My possible misunderstanding on a freely falling object

In summary: It implies that ##x_{A,f}=R+x_{B,f}##It implies that the final velocities won't change (how do I formulate that?)I arrived at the following equations $$v_{A,f}-v_{B,f}=v_{A,i}-v_{B,i}\\x_{A,f}-x_{B,f}-R=t(v_{A,i}-v_{B,i})$$And for that equality to happen we must have ##v_{A,i}-v_{B,i}=0## but this contradicts our assumption that ##v_{A,
  • #1
Anama Skout
53
13
Consider the famous kinematics equations $$x_f=x_i+v_i\cdot t+\frac12at^2$$ My question is: does this equation work for every point A & B along the parabola formed when dropping an object?
 
Physics news on Phys.org
  • #2
Yes. Each point is represented by a different t. To make that clear, I would write it as:

$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
 
  • Like
Likes Anama Skout
  • #3
DaleSpam said:
Yes. Each point is represented by a different t. To make that clear, I would write it as:

$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
 
  • #4
Anama Skout said:
Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
Why don't you try it and see for yourself?
 
  • #5
Doc Al said:
Why don't you try it and see for yourself?
I tried it with three points, always worked well (except the times when I made some mistakes -- of course), however I'm more interested into a rigorous proof.
 
  • #6
Anama Skout said:
however I'm more interested into a rigorous proof.
Should be doable with a bit of algebra. Go for it!
 
  • #7
I know it must be a simple thing I'm missing, but I can't figure it out, here's my try. Consider two points A & B such that ##x_{A,i}=R+x_{B,i}## and ##v_{A,i}\neq v_{B,i}## (for generality), then I should show that
  1. it implies that ##x_{A,f}=R+x_{B,f}##
  2. it implies that the final velocities won't change (how do I formulate that?)
I arrived at the following equations $$v_{A,f}-v_{B,f}=v_{A,i}-v_{B,i}\\x_{A,f}-x_{B,f}-R=t(v_{A,i}-v_{B,i})$$
And for that equality to happen we must have ##v_{A,i}-v_{B,i}=0## but this contradicts our assumption that ##v_{A,i}\neq v_{B,i}##, since clearly if -- for example -- you have an object falling onto the ground, if you take two distinct points in its trajectory, the velocity at each point won't be the same.

So what am I missing?

(here's a visualization of the situation)
 

Attachments

  • Snapshot.jpg
    Snapshot.jpg
    6.4 KB · Views: 411
  • #8
Anama Skout said:
Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
The given form of the equation only works for ##x_0## and ##v_0## defined at ##t=0##. At ##t=0## the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at ##t=0##.

In other words, unless ##A=B## the equations that you have written represent two distinct parabolas. I.e. the first equation passes through ##A## at ##t=0## and the second equation passes through ##B## at ##t=0##, they are not the same parabola even if both parabolas pass through ##A## and ##B##.

Also, I am not sure if you are thinking about parabolas through space or parabolas in spacetime. The equation you have written is for 1D motion, so it is only a parabola in spacetime.
 
  • #9
DaleSpam said:
The given form of the equation only works for ##x_0## and ##v_0## defined at ##t=0##. At ##t=0## the object is only at one point. If you use the same formula for a point which is on the parabola at a different time then you are going to get a different parabola which has that point at ##t=0##.

In other words, unless ##A=B## the equations that you have written represent two distinct parabolas. I.e. the first equation passes through ##A## at ##t=0## and the second equation passes through ##B## at ##t=0##, they are not the same parabola even if both parabolas pass through ##A## and ##B##.

Also, I am not sure if you are thinking about parabolas through space or parabolas in spacetime. The equation you have written is for 1D motion, so it is only a parabola in spacetime.

Ah, forgot about time, I should have added the following constraints ##t_B=T+t_A##, but again this doesn't yield much...
 
  • #10
So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?
 
  • #11
DaleSpam said:
So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?
I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)
 
Last edited:
  • #12
Why don't you treat each axis separately? I would think that would be simpler.
 
  • #13
Doc Al said:
Why don't you treat each axis separately? I would think that would be simpler.
I only treated the system w.r.t to a coordinate system in which the ##x##-axis is vertical.
 
  • #14
Anama Skout said:
I only treated the system w.r.t to a coordinate system in which the ##x##-axis is vertical.
Ah, so you are only talking about the vertical axis (as I first thought) despite the mention of parabolas. OK.
 
  • #15
Anama Skout said:
I'm thinking of parabolas in space parametrized w.r.t time. (like the one I drawed)
I am not quite sure why you want to put yourself through this, but OK. So a parabola in space can be defined by the parametric equations:
##x(\tau)=p \tau^2 + h##
##y(\tau)=2 p \tau + k##
##z(\tau)=0##
Where the vertex of the parabola is at ##(x,y,z)=(h,k,0)##, and ##p## indicates how flat the parabola is. The x direction is vertical, the y direction is horizontal along the arc of the parabola and the z direction is perpendicular to the parabola.

##\tau## is not time, it is a somewhat arbitrary parameter. To use this approach then you would need to solve for the relationship between ##\tau## and ##t##. That would involve some more free parameters which would allow you to have the same spatial parabola represented by the same equations in ##\tau## independently of when the object passes through any given points on the parabola.

I leave the details of figuring out the relationship between time and ##\tau## to you.
 
  • #16
DaleSpam said:
Yes. Each point is represented by a different t. To make that clear, I would write it as:

$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$
How would this equation work for parabola? This is a straight line equation isn't it?
 
  • #17
Hamza Abbasi said:
How would this equation work for parabola? This is a straight line equation isn't it?
The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.
 
  • #18
Anama Skout said:
Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
The confusion here is that you have defined too many variables. Point i or 0, is a chosen initial point on the trajectory. A and B are two other points on the same trajectory. So there is no separate initial position for A and a separate one for B.
 
  • #19
E
Doc Al said:
The motion is along a straight line. But expressed as a function of time, the equation describes a parabola.
Equation is of straight line but it explains parabola?? How come?
 
  • #20
Hamza Abbasi said:
Equation is of straight line but it explains parabola?? How come?
If you plot x as a function of time, it is not a straight line.
 
  • #21
Hamza Abbasi said:
How would this equation work for parabola? This is a straight line equation isn't it?
It is a straight line in space. It is a parabola in spacetime.
 
  • #22
DaleSpam said:
It is a straight line in space. It is a parabola in spacetime.
Wow! I didn't knew that :oldlove: , can you please post some pictorial representation?
 
  • #23
Hamza Abbasi said:
Wow! I didn't knew that :oldlove: , can you please post some pictorial representation?
You have the function. Plot it for yourself!
 
  • #24
Anama Skout said:
Thanks for your answer. But I have something different in mind. Suppose that the point A has ##x_{A,0}## and ##v_{A,i}##, then the equation will have the form $$x_A(t)=x_{A,i}+v_{A,i}\cdot t+\frac12at^2$$ and for a point B which has ##x_{B,0}## and ##v_{B,i}## then the equation will have the form $$x_B(t)=x_{B,i}+v_{B,i}\cdot t+\frac12at^2$$ Will this yield the same values in that for instance if we use them to determine what will happen after ##t=5## as long as we know what ##x_{A,0}##, ##v_{A,i}##, ##x_{B,0}##, and ##v_{B,i}## are (and the time difference between the two)?
If I understand you correctly, ##x_{B,i}## and ##v_{B,i}## are the position and velocity of the particle some time (say ##\Delta t##) after it passes through ##x_{A,i}##. So all you need to show is that the position given by the first equation at time ##t + \Delta t## is the same as the position given by the second equation at time ##t##. That's why I said it was just a bit of algebra.

Of course you'll need to express ##x_{B,i}## and ##v_{B,i}## in terms of ##x_{A,i}##, ##v_{A,i}##, and ##\Delta t##.
 
  • #25
It is not clear what you mean by your notation. You may be using it in a weird way.
For example, what is [itex]x_A[/itex]? If it is the position of point A, a point on the trajectory? It this is so, then it does not make sense to write [itex]x_A(t)[/itex]. The point is fixed, there is no time dependence for the position of that point.
Or A is some object moving along the trajectory and [itex]x_A(t)[/itex] is the position of this object.
Then B should be another object, starting from a different position? Is this what you mean?
You have two objects?
 
  • #26
Okay, sorry, it is all my fault for using a notation that I didn't properly explain. I'll use an example to illustrate what I want to show

A stone thrown from the top of a building is given an initial velocity of 20.0 m/s straight upward. The stone is launched 50.0 m above the ground, and the stone just misses the edge of the roof on its way down as shown in the figure below.
22q5A.png

In this example, I used the point A as the initial point. Now consider as if we're working with points B & C. All the values shown in the figure are relative to A. You can see that ##t_B=\rm{2.04 \;sec}## -- with that notation -- with respect to A, but it would be ##0\;\rm sec## with respect to B, or ##-2.04\;\rm sec## with respect to C.

Now what happens when the time passed from the initial time ##t=5\,\rm sec## ##\color{green}{\text{(with respect to A)}}##, i.e. what's the position & velocity at that time?

Well, we know that $$x_f=x_i+v_it+\frac12at^2,\tag{1}$$ we have that ##x_i=0{\rm\;sec},v_i=20.0\rm\;m/s## hence we get that $$\bbox[8pt,lightgreen,border:3px green solid]{\color{}{\large x_f=-22.5\rm\;m}}\tag{2}$$ Now, ##\color{blue}{\text{with respect to C}}##, we have that ##x_i={\rm\;0m},v_i=-20.0{\rm\;m/s},t=\big(5-4.08\big){\rm\;s}=0.92{\rm\;s}##, putting these values in ##(1)##, and sure enough: $$\bbox[8pt,skyblue,border:3px blue solid]{\color{}{\large x_f=-22.5\rm\;m}}\tag{3}$$ Now, ##\color{red}{\text{with respect to B}}## we have that ##x_i=0{\rm\;m},v_i=0{\rm\;m/s},t=\big(5-2.04\big){\rm\;m}=2.96{\rm\;m}##. Putting these things in ##(1)## yields $$\bbox[8pt,pink,border:3px red solid]{\color{}{\large x_f=-43\rm\;m}}\tag{4}$$ I assume these would lead us, under the correct transformation, to ##x_f## with respect to A.

But basically you got the idea, showing that no matter what points B & C we choose, it will always the same result, up to the change in the distance between them.
 
Last edited:
  • #27
Well, why use x when all your labels in the picture are y? But this is a minor point.
You did not make any mistake.
The position is -22.5 m in respect to A but -42.9 m in respect to B. Did you expect to get the same value? You are measuring (same) position from different origins. It will be unlikely to get the same value. If you look at the figure you can see that the distance from B to D is larger than from A to D.(or than C to D).
 
  • Like
Likes Anama Skout
  • #28
nasu said:
Well, why use x when all your labels in the picture are y? But this is a minor point.
You did not make any mistake.
The position is -22.5 m in respect to A but -42.9 m in respect to B. Did you expect to get the same value? You are measuring (same) position from different origins. It will be unlikely to get the same value. If you look at the figure you can see that the distance from B to D is larger than from A to D.(or than C to D).
Oups, another mistake, I wanted to say that we would get the same result between these points if we translate one to another. I'll edit the earlier post.
 
  • #29
I don't know what you mean by this "translate one to another". By changing the origin, the values of the coordinate will change and there is no mistake.
The positions will change but speed and acceleration won't change if you translate the origin to a different point along the same line.
 
  • Like
Likes Anama Skout
  • #30
nasu said:
I don't know what you mean by this "translate one to another". By changing the origin, the values of the coordinate will change and there is no mistake.
The positions will change but speed and acceleration won't change if you translate the origin to a different point along the same line.
In this case I translated A to B, the difference between the x-coordinate of the two is only 2.04 m, and doesn't help to get 22.5 m out of the 43 m, why?

Also, I agree that the change in the coordinate only amounts to a change in the origin, so there will be no differnece, okay. But why the speed won't change? I mean, conceptually, yes, it will be the same, but I'm looking for a rigorous derivation.
 
  • #31
No, the difference is 20.4 m.
A has 0 coordinate and B has 20.4 m. You are mixing coordinate and time.

Because the speed is the derivative of the position in respect to time. Adding a constant to the coordinate (this is what a translation means) does not change the derivative.
 
  • Like
Likes Anama Skout
  • #32
nasu said:
No, the difference is 20.4 m.
A has 0 coordinate and B has 20.4 m. You are mixing coordinate and time.

Because the speed is the derivative of the position in respect to time. Adding a constant to the coordinate (this is what a translation means) does not change the derivative.
Oh, sorry again for the mistake. Thanks for that, now it become all clear! :woot:
 
  • #33
But just one last thing, how can I show that the speed will be the same without using calculus?
 
  • #34
Well, look at finite changes. You can think about an average speed, which is Δx/Δt.
Calculate the change in position for two different origin.
For example, if the stone was at D and it moves 1m below D, by how much increases the distance between the stone and C or B?
 
  • #35
nasu said:
Well, look at finite changes. You can think about an average speed, which is Δx/Δt.
Calculate the change in position for two different origin.
For example, if the stone was at D and it moves 1m below D, by how much increases the distance between the stone and C or B?
I get this $$\overline{v}_1=\frac{\Delta x}{\Delta t}=\frac{(x_B-x_D)-(x_B-(x_D-1))}{t_{B;D}-t_{B;D'}}=-\frac1{t_{B;D}-t_{B;D'}},$$ where ##t_{B,D}## means the time the particle is at D relative to B, and D' stands for the point when the stone moves 1m below D. And with respect to C $$\overline{v}_2=\frac{\delta x}{\delta t}=\frac{(x_C-x_D)-(x_C-(x_D-1))}{t_{C;D}-t_{C;D'}}=-\frac1{t_{C;D}-t_{C;D'}},$$ so to show that these two things are equal one would require that $$t_{B;D}-t_{B;D'}=t_{C;D}-t_{C;D'}.\tag1$$ I'm stuck here...
 
<h2>1. What is a freely falling object?</h2><p>A freely falling object is an object that is only acted upon by the force of gravity. This means that it is not being pushed or pulled by any other forces, such as air resistance or friction.</p><h2>2. Why does a freely falling object accelerate?</h2><p>A freely falling object accelerates because of the force of gravity. As the object falls, it gains speed due to the constant acceleration caused by the pull of gravity.</p><h2>3. How does the mass of a freely falling object affect its acceleration?</h2><p>The mass of a freely falling object does not affect its acceleration. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. Since the only force acting on a freely falling object is gravity, its mass does not affect its acceleration.</p><h2>4. What is the acceleration of a freely falling object?</h2><p>The acceleration of a freely falling object is approximately 9.8 meters per second squared (m/s^2), which is the acceleration due to gravity on Earth. This value may vary slightly depending on the location and altitude.</p><h2>5. Can a freely falling object ever reach a constant speed?</h2><p>Yes, a freely falling object can reach a constant speed if there is no longer a net force acting on it. This can happen when the object reaches the ground or when it reaches terminal velocity, the maximum speed an object can reach due to air resistance.</p>

1. What is a freely falling object?

A freely falling object is an object that is only acted upon by the force of gravity. This means that it is not being pushed or pulled by any other forces, such as air resistance or friction.

2. Why does a freely falling object accelerate?

A freely falling object accelerates because of the force of gravity. As the object falls, it gains speed due to the constant acceleration caused by the pull of gravity.

3. How does the mass of a freely falling object affect its acceleration?

The mass of a freely falling object does not affect its acceleration. According to Newton's Second Law of Motion, the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass. Since the only force acting on a freely falling object is gravity, its mass does not affect its acceleration.

4. What is the acceleration of a freely falling object?

The acceleration of a freely falling object is approximately 9.8 meters per second squared (m/s^2), which is the acceleration due to gravity on Earth. This value may vary slightly depending on the location and altitude.

5. Can a freely falling object ever reach a constant speed?

Yes, a freely falling object can reach a constant speed if there is no longer a net force acting on it. This can happen when the object reaches the ground or when it reaches terminal velocity, the maximum speed an object can reach due to air resistance.

Similar threads

Replies
3
Views
862
Replies
5
Views
2K
Replies
18
Views
2K
Replies
26
Views
5K
Replies
30
Views
4K
Replies
11
Views
1K
  • Classical Physics
Replies
12
Views
1K
Replies
13
Views
4K
Replies
35
Views
3K
Back
Top