Can the average speed of a moving body be 0? [on hold]If 2 bodies are moving at equal speed parallel, and one...
Representing an indicator function: binary variables and "indicator constraints"
Subgroup generated by a subgroup and a conjugate of it
How to use the passive form to say "This flower was watered."
Why is su world executable?
Heyawacky: Ace of Cups
What if a restaurant suddenly cannot accept credit cards, and the customer has no cash?
Build a mob of suspiciously happy lenny faces ( ͡° ͜ʖ ͡°)
Output with the same length always
Are there any rules on how characters go from 0th to 1st level in a class?
What is the best way to use errors in mathematica M12?
Parse a simple key=value config file in C
Is this bar slide trick shown on Cheers real or a visual effect?
If it isn't [someone's name]!
Eric Andre had a dream
Regression when x and y each have uncertainties
Is a suspension needed to do wheelies?
Are there any OR challenges that are similar to kaggle's competitions?
Why does this image of cyclocarbon look like a nonagon?
The Shaman wandering spirit Lore's Arcane Enlightenment hex grants bonus arcane spells. Is there a spell level limit to what spells can be chosen?
μονάδαι as plural form of μονάς
Rotate List by K places
Icon is not displayed in lwc
Why do aircraft leave cruising altitude long before landing just to circle?
Do I need to start off my book by describing the character's "normal world"?
Can the average speed of a moving body be 0? [on hold]
If 2 bodies are moving at equal speed parallel, and one throws something at a right angle to itself, will the thing thrown hit the other body?Is it possible to have a rate of change of acceleration?If change in position over time is average velocity, why doesn't change in position over time squared equal average acceleration?Can an object *immediately* start moving at a high velocity?Physical meaning of average accelerationQuestion regarding average velocityAre my observations about the differences between average velocity and speed, in 1 dimension, correct?How can I understand if an object stay (zero velocity) or moving with constant velocity (zero acceleration)Is it possible to have a situation in which velocity of the particle is never 0 but its average velocity in an interval is 0?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
My book states that- "Average speed of a moving body can never be 0 except when time approaches infinity"
I understand that Average speed = (Total dist./ Total time) and that if time is infinity, average speed will be 0 (Anything divided by infinity gives 0)
But I would like to creatively and intuitively understand what does t=infinity mean.
kinematics
New contributor
$endgroup$
put on hold as unclear what you're asking by Bill N, Jon Custer, ZeroTheHero, John Rennie, stafusa 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 6 more comments
$begingroup$
My book states that- "Average speed of a moving body can never be 0 except when time approaches infinity"
I understand that Average speed = (Total dist./ Total time) and that if time is infinity, average speed will be 0 (Anything divided by infinity gives 0)
But I would like to creatively and intuitively understand what does t=infinity mean.
kinematics
New contributor
$endgroup$
put on hold as unclear what you're asking by Bill N, Jon Custer, ZeroTheHero, John Rennie, stafusa 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
4
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
5
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
2
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
1
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago
|
show 6 more comments
$begingroup$
My book states that- "Average speed of a moving body can never be 0 except when time approaches infinity"
I understand that Average speed = (Total dist./ Total time) and that if time is infinity, average speed will be 0 (Anything divided by infinity gives 0)
But I would like to creatively and intuitively understand what does t=infinity mean.
kinematics
New contributor
$endgroup$
My book states that- "Average speed of a moving body can never be 0 except when time approaches infinity"
I understand that Average speed = (Total dist./ Total time) and that if time is infinity, average speed will be 0 (Anything divided by infinity gives 0)
But I would like to creatively and intuitively understand what does t=infinity mean.
kinematics
kinematics
New contributor
New contributor
New contributor
asked 2 days ago
Happy UnicornHappy Unicorn
224 bronze badges
224 bronze badges
New contributor
New contributor
put on hold as unclear what you're asking by Bill N, Jon Custer, ZeroTheHero, John Rennie, stafusa 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Bill N, Jon Custer, ZeroTheHero, John Rennie, stafusa 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Bill N, Jon Custer, ZeroTheHero, John Rennie, stafusa 17 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
4
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
5
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
2
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
1
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago
|
show 6 more comments
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
4
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
5
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
2
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
1
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
4
4
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
5
5
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
2
2
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
1
1
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago
|
show 6 more comments
5 Answers
5
active
oldest
votes
$begingroup$
$t=infty$ is a not a useful statement in this context and should be replaced by the time tends towards infinity, $tto infty$, and then the average speed tends towards zero.
So the average speed can get smaller and smaller and smaller but given that you will always be dividing a number (distance travelled) by a very much larger number (time taken) the average speed can never be exactly equal to zero.
$endgroup$
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
add a comment |
$begingroup$
First of all, as others have pointed out, $ t=infty$ should be replaced by 'as time approaches infinity' ($t rightarrow infty$).
Now, the average speed will approach zero if the total distance covered doesn't approach infinity as time approaches infinity. For example, if something is moving at a constant speed then as time approaches infinity the distance covered will also approach infinity and the average speed will remain that constant, not zero.
But for the case when total distance covered is finite even in infinite time then surely, it will be moving at a very, very slow speed. Hence it's average speed is also very small, approaching zero.
EDIT: as people have pointed out in the comments, the average speed may approach zero even if distance approach infinity, if it's approaching it slower than time. E.g. $s=sqrt t$
$endgroup$
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
add a comment |
$begingroup$
There are a number of ways to answer this, and the answers would depend on how we interpret specific terms. Although we normally use precise meanings sometimes we are not always consistent.
In the book's statement it is clear that if an object moves with constant velocity (being the same speed and direction) then it's average speed will always be the same irrespective of how much time has passed.
So a more precise way of stating what the book means would be to say:
The average speed of an object that has moved can never be 0, but will approach 0 as the time over which we are taking the average speed approaches inifinty.
How are we measuring the time and the distance to identify the average speed? If we have an object in linear motion on the x axis with the displacement from its starting point given by $sin(t)$ then every $pi$ seconds it will be back where it started and at that instant we might legitimately say that it's average speed is 0.
Normally we talk about velocity rather than speed (velocity has a direction (vector value), speed is just an amount (scalar value)) and as velocity (in a linear direction) can be negative it's clear that an average velocity can be zero.
We also might measure distance from the starting point at a particular instant, rather than considering the total distance travelled. In the example where the position is found from $sin(t)$ the position may be the same as the starting point, but the total distance travelled will never be zero.
So an over-simplified statement often raises more questions than it attempts to answer.
$endgroup$
add a comment |
$begingroup$
Consider an ideal pendulum or an orbit, which both have symmetrical motions. If direction is taken into account (velocity), then everything cancels/balances out. The pendulum bob or planet have a fixed average position. Neither really "goes anywhere" over time, but both certainly have a non-zero average speed.
Velocity is the derivative of displacement (position), and speed is the derivative of the absolute displacement (distance). Taking the absolute value prevents the "balancing" part, so any object that has ever moved will have a non-zero speed from that time forward.
Look at it another way: If an object has moved, when will its average speed be zero? Mathematically, when time approaches infinity. Interpreting the mathematical result as a physical result, you can say that the answer is "never"
New contributor
$endgroup$
add a comment |
$begingroup$
A few basic things before we play with the limit of $ttoinfty$. If an object ever moves then the distance traveled by the object has to be a positive finite number, let's say, $s$. By convention, we take that we are analyzing the motion over a period of time starting $t=0$ up until some time $t$. Now, by definition, the average speed is $frac{s}{t}$. As already pointed out by another answer, in an actual experiment, we always deal with a finite value of $t$. Thus, $frac{s}{t}$ is bound to be a finite positive quantity.
However, in physics, there are several situations in which the time over which we observe a system can be very very large as compared to the typical/intrinsic timescale associated with the situation at hand. This motivates us to think carefully about the limit $ttoinfty$ as it can often simplify the problem greatly and/or can reveal interesting features of the system. In short, while it doesn't really make any physical sense to say that $t=infty$, there exist very many motivations (both theoretical and pragmatic) to study the limit $ttoinfty$. So, let's try to analyze what can happen to the average speed, $frac{s}{t}$ in this limit $ttoinfty$.
Case I: $s$ remains finite as $ttoinfty$
This is an easily imaginable case. The particle can either simply stop moving after a certain instance of time, trivially leading to a finite distance traveled even as $ttoinfty$. Or, a particle's motion can be such that it keeps moving more and more slowly as time increases in such a fashion that the distance covered is finite even as $ttoinfty$. For example, if a particle moves according to the equation $x=tan^{-1}t$ then the total distance traveled as $ttoinfty$ is only $frac{pi}{2}$. Thus, the average speed tends to $0$ as $ttoinfty$. Notice that this particle never stops but its speed, $frac{d}{dt}tan^{-1}t=frac{1}{t^2+1}$ keeps decreasing as $ttoinfty$ as expected.
Case II: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}to 0$
The distance traveled by the particle can also approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "slower" than $t$ approaches $infty$ then the ratio $frac{s}{t}$ would approach zero. For example, if the motion of a particle is described by $x=sqrt{t}$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, since a square-root growth is slower than linear growth, the ratio $frac{s}{t}=frac{sqrt{t}}{t}=frac{1}{sqrt{t}}to 0$ as $ttoinfty$.
Case III: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}$ remains finite
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "at a comparable rate" to with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would remain finite. For example, if the motion of a particle is described by $x=3t$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, both $s$ and $t$ grow linearly in $t$, the ratio $frac{s}{t}=frac{{3t}}{t}=3to 3$ as $ttoinfty$.
Case IV: $s$ also $toinfty$ as $ttoinfty$ and $frac{s}{t}toinfty$
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ in such a way that it approaches $infty$ "at a faster rate" than with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would also diverge. For example, if the motion of a particle is described by $x=t^2$ then clearly the distance traveled by the particle diverges as $ttoinfty$. And, since a quadratic growth is faster than a linear growth, the ratio $frac{s}{t}=frac{{t^2}}{t}=tto infty$ as $ttoinfty$.
So, as it happens in Case I and Case II, the average speed can tend to zero as time approaches infinity. But, for any finite interval of time, the average speed cannot be zero as long as the particle has ever moved. Moreover, notice that even in the limit of time approaching infinity, the average speed can also remain finite as in Case III or it can also diverge as in Case IV. Which of these cases will be realized depends on the nature of the motion (trajectory).
$endgroup$
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$t=infty$ is a not a useful statement in this context and should be replaced by the time tends towards infinity, $tto infty$, and then the average speed tends towards zero.
So the average speed can get smaller and smaller and smaller but given that you will always be dividing a number (distance travelled) by a very much larger number (time taken) the average speed can never be exactly equal to zero.
$endgroup$
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
add a comment |
$begingroup$
$t=infty$ is a not a useful statement in this context and should be replaced by the time tends towards infinity, $tto infty$, and then the average speed tends towards zero.
So the average speed can get smaller and smaller and smaller but given that you will always be dividing a number (distance travelled) by a very much larger number (time taken) the average speed can never be exactly equal to zero.
$endgroup$
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
add a comment |
$begingroup$
$t=infty$ is a not a useful statement in this context and should be replaced by the time tends towards infinity, $tto infty$, and then the average speed tends towards zero.
So the average speed can get smaller and smaller and smaller but given that you will always be dividing a number (distance travelled) by a very much larger number (time taken) the average speed can never be exactly equal to zero.
$endgroup$
$t=infty$ is a not a useful statement in this context and should be replaced by the time tends towards infinity, $tto infty$, and then the average speed tends towards zero.
So the average speed can get smaller and smaller and smaller but given that you will always be dividing a number (distance travelled) by a very much larger number (time taken) the average speed can never be exactly equal to zero.
answered 2 days ago
FarcherFarcher
55.6k3 gold badges45 silver badges120 bronze badges
55.6k3 gold badges45 silver badges120 bronze badges
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
add a comment |
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
2
2
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Certainly $ttoinfty$ isn't sufficient for the average velocity to go to $0$?
$endgroup$
– Aaron Stevens
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
$begingroup$
Why assume that time taken is "very much larger" than distance traveled? Wouldn't that depend entire on the average velocity during that infinite time? This seems to pre-suppose that the distance traveled doesn't also approach infinity, which would really be pre-supposing that average speed tends towards zero to conclude that it tends towards zero.
$endgroup$
– JMac
2 days ago
5
5
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@JMac I am trying to explain is that physically there is no such thing as an infinite time or an infinite distance. So as long as long as a body moves it cannot have an exactly zero average speed. So one finite number will be divided by another finite number and even if the rate at which the denominator increase is much greater than the rate at which the numerator increases the fraction can never equal zero.
$endgroup$
– Farcher
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
$begingroup$
@Farcher "Can get smaller and smaller." not "Gets smaller and smaller". Yup, I see how I misinterpreted it. Good explanation.
$endgroup$
– JMac
2 days ago
1
1
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
$begingroup$
→ is the symbol for limit; the only way in which "infinity" is useful. - Does an equation containing infinity not equal 0 or infinity exist?
$endgroup$
– Mazura
2 days ago
add a comment |
$begingroup$
First of all, as others have pointed out, $ t=infty$ should be replaced by 'as time approaches infinity' ($t rightarrow infty$).
Now, the average speed will approach zero if the total distance covered doesn't approach infinity as time approaches infinity. For example, if something is moving at a constant speed then as time approaches infinity the distance covered will also approach infinity and the average speed will remain that constant, not zero.
But for the case when total distance covered is finite even in infinite time then surely, it will be moving at a very, very slow speed. Hence it's average speed is also very small, approaching zero.
EDIT: as people have pointed out in the comments, the average speed may approach zero even if distance approach infinity, if it's approaching it slower than time. E.g. $s=sqrt t$
$endgroup$
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
add a comment |
$begingroup$
First of all, as others have pointed out, $ t=infty$ should be replaced by 'as time approaches infinity' ($t rightarrow infty$).
Now, the average speed will approach zero if the total distance covered doesn't approach infinity as time approaches infinity. For example, if something is moving at a constant speed then as time approaches infinity the distance covered will also approach infinity and the average speed will remain that constant, not zero.
But for the case when total distance covered is finite even in infinite time then surely, it will be moving at a very, very slow speed. Hence it's average speed is also very small, approaching zero.
EDIT: as people have pointed out in the comments, the average speed may approach zero even if distance approach infinity, if it's approaching it slower than time. E.g. $s=sqrt t$
$endgroup$
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
add a comment |
$begingroup$
First of all, as others have pointed out, $ t=infty$ should be replaced by 'as time approaches infinity' ($t rightarrow infty$).
Now, the average speed will approach zero if the total distance covered doesn't approach infinity as time approaches infinity. For example, if something is moving at a constant speed then as time approaches infinity the distance covered will also approach infinity and the average speed will remain that constant, not zero.
But for the case when total distance covered is finite even in infinite time then surely, it will be moving at a very, very slow speed. Hence it's average speed is also very small, approaching zero.
EDIT: as people have pointed out in the comments, the average speed may approach zero even if distance approach infinity, if it's approaching it slower than time. E.g. $s=sqrt t$
$endgroup$
First of all, as others have pointed out, $ t=infty$ should be replaced by 'as time approaches infinity' ($t rightarrow infty$).
Now, the average speed will approach zero if the total distance covered doesn't approach infinity as time approaches infinity. For example, if something is moving at a constant speed then as time approaches infinity the distance covered will also approach infinity and the average speed will remain that constant, not zero.
But for the case when total distance covered is finite even in infinite time then surely, it will be moving at a very, very slow speed. Hence it's average speed is also very small, approaching zero.
EDIT: as people have pointed out in the comments, the average speed may approach zero even if distance approach infinity, if it's approaching it slower than time. E.g. $s=sqrt t$
edited yesterday
answered 2 days ago
ScienciscoSciencisco
37610 bronze badges
37610 bronze badges
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
add a comment |
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
$begingroup$
The total distance covered doesn't need to be finite for the average speed to be zero in the limit $ttoinfty$. It can just approach infinity slower than $t$. For example, the average speed for $x=sqrt{t}$ as $ttoinfty$ is zero despite the fact that $x$ also approaches $infty$ as $ttoinfty$.
$endgroup$
– Dvij Mankad
yesterday
add a comment |
$begingroup$
There are a number of ways to answer this, and the answers would depend on how we interpret specific terms. Although we normally use precise meanings sometimes we are not always consistent.
In the book's statement it is clear that if an object moves with constant velocity (being the same speed and direction) then it's average speed will always be the same irrespective of how much time has passed.
So a more precise way of stating what the book means would be to say:
The average speed of an object that has moved can never be 0, but will approach 0 as the time over which we are taking the average speed approaches inifinty.
How are we measuring the time and the distance to identify the average speed? If we have an object in linear motion on the x axis with the displacement from its starting point given by $sin(t)$ then every $pi$ seconds it will be back where it started and at that instant we might legitimately say that it's average speed is 0.
Normally we talk about velocity rather than speed (velocity has a direction (vector value), speed is just an amount (scalar value)) and as velocity (in a linear direction) can be negative it's clear that an average velocity can be zero.
We also might measure distance from the starting point at a particular instant, rather than considering the total distance travelled. In the example where the position is found from $sin(t)$ the position may be the same as the starting point, but the total distance travelled will never be zero.
So an over-simplified statement often raises more questions than it attempts to answer.
$endgroup$
add a comment |
$begingroup$
There are a number of ways to answer this, and the answers would depend on how we interpret specific terms. Although we normally use precise meanings sometimes we are not always consistent.
In the book's statement it is clear that if an object moves with constant velocity (being the same speed and direction) then it's average speed will always be the same irrespective of how much time has passed.
So a more precise way of stating what the book means would be to say:
The average speed of an object that has moved can never be 0, but will approach 0 as the time over which we are taking the average speed approaches inifinty.
How are we measuring the time and the distance to identify the average speed? If we have an object in linear motion on the x axis with the displacement from its starting point given by $sin(t)$ then every $pi$ seconds it will be back where it started and at that instant we might legitimately say that it's average speed is 0.
Normally we talk about velocity rather than speed (velocity has a direction (vector value), speed is just an amount (scalar value)) and as velocity (in a linear direction) can be negative it's clear that an average velocity can be zero.
We also might measure distance from the starting point at a particular instant, rather than considering the total distance travelled. In the example where the position is found from $sin(t)$ the position may be the same as the starting point, but the total distance travelled will never be zero.
So an over-simplified statement often raises more questions than it attempts to answer.
$endgroup$
add a comment |
$begingroup$
There are a number of ways to answer this, and the answers would depend on how we interpret specific terms. Although we normally use precise meanings sometimes we are not always consistent.
In the book's statement it is clear that if an object moves with constant velocity (being the same speed and direction) then it's average speed will always be the same irrespective of how much time has passed.
So a more precise way of stating what the book means would be to say:
The average speed of an object that has moved can never be 0, but will approach 0 as the time over which we are taking the average speed approaches inifinty.
How are we measuring the time and the distance to identify the average speed? If we have an object in linear motion on the x axis with the displacement from its starting point given by $sin(t)$ then every $pi$ seconds it will be back where it started and at that instant we might legitimately say that it's average speed is 0.
Normally we talk about velocity rather than speed (velocity has a direction (vector value), speed is just an amount (scalar value)) and as velocity (in a linear direction) can be negative it's clear that an average velocity can be zero.
We also might measure distance from the starting point at a particular instant, rather than considering the total distance travelled. In the example where the position is found from $sin(t)$ the position may be the same as the starting point, but the total distance travelled will never be zero.
So an over-simplified statement often raises more questions than it attempts to answer.
$endgroup$
There are a number of ways to answer this, and the answers would depend on how we interpret specific terms. Although we normally use precise meanings sometimes we are not always consistent.
In the book's statement it is clear that if an object moves with constant velocity (being the same speed and direction) then it's average speed will always be the same irrespective of how much time has passed.
So a more precise way of stating what the book means would be to say:
The average speed of an object that has moved can never be 0, but will approach 0 as the time over which we are taking the average speed approaches inifinty.
How are we measuring the time and the distance to identify the average speed? If we have an object in linear motion on the x axis with the displacement from its starting point given by $sin(t)$ then every $pi$ seconds it will be back where it started and at that instant we might legitimately say that it's average speed is 0.
Normally we talk about velocity rather than speed (velocity has a direction (vector value), speed is just an amount (scalar value)) and as velocity (in a linear direction) can be negative it's clear that an average velocity can be zero.
We also might measure distance from the starting point at a particular instant, rather than considering the total distance travelled. In the example where the position is found from $sin(t)$ the position may be the same as the starting point, but the total distance travelled will never be zero.
So an over-simplified statement often raises more questions than it attempts to answer.
edited yesterday
answered 2 days ago
Rob LambdenRob Lambden
1663 bronze badges
1663 bronze badges
add a comment |
add a comment |
$begingroup$
Consider an ideal pendulum or an orbit, which both have symmetrical motions. If direction is taken into account (velocity), then everything cancels/balances out. The pendulum bob or planet have a fixed average position. Neither really "goes anywhere" over time, but both certainly have a non-zero average speed.
Velocity is the derivative of displacement (position), and speed is the derivative of the absolute displacement (distance). Taking the absolute value prevents the "balancing" part, so any object that has ever moved will have a non-zero speed from that time forward.
Look at it another way: If an object has moved, when will its average speed be zero? Mathematically, when time approaches infinity. Interpreting the mathematical result as a physical result, you can say that the answer is "never"
New contributor
$endgroup$
add a comment |
$begingroup$
Consider an ideal pendulum or an orbit, which both have symmetrical motions. If direction is taken into account (velocity), then everything cancels/balances out. The pendulum bob or planet have a fixed average position. Neither really "goes anywhere" over time, but both certainly have a non-zero average speed.
Velocity is the derivative of displacement (position), and speed is the derivative of the absolute displacement (distance). Taking the absolute value prevents the "balancing" part, so any object that has ever moved will have a non-zero speed from that time forward.
Look at it another way: If an object has moved, when will its average speed be zero? Mathematically, when time approaches infinity. Interpreting the mathematical result as a physical result, you can say that the answer is "never"
New contributor
$endgroup$
add a comment |
$begingroup$
Consider an ideal pendulum or an orbit, which both have symmetrical motions. If direction is taken into account (velocity), then everything cancels/balances out. The pendulum bob or planet have a fixed average position. Neither really "goes anywhere" over time, but both certainly have a non-zero average speed.
Velocity is the derivative of displacement (position), and speed is the derivative of the absolute displacement (distance). Taking the absolute value prevents the "balancing" part, so any object that has ever moved will have a non-zero speed from that time forward.
Look at it another way: If an object has moved, when will its average speed be zero? Mathematically, when time approaches infinity. Interpreting the mathematical result as a physical result, you can say that the answer is "never"
New contributor
$endgroup$
Consider an ideal pendulum or an orbit, which both have symmetrical motions. If direction is taken into account (velocity), then everything cancels/balances out. The pendulum bob or planet have a fixed average position. Neither really "goes anywhere" over time, but both certainly have a non-zero average speed.
Velocity is the derivative of displacement (position), and speed is the derivative of the absolute displacement (distance). Taking the absolute value prevents the "balancing" part, so any object that has ever moved will have a non-zero speed from that time forward.
Look at it another way: If an object has moved, when will its average speed be zero? Mathematically, when time approaches infinity. Interpreting the mathematical result as a physical result, you can say that the answer is "never"
New contributor
New contributor
answered 2 days ago
Thomas ZwaagstraThomas Zwaagstra
1011 bronze badge
1011 bronze badge
New contributor
New contributor
add a comment |
add a comment |
$begingroup$
A few basic things before we play with the limit of $ttoinfty$. If an object ever moves then the distance traveled by the object has to be a positive finite number, let's say, $s$. By convention, we take that we are analyzing the motion over a period of time starting $t=0$ up until some time $t$. Now, by definition, the average speed is $frac{s}{t}$. As already pointed out by another answer, in an actual experiment, we always deal with a finite value of $t$. Thus, $frac{s}{t}$ is bound to be a finite positive quantity.
However, in physics, there are several situations in which the time over which we observe a system can be very very large as compared to the typical/intrinsic timescale associated with the situation at hand. This motivates us to think carefully about the limit $ttoinfty$ as it can often simplify the problem greatly and/or can reveal interesting features of the system. In short, while it doesn't really make any physical sense to say that $t=infty$, there exist very many motivations (both theoretical and pragmatic) to study the limit $ttoinfty$. So, let's try to analyze what can happen to the average speed, $frac{s}{t}$ in this limit $ttoinfty$.
Case I: $s$ remains finite as $ttoinfty$
This is an easily imaginable case. The particle can either simply stop moving after a certain instance of time, trivially leading to a finite distance traveled even as $ttoinfty$. Or, a particle's motion can be such that it keeps moving more and more slowly as time increases in such a fashion that the distance covered is finite even as $ttoinfty$. For example, if a particle moves according to the equation $x=tan^{-1}t$ then the total distance traveled as $ttoinfty$ is only $frac{pi}{2}$. Thus, the average speed tends to $0$ as $ttoinfty$. Notice that this particle never stops but its speed, $frac{d}{dt}tan^{-1}t=frac{1}{t^2+1}$ keeps decreasing as $ttoinfty$ as expected.
Case II: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}to 0$
The distance traveled by the particle can also approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "slower" than $t$ approaches $infty$ then the ratio $frac{s}{t}$ would approach zero. For example, if the motion of a particle is described by $x=sqrt{t}$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, since a square-root growth is slower than linear growth, the ratio $frac{s}{t}=frac{sqrt{t}}{t}=frac{1}{sqrt{t}}to 0$ as $ttoinfty$.
Case III: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}$ remains finite
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "at a comparable rate" to with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would remain finite. For example, if the motion of a particle is described by $x=3t$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, both $s$ and $t$ grow linearly in $t$, the ratio $frac{s}{t}=frac{{3t}}{t}=3to 3$ as $ttoinfty$.
Case IV: $s$ also $toinfty$ as $ttoinfty$ and $frac{s}{t}toinfty$
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ in such a way that it approaches $infty$ "at a faster rate" than with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would also diverge. For example, if the motion of a particle is described by $x=t^2$ then clearly the distance traveled by the particle diverges as $ttoinfty$. And, since a quadratic growth is faster than a linear growth, the ratio $frac{s}{t}=frac{{t^2}}{t}=tto infty$ as $ttoinfty$.
So, as it happens in Case I and Case II, the average speed can tend to zero as time approaches infinity. But, for any finite interval of time, the average speed cannot be zero as long as the particle has ever moved. Moreover, notice that even in the limit of time approaching infinity, the average speed can also remain finite as in Case III or it can also diverge as in Case IV. Which of these cases will be realized depends on the nature of the motion (trajectory).
$endgroup$
add a comment |
$begingroup$
A few basic things before we play with the limit of $ttoinfty$. If an object ever moves then the distance traveled by the object has to be a positive finite number, let's say, $s$. By convention, we take that we are analyzing the motion over a period of time starting $t=0$ up until some time $t$. Now, by definition, the average speed is $frac{s}{t}$. As already pointed out by another answer, in an actual experiment, we always deal with a finite value of $t$. Thus, $frac{s}{t}$ is bound to be a finite positive quantity.
However, in physics, there are several situations in which the time over which we observe a system can be very very large as compared to the typical/intrinsic timescale associated with the situation at hand. This motivates us to think carefully about the limit $ttoinfty$ as it can often simplify the problem greatly and/or can reveal interesting features of the system. In short, while it doesn't really make any physical sense to say that $t=infty$, there exist very many motivations (both theoretical and pragmatic) to study the limit $ttoinfty$. So, let's try to analyze what can happen to the average speed, $frac{s}{t}$ in this limit $ttoinfty$.
Case I: $s$ remains finite as $ttoinfty$
This is an easily imaginable case. The particle can either simply stop moving after a certain instance of time, trivially leading to a finite distance traveled even as $ttoinfty$. Or, a particle's motion can be such that it keeps moving more and more slowly as time increases in such a fashion that the distance covered is finite even as $ttoinfty$. For example, if a particle moves according to the equation $x=tan^{-1}t$ then the total distance traveled as $ttoinfty$ is only $frac{pi}{2}$. Thus, the average speed tends to $0$ as $ttoinfty$. Notice that this particle never stops but its speed, $frac{d}{dt}tan^{-1}t=frac{1}{t^2+1}$ keeps decreasing as $ttoinfty$ as expected.
Case II: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}to 0$
The distance traveled by the particle can also approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "slower" than $t$ approaches $infty$ then the ratio $frac{s}{t}$ would approach zero. For example, if the motion of a particle is described by $x=sqrt{t}$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, since a square-root growth is slower than linear growth, the ratio $frac{s}{t}=frac{sqrt{t}}{t}=frac{1}{sqrt{t}}to 0$ as $ttoinfty$.
Case III: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}$ remains finite
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "at a comparable rate" to with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would remain finite. For example, if the motion of a particle is described by $x=3t$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, both $s$ and $t$ grow linearly in $t$, the ratio $frac{s}{t}=frac{{3t}}{t}=3to 3$ as $ttoinfty$.
Case IV: $s$ also $toinfty$ as $ttoinfty$ and $frac{s}{t}toinfty$
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ in such a way that it approaches $infty$ "at a faster rate" than with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would also diverge. For example, if the motion of a particle is described by $x=t^2$ then clearly the distance traveled by the particle diverges as $ttoinfty$. And, since a quadratic growth is faster than a linear growth, the ratio $frac{s}{t}=frac{{t^2}}{t}=tto infty$ as $ttoinfty$.
So, as it happens in Case I and Case II, the average speed can tend to zero as time approaches infinity. But, for any finite interval of time, the average speed cannot be zero as long as the particle has ever moved. Moreover, notice that even in the limit of time approaching infinity, the average speed can also remain finite as in Case III or it can also diverge as in Case IV. Which of these cases will be realized depends on the nature of the motion (trajectory).
$endgroup$
add a comment |
$begingroup$
A few basic things before we play with the limit of $ttoinfty$. If an object ever moves then the distance traveled by the object has to be a positive finite number, let's say, $s$. By convention, we take that we are analyzing the motion over a period of time starting $t=0$ up until some time $t$. Now, by definition, the average speed is $frac{s}{t}$. As already pointed out by another answer, in an actual experiment, we always deal with a finite value of $t$. Thus, $frac{s}{t}$ is bound to be a finite positive quantity.
However, in physics, there are several situations in which the time over which we observe a system can be very very large as compared to the typical/intrinsic timescale associated with the situation at hand. This motivates us to think carefully about the limit $ttoinfty$ as it can often simplify the problem greatly and/or can reveal interesting features of the system. In short, while it doesn't really make any physical sense to say that $t=infty$, there exist very many motivations (both theoretical and pragmatic) to study the limit $ttoinfty$. So, let's try to analyze what can happen to the average speed, $frac{s}{t}$ in this limit $ttoinfty$.
Case I: $s$ remains finite as $ttoinfty$
This is an easily imaginable case. The particle can either simply stop moving after a certain instance of time, trivially leading to a finite distance traveled even as $ttoinfty$. Or, a particle's motion can be such that it keeps moving more and more slowly as time increases in such a fashion that the distance covered is finite even as $ttoinfty$. For example, if a particle moves according to the equation $x=tan^{-1}t$ then the total distance traveled as $ttoinfty$ is only $frac{pi}{2}$. Thus, the average speed tends to $0$ as $ttoinfty$. Notice that this particle never stops but its speed, $frac{d}{dt}tan^{-1}t=frac{1}{t^2+1}$ keeps decreasing as $ttoinfty$ as expected.
Case II: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}to 0$
The distance traveled by the particle can also approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "slower" than $t$ approaches $infty$ then the ratio $frac{s}{t}$ would approach zero. For example, if the motion of a particle is described by $x=sqrt{t}$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, since a square-root growth is slower than linear growth, the ratio $frac{s}{t}=frac{sqrt{t}}{t}=frac{1}{sqrt{t}}to 0$ as $ttoinfty$.
Case III: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}$ remains finite
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "at a comparable rate" to with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would remain finite. For example, if the motion of a particle is described by $x=3t$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, both $s$ and $t$ grow linearly in $t$, the ratio $frac{s}{t}=frac{{3t}}{t}=3to 3$ as $ttoinfty$.
Case IV: $s$ also $toinfty$ as $ttoinfty$ and $frac{s}{t}toinfty$
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ in such a way that it approaches $infty$ "at a faster rate" than with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would also diverge. For example, if the motion of a particle is described by $x=t^2$ then clearly the distance traveled by the particle diverges as $ttoinfty$. And, since a quadratic growth is faster than a linear growth, the ratio $frac{s}{t}=frac{{t^2}}{t}=tto infty$ as $ttoinfty$.
So, as it happens in Case I and Case II, the average speed can tend to zero as time approaches infinity. But, for any finite interval of time, the average speed cannot be zero as long as the particle has ever moved. Moreover, notice that even in the limit of time approaching infinity, the average speed can also remain finite as in Case III or it can also diverge as in Case IV. Which of these cases will be realized depends on the nature of the motion (trajectory).
$endgroup$
A few basic things before we play with the limit of $ttoinfty$. If an object ever moves then the distance traveled by the object has to be a positive finite number, let's say, $s$. By convention, we take that we are analyzing the motion over a period of time starting $t=0$ up until some time $t$. Now, by definition, the average speed is $frac{s}{t}$. As already pointed out by another answer, in an actual experiment, we always deal with a finite value of $t$. Thus, $frac{s}{t}$ is bound to be a finite positive quantity.
However, in physics, there are several situations in which the time over which we observe a system can be very very large as compared to the typical/intrinsic timescale associated with the situation at hand. This motivates us to think carefully about the limit $ttoinfty$ as it can often simplify the problem greatly and/or can reveal interesting features of the system. In short, while it doesn't really make any physical sense to say that $t=infty$, there exist very many motivations (both theoretical and pragmatic) to study the limit $ttoinfty$. So, let's try to analyze what can happen to the average speed, $frac{s}{t}$ in this limit $ttoinfty$.
Case I: $s$ remains finite as $ttoinfty$
This is an easily imaginable case. The particle can either simply stop moving after a certain instance of time, trivially leading to a finite distance traveled even as $ttoinfty$. Or, a particle's motion can be such that it keeps moving more and more slowly as time increases in such a fashion that the distance covered is finite even as $ttoinfty$. For example, if a particle moves according to the equation $x=tan^{-1}t$ then the total distance traveled as $ttoinfty$ is only $frac{pi}{2}$. Thus, the average speed tends to $0$ as $ttoinfty$. Notice that this particle never stops but its speed, $frac{d}{dt}tan^{-1}t=frac{1}{t^2+1}$ keeps decreasing as $ttoinfty$ as expected.
Case II: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}to 0$
The distance traveled by the particle can also approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "slower" than $t$ approaches $infty$ then the ratio $frac{s}{t}$ would approach zero. For example, if the motion of a particle is described by $x=sqrt{t}$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, since a square-root growth is slower than linear growth, the ratio $frac{s}{t}=frac{sqrt{t}}{t}=frac{1}{sqrt{t}}to 0$ as $ttoinfty$.
Case III: $s$ also $toinfty$ as $ttoinfty$ but $frac{s}{t}$ remains finite
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ but if it approaches $infty$ "at a comparable rate" to with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would remain finite. For example, if the motion of a particle is described by $x=3t$ then clearly the distance traveled by the particle also diverges as $ttoinfty$. But, both $s$ and $t$ grow linearly in $t$, the ratio $frac{s}{t}=frac{{3t}}{t}=3to 3$ as $ttoinfty$.
Case IV: $s$ also $toinfty$ as $ttoinfty$ and $frac{s}{t}toinfty$
The distance traveled by the particle can approach $infty$ in the limit $ttoinfty$ in such a way that it approaches $infty$ "at a faster rate" than with which $t$ approaches $infty$ then the ratio $frac{s}{t}$ would also diverge. For example, if the motion of a particle is described by $x=t^2$ then clearly the distance traveled by the particle diverges as $ttoinfty$. And, since a quadratic growth is faster than a linear growth, the ratio $frac{s}{t}=frac{{t^2}}{t}=tto infty$ as $ttoinfty$.
So, as it happens in Case I and Case II, the average speed can tend to zero as time approaches infinity. But, for any finite interval of time, the average speed cannot be zero as long as the particle has ever moved. Moreover, notice that even in the limit of time approaching infinity, the average speed can also remain finite as in Case III or it can also diverge as in Case IV. Which of these cases will be realized depends on the nature of the motion (trajectory).
answered yesterday
Dvij MankadDvij Mankad
6,7754 gold badges28 silver badges80 bronze badges
6,7754 gold badges28 silver badges80 bronze badges
add a comment |
add a comment |
$begingroup$
Look at this equation for the average velocity $overline {nu }=dfrac {1}{n}sum ^{n}_{i}dfrac {Delta s_{i}}{Delta t_{i}}$ delta t and delta s are for all intervals positive, so you never get average velocity zero
$endgroup$
– Eli
2 days ago
4
$begingroup$
As $tto infty$ doesn't total distance get larger?
$endgroup$
– jim
2 days ago
5
$begingroup$
Your title asks one question but in the text you ask another: "I would like to creatively and intuitively understand what does t=infinity mean." What's your real question?
$endgroup$
– Bill N
2 days ago
2
$begingroup$
@Eli If you are always moving in the same direction then the path won't be negative, but if you change direction then you could back track on yourself. The book mentioned a moving object but not how it's moving. If an object travels backwards and forwards in a straight line it's moving, but not always in the same direction. Velocity takes into account the direction, as well as the speed. If direction changes then you may find yout delta s becomes negative.
$endgroup$
– Rob Lambden
2 days ago
1
$begingroup$
The title's easy (from a frame of reference of zero velocity, it cannot not be); I got no clue what T equals though.
$endgroup$
– Mazura
2 days ago