My possible misunderstanding on a freely falling object

[Anama Skout]

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?

 

[Dale]

Yes. Each point is represented by a different t. To make that clear, I would write it as:

$$x(t)=x_0+v_0 t+\frac{1}{2}a t^2$$

 

[Anama Skout]

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)?

 

[Doc Al]

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?

 

[Anama Skout]

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.

 

[Doc Al]

however I'm more interested into a rigorous proof.

Should be doable with a bit of algebra. Go for it!

 

[Anama Skout]

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)

 

[Dale]

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.

 

[Anama Skout]

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...

 

[Dale]

So are you thinking of parabolas in space parametrized by time or are you thinking of parabolas in spacetime?

 

[Anama Skout]

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)

 

[Doc Al]

Why don't you treat each axis separately? I would think that would be simpler.

 

[Anama Skout]

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.

 

[Doc Al]

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.

 

[Dale]

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.

 

[Hamza Abbasi]

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?

 

[Doc Al]

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.

 

[Chandra Prayaga]

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.

 

[Hamza Abbasi]

E

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?

 

[Doc Al]

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.

 

[Dale]

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.

 

[Hamza Abbasi]

It is a straight line in space. It is a parabola in spacetime.

Wow! I didn't knew that , can you please post some pictorial representation?

 

[Doc Al]

Wow! I didn't knew that , can you please post some pictorial representation?

You have the function. Plot it for yourself!

 

[Doc Al]

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 $\Delta t$.

 

[nasu]

It is not clear what you mean by your notation. You may be using it in a weird way.
For example, what is $x_A$? 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 $x_A(t)$. 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 $x_A(t)$ 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?

 

[Anama Skout]

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.
​

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.

 

[nasu]

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).

 

[Anama Skout]

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.

 

[nasu]

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.

 

[Anama Skout]

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.

 

[nasu]

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.

 

[Anama Skout]

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!

 

[Anama Skout]

But just one last thing, how can I show that the speed will be the same without using calculus?

 

[nasu]

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?

 

[Anama Skout]

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...