

However, instead of being a descriptor of a constant velocity, this vector is a descriptor of how the velocity is changing over time.

 

Just like you can have a normal distance divided by time, you can also have a vector divided by time. It is possible to multiply vectors (and therefore divide as well) by a scalar (a number without a direction). This means it is possible to have a velocity divided by time, and since a velocity contains a distance (its magnitude), then this becomes a rate of change, or in other words an acceleration.

 

Distance divided by time is Speed. Speed divided by time is acceleration. A velocity already has a Speed, so dividing it by time divides the Speed by time.

 

In the details of the scenario we are not told the rate at such the train is accelerating, nor are we given the details necessary to compute it. To do so we need to know over what time this acceleration happened (as we already know the velocity of the train, we could use that and the time over which the train accelerated to calculate the rate of acceleration). As we have not been given these details, we can assume that they are irrelevant to the scenario in question.

 

What we can say, however, is that the train was at one point travelling at 20m/s down the tracks relative to the tracks, and at another, later, point it was travelling at 0m/s down the tracks relative to the tracks (using the default direction of the trains facing based on common linguistic convention, not necessarily a mathematical one).

 

This means that its velocity relative to the tracks has decreased. This could be due to one of two things. First, the train might have applied its acceleration direction up the tracks until such time as the relative velocity between the train and the tracks become 0m/s, or that the tracks accelerated down the tracks relative to the first frame of reference until the relative velocities of the the train and the tracks become 0m/s.

 

There are many other descriptions of this event depending on the frame of reference you use. We could have used the ball, the station or even a frame of reference of the train prior to crossing the starting line. However, linguistically, we tend to use the description of the event that is easiest to communicate.

 

In such descriptions, we must always remember, the description is meant to be communicated easily, not necessarily precisely (as this post should attest as I am trying to communicate accurately rather than easily and it is taking a lot more time).

 

Because it is easier linguistically, to use terms such as "decelerate", or "negative velocity", they aren't strictly speaking mathematically correct.

 

Now, as the frame of reference of the train contains a changing velocity and we can't calculate the rate of change in velocity, then we can't actually state what the velocity of the ball is relative to the train. However, as the ball is not the object that is experiencing an acceleration, we can easily give it's velocity relative to another non accelerating frame of reference, say the track for instance.

 

In the scenario the ball is stated to have a velocity of 0m/s relative to the tracks, so therefore we can state that the ball has a relative velocity of 0m/s relative to the tracks.

 

But, as you want a velocity relative to the train, we can't actually easily do this as you have not provided enough information in your scenario description to allow us to do this.

 

Finally, for the third frame of reference (3), the train is at rest relative to the tracks. This means its velocity is 0m/s down the tracks (again, using the linguistic default of the train as explained earlier).

 

We also know that at this point the velocity of the ball relative to the tracks is also 0m/s (with the direction unimportant at this point of the scenario).

 

It is a simple matter of vector addition to arrive at the solution of 0m/s down the track relative to the train.

 

So to answer your question: "What is the ball's velocity relative to the train?"

 

Before the ball is thrown it has a relative velocity of 0m/s down the track relative to the train. After the ball is thrown and before the train passes the finish line, the relative velocity of the ball to the train is 20m/s up the track relative to the train. While the train is slowing down (relative to the track) we don't have enough information needed within the scenario description to give an exact answer to this situation. And, after the train has reached a velocity of 0m/s down the track relative to the track, the ball has a relative velocity to the train of 0m/s down the track relative to the train.

[Sisyphus]

Edtharan, that's an extremely long-winded explanation, which though correct (as far as I can see, I just skimmed), might lead to some problems. I have a *hunch* that despite the small velocities, this is actually somehow a relativity question, meaning you shouldn't be adding vectors like that. As it happens, there are really only two reference frames in the problem (tracks and train, the ball stated explicitly as belonging to the track frame), so it happens to work out the same.

[ydoaPs]

What is the future?

[Motor Daddy]

What is the future?

 

Anticipated motion that has yet to occur.

[ydoaPs]

What does it mean to occur?

[Sayonara]

You can't use "anticipated" in the definition of future, because its own definition is dependent on the concept already being available.

[Motor Daddy]

You can't use "anticipated" in the definition of future, because its own definition is dependent on the concept already being available.

 

Example: I am on a train traveling to NY that has already departed Baltimore. I am anticipating the arrival at the NY station, but I have yet to arrive.

 

Is this a word game now? Do you have an answer to my original post?

 

What does it mean to occur?

 

To happen.

[Klaynos]

Example: I am on a train traveling to NY that has already departed Baltimore. I am anticipating the arrival at the NY station, but I have yet to arrive.

 

Is this a word game now? Do you have an answer to my original post?

 

 

 

To happen.

 

You've not replied to anyone else who's answered why should anyone else bother? And can anyone else say irony?

[ydoaPs]

What does it mean to happen?

[Motor Daddy]

You've not replied to anyone else who's answered why should anyone else bother? And can anyone else say irony?

 

I was waiting for a larger response. I want to see different individual's answer before I start my explanation.

 

It seems some people disregard the original post, and wait for a chance to pounce. Almost like an ambush. What is your answer to the original question?

 

What does it mean to happen?

 

I have no idea.

[Sayonara]

Example: I am on a train traveling to NY that has already departed Baltimore. I am anticipating the arrival at the NY station, but I have yet to arrive.

The definition of anticipation refers to events in the future, so the word itself can't be used to define the word "future".

 

Stop acting like an idiot and just say "yes, I should have chosen my words more carefully. My mistake."

 

Is this a word game now?

Seems to be your speciality.

 

Do you have an answer to my original post?

No, since I have no intention of addressing your original post at this time, irrespective of the fact that I will post whatever objections I see fit as and when you say something which is patently false.

[Klaynos]

Any reply I make would be broadly similar to the first 3 replies but I suspect you're going to try a little game so I honestly can't be bothered to work it through myself... I suspect you'll do frame mixing or create errors where the acceleration is.

[Motor Daddy]

The definition of anticipation refers to events in the future, so the word itself can't be used to define the word "future".

 

Stop acting like an idiot and just say "yes, I should have chosen my words more carefully. My mistake."

 

 

Seems to be your speciality.

 

 

No, since I have no intention of addressing your original post at this time, irrespective of the fact that I will post whatever objections I see fit as and when you say something which is patently false.

 

My bad.

 

Yes, I should have chosen my words more carefully. My mistake.

[ydoaPs]

How are you doing on my word problem?

[Motor Daddy]

How are you doing on my word problem?

 

I don't understand the meanings of the words, so I can't do it.

 

And can anyone else say irony?

 

How's that for irony?

Edited June 25, 2008 by Motor Daddy

[Edtharan]

I have a *hunch* that despite the small velocities, this is actually somehow a relativity question, meaning you shouldn't be adding vectors like that.

You must have skimmed it . And you are right, however I stated that I was only using classical mechanics even though I should be using Relativistic mechanics, but due to the small speeds, the differences would not be too drastic.

 

Anticipated motion that has yet to occur.

You did not state this as part of the scenario or as part of the question.

 

Linguistically, the way you have written that scenario and the question, the answer should be 20m/s up the tracks relative to the train, as the question about what speed the ball is relative to the train occurs after the scenario stated that the train was moving 20m/s down the tracks relative to the tracks and the ball was released at 0m/s relative to the tracks.

 

It seems some people disregard the original post, and wait for a chance to pounce. Almost like an ambush. What is your answer to the original question?

I gave my response to your question and answered quite extensively. From teh thread title and the working of your question (the thread title is something about the future, but the wording of the question has nothing in it about the future), I was uncertain of how they linked together. So I answered the question and also included temporal references so that you could use them better.

 

So if you want an answer that is supposed to be a prediction of the future, then: Once the train has arrived at the station and reached a relative speed of 0m/s relative to the tracks, then if the train or ball does not accelerate in any direction, then the relative speeds between them will remain at 0m/s.

[Motor Daddy]

You must have skimmed it . And you are right, however I stated that I was only using classical mechanics even though I should be using Relativistic mechanics, but due to the small speeds, the differences would not be too drastic.

 

 

You did not state this as part of the scenario or as part of the question.

 

Linguistically, the way you have written that scenario and the question, the answer should be 20m/s up the tracks relative to the train, as the question about what speed the ball is relative to the train occurs after the scenario stated that the train was moving 20m/s down the tracks relative to the tracks and the ball was released at 0m/s relative to the tracks.

 

 

I gave my response to your question and answered quite extensively. From teh thread title and the working of your question (the thread title is something about the future, but the wording of the question has nothing in it about the future), I was uncertain of how they linked together. So I answered the question and also included temporal references so that you could use them better.

 

So if you want an answer that is supposed to be a prediction of the future, then: Once the train has arrived at the station and reached a relative speed of 0m/s relative to the tracks, then if the train or ball does not accelerate in any direction, then the relative speeds between them will remain at 0m/s.

 

Thanks for the answer. I appreciate it.