Deriving kinematic equation for position

[annamal]

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time.
$v = v_i + a(t_f - t_i)$
$dx = v_idt + at_fdt - atdt$
integrating
$x_f - x_i = v_i(t_f - t_i) + at_f(t_f - t) - \frac{at^2}{2}|_{t}^{t_f}$

If this is not the way to derive an equation for position with initial time as variable, what is?

 

[bob012345]

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time.

Yes, but it is really only the difference that matters.

 

[annamal]

Yes, but it is really only the difference that matters.

Ok, did I derive it correctly. The way I derived it doesn't generate the correct answer for position though

 

[nasu]

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time.
$v = v_i + a(t_f - t_i)$
$dx = v_idt + at_fdt - atdt$
integrating
$x_f - x_i = v_i(t_f - t_i) + at_f(t_f - t) - \frac{at^2}{2}|_{t}^{t_f}$

If this is not the way to derive an equation for position with initial time as variable, what is?

Both final time and initial time are values of the variable time (t). There is only one variable, the time. To say that initial value or final value are variable does not make much sense. In the normal derivations the initial and final values determine the integration limits. Your first equation does not contain any variable as it is. It just gives one single value of velocity for a given acceleration and specific time interval. There is nothing to integrate here. You have one constant equal to another constant. You are just playing with "integrals" without thiking if it makes sense what you are doing.

 

[bob012345]

Both final time and initial time are values of the variable time (t). There is only one variable, the time. To say that initial value or final value are variable does not make much sense. In the normal derivations the initial and final values determine the integration limits. Your first equation does not contain any variable as it is. It just gives one single value of velocity for a given acceleration and specific time interval. There is nothing to integrate here. You have one constant equal to another constant. You are just playing with "integrals" without thiking if it makes sense what you are doing.

That's true but usually, at least as I have seen it, one just leaves $t_f=t$ and $t_i=0$ but that does not have to be the case. The OP seems to want to use $t_i=t$ and $t_f$ as a constant.

 

[bob012345]

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time.
$v = v_i + a(t_f - t_i)$
$dx = v_idt + at_fdt - atdt$
integrating
$x_f - x_i = v_i(t_f - t_i) + at_f(t_f - t) - \frac{at^2}{2}|_{t}^{t_f}$

If this is not the way to derive an equation for position with initial time as variable, what is?

The mistake starts in the first equations which should be
$v = v_i + a(t_f - t)$

then the last equation becomes
$x_f - x_i = v_i(t_f - t ) + at_f(t_f - t) - \frac{\large at^2}{2}|_{t}^{t_f}$

just finish evaluating the limits

$x_f - x_i = v_i(t_f - t ) + at_f(t_f - t) - \frac{\large a}{2}(t_f^2 - t^2)$

after some algebra you end up with

$x_f = x_i + v_i(t_f - t ) + \frac{\large a}{2}(t_f - t)^2$

I think this is Ok as long as you understand that the problem stated this way is solving for some unknown starting time in the past and $t_f > t$ but as I said earlier, only the difference in time matters.

 

[annamal]

The mistake starts in the first equations which should be
$v = v_i + a(t_f - t)$

then the last equation becomes
$x_f - x_i = v_i(t_f - t ) + at_f(t_f - t) - \frac{\large at^2}{2}|_{t}^{t_f}$

just finish evaluating the limits

$x_f - x_i = v_i(t_f - t ) + at_f(t_f - t) - \frac{\large a}{2}(t_f^2 - t^2)$

after some algebra you end up with

$x_f = x_i + v_i(t_f - t ) + \frac{\large a}{2}(t_f - t)^2$

I think this is Ok as long as you understand that the problem stated this way is solving for some unknown starting time in the past and $t_f > t$ but as I said earlier, only the difference in time matters.

Oh yes, I forgot to mention something. I noticed it breaks down with unconstant acceleration. If we started with $a = \alpha t$ where $\alpha = 1.3t$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^2}{2}|_t^{t_f}$

 

[nasu]

Of course it breaks down. The equation in this form is valid only for constant accekeration. It was obtained with the condition that acceleration is constant. Physics is about understanding what actaully happens. Just playing blindly with equations is not physics. I am not sure if it even math.

 

[nasu]

That's true but usually, at least as I have seen it, one just leaves $t_f=t$ and $t_i=0$ but that does not have to be the case. The OP seems to want to use $t_i=t$ and $t_f$ as a constant.

And what is t? I mean, measured from when? From the t _f but going backwards? And what is v _i in this case? The final velocity? It cannot be initial because it corresponds to the final time, right?

 

[bob012345]

And what is t? I mean, measured from when? From the t _f but going backwards? And what is v _i in this case? The final velocity? It cannot be initial because it corresponds to the final time, right?

No, everything is the same. The usual kinematic equation is

$$x_f = x_i + v_i(t_f - t_i ) + \frac{\large a}{2}(t_f - t_i)^2$$

where we usually let $t_f$ be the variable $t$ and set $t_i=0$ but as I explained above we don't have to. I could ask something like given acceleration $a$ we have at $t_f$ the position is $x_f$, at what time in the past was the position $x_i$ and the velocity $v_i$?

 

[Delta2]

I think the OP has done a chaotic mess here , regarding what are the constants, what are the variables and how exactly we integrate.

Sorry @annamal if I offended you that way, but I think there is some chaos in your mind.
Having said that, I also have to say that sometimes geniuses think in a chaotic way.

 

[PeroK]

It's perfectly valid to play about with equations as long as you are clear about what you are doing. There's nothing to stop you considering $t_i$ and $t_f$ as variables.

It's simply convention to treat them as fixed initial and final values of the variable $t$. It's certainly mathematically valid to consider displacement as a function of the two variables, as in the case of constant velocity:$$s = v(t_f - t_ i)$$Physically this would tend to be seen as a series of experiments with different start and finish times. Or, $v_f$ could be seen as equivalent to the usual variable $t$.

 

[annamal]

Of course it breaks down. The equation in this form is valid only for constant accekeration. It was obtained with the condition that acceleration is constant. Physics is about understanding what actaully happens. Just playing blindly with equations is not physics. I am not sure if it even math.

But I don't understand why it is breaking down because I am integrating the equations properly

 

[annamal]

And what is t? I mean, measured from when? From the t _f but going backwards? And what is v _i in this case? The final velocity? It cannot be initial because it corresponds to the final time, right?

t is the initial time. Instead of a variable final time, I use a variable initial time and constant final time. vi is the initial velocity...

 

[Delta2]

$v_i$ is constant or variable that is $v_i(t_i)$?

 

[bob012345]

$v_i$ is constant or variable that is $v_i(t_i)$?

There is no rule that $v_i$ has to be unknown also if you treat the start time as the unknown. It depends on the givens of the problem.

 

[Delta2]

There is no rule that $v_i$ has to be unknown also if you treat the start time as the unknown. It depends on the givens of the problem.

So your take is that there is no problem it can be considered as constant?

 

[bob012345]

Oh yes, I forgot to mention something. I noticed it breaks down with unconstant acceleration. If we started with $a = \alpha t$ where $\alpha = 1.3t$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^2}{2}|_t^{t_f}$

Doing this the usual way one gets

$$x_f = x_i + v_i t+ \frac{\alpha t ^3 }{6} $$

So you need a $t^3$ in there.

 

[bob012345]

So your take is that there is no problem it can be considered as constant?

I think so. Suppose the problem was to find the start time when the velocity was such and such. In general any combination might be the knowns.

 

[nasu]

But I don't understand why it is breaking down because I am integrating the equations properly

No, you are not. If the acceleration is a function of time you need to know this function and it is under the integral. This, besides the fact that you start with an equation which is wrong for non-constant acceleration. You only can write dv=a dt and integrate it properly. But not an equation with finite quantities which results from different conditions.

 

[pnachtwey]

We usually have an initial time and then find an equation for the variable final time. Can we derive a formula to calculate position with final time and variable initial time.
$v = v_i + a(t_f - t_i)$
$dx = v_idt + at_fdt - atdt$
integrating
$x_f - x_i = v_i(t_f - t_i) + at_f(t_f - t) - \frac{at^2}{2}|_{t}^{t_f}$

If this is not the way to derive an equation for position with initial time as variable, what is?

I don't agree with the at_f(t_f-t) part.
$\Delta x=v\cdot\Delta t+\frac{a\cdot\Delta t^2}{2}$
Get a program that can integrate and differentiate polynomials.
wxMaxima will do.

 

[SammyS]

I don't agree with the at_f(t_f-t) part.
$\Delta x=v\cdot\Delta t+\frac{a\cdot\Delta t^2}{2}$
Get a program that can integrate and differentiate polynomials.
wxMaxima will do.

Just to make thus clear to OP:
at_f(t_f-t) in LaTeX is: $at_f(t_f-t)$ .

And the $\displaystyle \Delta t^2 \text{ is } \Delta (t^2) = (t_f^2 - t^2) ~~ .$

 

[annamal]

But not an equation with finite quantities which results from different conditions.

What do you mean by an equation with finite quantities which results from different conditions?

 

[annamal]

Doing this the usual way one gets

$$x_f = x_i + v_i t+ \frac{\alpha t ^3 }{6} $$

So you need a $t^3$ in there.

Yes, I meant
If we started with $a = \alpha t$ where $\alpha = 1.3t$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$,
this equation for $x_f$ is not correct (constant $t_f$ but variable initial time). But why?

 

[SammyS]

Yes, I meant
If we started with $a = \alpha t$ where $\alpha = 1.3t$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$,
this equation for $x_f$ is not correct (constant $t_f$ but variable initial time). But why?

So, you mean that $a=1.3 t^2$.

 

[nasu]

What do you mean by an equation with finite quantities which results from different conditions?

I mean the equation $v_f=v_i+a(t_f-t_i)$. (1)
The starting point is the equation with differential quantities
$dv= a dt$. (2)
This (equation number 2) is true for any type of acceleration as it follow from the definition of acceleration ($a= \frac{dv}{dt}$) .
If you integrate this equation assuming constant acceleration and the integration limits for time as $t_i$ and $t_f$ then (and only then) you get equation (1).
For any other type of acceleration (other than constant) you need to start again wih equation (2) and see what you get. But you need to know in which way is the acceleratio non-constant. This means to know the function of time, a(t).

 

[pnachtwey]

This is the blind leading the blind. I write code for motion controllers and sell them around the world.
Below is A SIMPLE example of using seven third order polynomials to make a whole motion from one point to another.
https://deltamotion.com/peter/wxMaxima/Seg1234567.html
The code is here
https://deltamotion.com/peter/wxMaxima/Seg1234567.wxmx
wxMaxima can be downloaded from here
https://wxmaxima-developers.github.io/wxmaxima/download.html

Most motion profiles use third order polynomials and they are pieced together to make a whole motion profile.
It is possible to make the whole move with just one fifth order polynomial. A fifth order polynomial has six coefficients. The first three are known because they are the initial speed, velocity and acceleration/2. The next higher coefficients must be calculated using a system of 3 equations and 3 unknowns.

 

[nasu]

Thank you for showing the light to the blind.

 

[SammyS]

Thank you for showing the light to the blind.

Thought that maybe @pnachtwey was simply being self referencing as he/she tried to guide OP.

 

[pnachtwey]

No, my post #21 was accurate. There have been too many posts that don't show basic knowledge of a third order equation applied to motion or even how to integrate and differentiate basic polynomials. My link above shows how it is done. Now do you want to dispute what I posted in #27 ?

You guys really should get wxMaxima if you are cheap. wzMaxima is a free download. Mathematica or Maple is better but costs $$$$.
For some reason these math packages are called CAS or computer algebra systems but they do much more than algebra.

 

[SammyS]

I apologize if I insulted you.

OP has started two similar threads recently. Both are rather unclear in regards to details. Seems he wants to obtain kinematic equations, but have them expressed in terms of some final values rather than initial values, but he's not that clear about the details. Then somewhere along in the thread, he throws in a comment about acceleration not being constant - in this thread it's Post #7.
Even there, he's got contradictory information.
He says $a=\alpha t$. (I assume that $a$ is acceleration.) Seems like he treats $a$ as being linear in time, $t$ and $\alpha$ is constant. Then he turns right around and says $\alpha=1.3\,t$.

And, by the way, we are not unfamiliar with using higher order polynomials to describe motion.

 

[Delta2]

And, by the way, we are not unfamiliar with using higher order polynomials to describe motion.

I am, best I know is suvat equations with constant acceleration which result in second order polynomial with respect to t. I guess higher order polynomials come if the acceleration is a generic polynomial of time.

 

[bob012345]

No, my post #21 was accurate. There have been too many posts that don't show basic knowledge of a third order equation applied to motion or even how to integrate and differentiate basic polynomials. My link above shows how it is done. Now do you want to dispute what I posted in #27 ?

I assume everything you posted is correct but how does it help the OP do what they asked which is to reverse the usual order of time in the basic kinematic equation?

You guys really should get wxMaxima if you are cheap. wzMaxima is a free download. Mathematica or Maple is better but costs $$$$.
For some reason these math packages are called CAS or computer algebra systems but they do much more than algebra.

Again, the point of this thread is not numerical calculation techniques. It is deriving the kinematic equation under a different assumption of time order.

 

[annamal]

So, you mean that $a=1.3 t^2$.

I made another typo. I meant
If we started with $a = \alpha t$ where $\alpha = 1.3$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$,
this equation for $x_f$ is not correct (constant $t_f$ but variable initial time). But why?

 

[Steve4Physics]

I made another typo. I meant
If we started with $a = \alpha t$ where $\alpha = 1.3$, $v_i = 5.8$, $t_f = 4$, t = 2, $x_i = 9.83$
$v = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$,
this equation for $x_f$ is not correct (constant $t_f$ but variable initial time). But why?

I’d like to have a try too…
_______

Why have you given the values of $\alpha = 1.3t, v_i = 5.8,$ etc. when these values are not used/required?
______

Your equation:
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^2}{2}|_t^{t_f}$
is wrong. @bob012345 has already given you the correct equation in Post #18:
$x_f = x_i + v_i t+ \frac{\alpha t ^3 }{6}$

This could be written
$x_f = x_i + v_i (t_f-t_i)+ \frac{\alpha (t_f-t_i) ^3 }{6}$

If you want the initial time as a variable (and, correspondingly, initial position as a variable) then simply rewrite $t_i$ as $t$ and $x_i$ as $x$ to give:
$x_f = x + v_i (t_f-t)+ \frac{\alpha (t_f-t) ^3 }{6}$

If you want rearrange this to make x the subject: you now have an equation which gives x(t), i.e. initial position, as a function of initial time. (As opposed to the more conventional version where we give final position as a function of final time.)
__________

Also, in your (incorrect) equation, the last term, $- \frac{\alpha t^3}{6}|_t^{t_f}$, is negative. But over time (for positive values of $\alpha$) you expect $x_f$ to keep increasing in the positive x-direction. So the negative sign can’t be right.