Describing motion of a particle qualitatively

In summary: A good example is the function ##y = \ln x## where ##\dfrac{dy}{dx} = \dfrac 1 x##.So I was correct to find the angle goes to zero for the angle the finial velocity vector makes with the horizontal?Yes, that's what you meant.The angle that the finial position vector makes with the horizontal is the same as the angle the particle's final velocity vector makes with the horizontal.
  • #1
ChiralSuperfields
1,203
132
Homework Statement
See below
Relevant Equations
Position vector is the same as integral of velocity vector
For part (a) and (b) of this problem,
1670124771990.png

The solution is,
1670124812325.png

However, how did they arrive at their conclusion in part(b)?

As you can't graph it on a GC, I decide to imagine plugging in values for t, which I see that the 2t^3 grows quicker than the t^2 which is why I think they said that the particle will asymptote with the x-axis.

Anybody else have a different way at arriving at the conclusion given in part(b)? Feel free to let me know!

Many thanks!
 

Attachments

  • 1670124766276.png
    1670124766276.png
    8.1 KB · Views: 65
Physics news on Phys.org
  • #2
can you calculate the velocity vector at t=0?

Define the angle that the particle moves in the x-y plane as ##\theta## you have ## \tan \theta = \dfrac{v_y}{v_x}##. Now what is the limit of this angle as ## t \to \infty##?
 
  • Like
Likes ChiralSuperfields
  • #3
1670134241597.png
Thanks!
 
  • #4
  • Like
Likes ChiralSuperfields
  • #5
Callumnc1 said:
As you can't graph it on a GC, I decide to imagine plugging in values for t, which I see that the 2t^3 grows quicker than the t^2 which is why I think they said that the particle will asymptote with the x-axis.
You have the right idea, but it's not correct to say it "will asymptote with the x-axis" as that implies ##y \to 0## as ##x \to \infty##. In fact, there is no asymptote since both ##x## and ##y## grow without bound as ##t \to \infty##.
 
  • Like
Likes jbriggs444, ChiralSuperfields, PeroK and 1 other person
  • #6
Thanks for your replies @malawi_glenn and @vela! Sorry I think I evaluated the limit incorrectly, is this correct now?
1670139030980.png

Since this is the angle between the horizontal and the velocity vector of the particle after an infinite time, could someone please explain to me how the particle dose not asymptote with the x-axis if my equation is correct? I thought that the x-component of the velocity would increase faster than the y-component of the velocity since it is squared.

Many thanks!
 
  • #7
Callumnc1 said:
could someone please explain to me how the particle dose not asymptote with the x-axis if my equation is correct?

See velas reply.

(x(t), y(t)) = (2t^3 , t^2)

Can you draw this curve in the x-y plane?
 
  • Like
Likes ChiralSuperfields
  • #8
malawi_glenn said:
See velas reply.

(x(t), y(t)) = (2t^3 , t^2)

Can you draw this curve in the x-y plane?
Yes, you just in terms of the third variable t?
 
  • #9
Callumnc1 said:
Yes, you just in terms of the third variable t?
Hint: solve t in terms of x first, then find y in terms of x.
 
  • Like
Likes ChiralSuperfields
  • #10
malawi_glenn said:
Hint: solve t in terms of x first, then find y in terms of x.
Thanks for your reply @malawi_glenn ,
1670141286885.png
,

I put it into Desmos I can see that I dose not really asymptote with the x-axis. https://www.desmos.com/calculator/habtdbe3mt

However, what was wrong with the way I found the angle previously?

Many thanks!
 
  • #11
Callumnc1 said:
However, what was wrong with the way I found the angle previously?

You posted one attempt first, I said it was wrong. Then you gave the correct one.
 
  • Like
Likes ChiralSuperfields
  • #12
That the angle goes to zero does not mean that the x-axis becomes an asymptote. The simplest counter example would be the constant function y(x) = k with k different from 0. This clearly has y = k as the asymptote but the angle approaches 0 as x goes to infinity.
 
  • Like
Likes ChiralSuperfields
  • #13
Callumnc1 said:
Since this is the angle between the horizontal and the velocity vector of the particle after an infinite time, could someone please explain to me how the particle dose not asymptote with the x-axis if my equation is correct?
First, asymptote is a noun, not a verb. :) I think you mean the curve gets closer and closer to becoming parallel to the x-axis, which you showed by evaluating the limit, not that it's asymptotic to the x-axis.

Second, for the x-axis to be an asymptote, the particle's trajectory would have to get arbitrarily close the the x-axis without crossing it as ##x \to \infty##. Your results, however, shows that the particle gets farther from the x-axis as it moves.

It may seem counterintuitive that the curve can approach becoming horizontal yet still remain unbounded. It's one of the things you have to be careful about when dealing with infinity.
 
  • Like
Likes ChiralSuperfields and PeroK
  • #14
vela said:
Second, for the x-axis to be an asymptote, the particle's trajectory would have to get arbitrarily close the the x-axis without crossing it as x→∞.
My emphasis.

There is nothing stopping a function from crossing its asymptote. Consider ##\sin(x)/x## which has the ##x## axis as an asymptote yet crosses it an infinite number of times.
 
  • Like
Likes vela, ChiralSuperfields, SammyS and 1 other person
  • #15
vela said:
It may seem counterintuitive that the curve can approach becoming horizontal yet still remain unbounded.
A good example is the function ##y = \ln x## where ##\dfrac{dy}{dx} = \dfrac 1 x##.
 
  • Like
Likes ChiralSuperfields
  • #16
So I was correct to find the angle goes to zero for the angle the finial velocity vector makes with the horizontal? However, my interpretation of what that angle meant was wrong? What if I found the angle that the finial position vector makes with the horizontal?

Many thanks!
 
  • #17
Here is a parametric plot that says it all. Consider a point on the solid line to represent the position of the particle as it moves in the xy-plane. Describe the motion.

DescribetheMotion.png
 
  • Like
Likes ChiralSuperfields
  • #18
kuruman said:
Here is a parametric plot that says it all. Consider a point on the solid line to represent the position of the particle as it moves in the xy-plane. Describe the motion.

View attachment 318214
Thanks for your reply @kuruman ! The motion of the particle looks to me like half of a horizontal parabola. At small time values there is a rapid increase in y-values however, after larger time values the y and x position increase at the same rate. It looks like after large time values the parametric can be extrapolated as a straight line that makes an angle of 45 degrees with the horizontal.

Are you trying to say that I calculated the angle of the velocity vector after an infinite time interval incorrectly (post #6)?

Many thanks!
 
Last edited:
  • #19
Callumnc1 said:
Are you trying to say that I calculated the angle of the velocity vector after an infinite time interval incorrectly (post #6)?
Yes, I am. Just look at the velocity vector, ##\mathbf{v}=6t^2~\mathbf{\hat i}+2t~\mathbf{\hat j}.## In order for the particle to move parallel to the x-axis, the y-component of the velocity must decrease and approach zero as time increases. Actually, the y-component increases linearly with increasing time, so it will never go to zero. Of course, the x-component increases faster than the y-component which means, as the solution says, that the particle will turn to move closer to being parallel to the x-axis. However, it will never be parallel to the x-axis, even after infinite time has elapsed.
 
  • Like
Likes ChiralSuperfields
  • #20
kuruman said:
Yes, I am. Just look at the velocity vector, ##\mathbf{v}=6t^2~\mathbf{\hat i}+2t~\mathbf{\hat j}.## In order for the particle to move parallel to the x-axis, the y-component of the velocity must decrease and approach zero as time increases. Actually, the y-component increases linearly with increasing time, so it will never go to zero. Of course, the x-component increases faster than the y-component which means, as the solution says, that the particle will turn to move closer to being parallel to the x-axis. However, it will never be parallel to the x-axis, even after infinite time has elapsed.
Thanks for your reply @kuruman! How do you calculate the angle correctly (is it 45 degrees)? I should be able to find it using the velocity vector or position vector method?

Many thanks!
 
  • #21
If you are asking about the angle ##\theta## of the velocity vector relative to the x-axis, ##\tan\theta=\dfrac{v_y}{v_x}=\dfrac{1}{3t}.## It is equal to 45° at ##t=\frac{1}{3}## time units.
 
  • Like
Likes ChiralSuperfields
  • #22
kuruman said:
If you are asking about the angle ##\theta## of the velocity vector relative to the x-axis, ##\tan\theta=\dfrac{v_y}{v_x}=\dfrac{1}{3t}.## It is equal to 45° at ##t=\frac{1}{3}## time units.
Sorry @kuruman, I was asking what the velocity vector would be at t = infinity time units.

Many thanks!
 
  • #23
Callumnc1 said:
Sorry @kuruman, I was asking what the velocity vector would be at t = infinity time units.

Many thanks!
You know that ##\mathbf{v}=6t^2~\mathbf{\hat i}+2t~\mathbf{\hat j}.## Put in ##t=\infty## and see what you get.
 
  • Like
Likes ChiralSuperfields
  • #24
kuruman said:
You know that ##\mathbf{v}=6t^2~\mathbf{\hat i}+2t~\mathbf{\hat j}.## Put in ##t=\infty## and see what you get.
Sorry @kuruman, I think I have confused myself. I am not asking what the velocity vector is at time at t = infinity, but the angle the velocity vector makes with the horizontal at t = infinity.

Is this correct?
1670208349813.png

Manyh thanks!
 
  • #25
The fact remains that as long as the velocity has a non-zero y-component, the angle will never be zero but will have some finite value. You can argue that $$\lim_{t \rightarrow \infty} \left({\frac {1} {3t}}\right)=0$$all you want, but you cannot argue that the "##1##" in the numerator of the fraction is zero. It doesn't happen, not even for small values of ##1##.
 
  • Like
Likes ChiralSuperfields
  • #26
kuruman said:
The fact remains that as long as the velocity has a non-zero y-component, the angle will never be zero but will have some finite value. You can argue that $$\lim_{t \rightarrow \infty} \left({\frac {1} {3t}}\right)=0$$all you want, but you cannot argue that the "##1##" in the numerator of the fraction is zero. It doesn't happen, not even for small values of ##1##.
Thanks for your reply @kuruman!

So we are not really looking for the behavior of the position function for very larger values of time. So when they said 'describe the motion quantitively' they were really talking about for some values of t, in which case the angle that the position vector at some later time t is non-zero.

Many thanks!
 
  • #27
Callumnc1 said:
Thanks for your reply @kuruman!

So we are not really looking for the behavior of the position function for very larger values of time. So when they said 'describe the motion quantitively' they were really talking about for some values of t, in which case the angle that the position vector at some later time t is non-zero.

Many thanks!
Yes. The boxed answer in the solution for part (b) is what I would have written. I repeat, the angle is always non-zero for reasons I have already explained. Also, the statement of the problem says "describe the motion qualitatively" (emphasis mine), not quantitively.
 
  • Like
Likes ChiralSuperfields
  • #28
kuruman said:
Yes. The boxed answer in the solution for part (b) is what I would have written. I repeat, the angle is always non-zero for reasons I have already explained. Also, the statement of the problem says "describe the motion qualitatively" (emphasis mine), not quantitively.
Ok thanks @kuruman ! I will also see what happens when I find the angle the position vector makes with the horizontal for various time values.

Many thanks!
 
  • #29
Callumnc1 said:
Ok thanks @kuruman ! I will also see what happens when I find the angle the position vector makes with the horizontal for various time values.

Many thanks!
Ok thank you very much @kuruman !
 

1. What is meant by "motion of a particle"?

The term "motion of a particle" refers to the movement or change in position of a single object or point in space over a period of time.

2. How is the motion of a particle described qualitatively?

The motion of a particle can be described qualitatively by its speed, direction, and acceleration. This can include whether the particle is moving at a constant speed, changing direction, or speeding up or slowing down.

3. What is the difference between qualitative and quantitative descriptions of motion?

Qualitative descriptions of motion focus on the overall characteristics and changes in movement, while quantitative descriptions involve precise measurements of distance, time, and speed.

4. Why is it important to describe motion of a particle qualitatively?

Describing motion qualitatively allows us to understand the general behavior and patterns of a particle's movement, which can help us make predictions and analyze more complex systems.

5. What are some common examples of qualitative descriptions of motion?

Some common examples of qualitative descriptions of motion include describing the movement of a pendulum, a car accelerating from a stop sign, or a rollercoaster going up and down a series of hills.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
256
  • Introductory Physics Homework Help
Replies
3
Views
700
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
734
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
17
Views
362
Back
Top