I Deriving kinematic equation for position

[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.

 

[PeroK]

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?

This looks like a muddle to me. It's either $t$ and $t_i$; or, $t_f$ and $t_i$. But, $t_f$ and $t$ is confused, if not plain wrong.

 

[SammyS]

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?

You need to go through this one step at a time. You need to be consistent. For instance, velocity, $v$, is a function of time. If final time, $t_f$, is a fixed quantity and initial time, $t$, is the independent variable, then initial velocity is $v(t)$, which depends on $t$, and final velocity is $v(t_f)$. In other words, also final velocity which is fixed: $v_f=v(t_f)$ .

Probably go back and look at acceleration. (By the way: Lower case alpha is a poor choice for the time derivative of acceleration which by the way has a name - or two - known as 'jerk' or 'jolt'.) $\ \alpha$ looks a lot like $a$.

 

[bob012345]

Ok, for a non-linear acceleration, the limits of integration matter not just the $\Delta{t}$. Integrating between 2 and 4 seconds is not the same as between 0 and 2 seconds. The only way not to have a mess is to let the lower limit or upper limit be zero so the time is the same as the $\Delta{t}$.

 

[pnachtwey]

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?

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

$\Delta t$ can be negative. The trick part calculating the coefficients and it seems people have a hard time doing that. If you assume $\Delta t$ can be negative then it is possible to calculate a position, velocity and acceleration before time 0 for that particular polynomial. Notice when $\Delta t$ is negative then the the sign of all the odd powers of $\Delta t$ are negative. The OP would be better served if he explained what he is trying to do in the real world.

 

[bob012345]

$\Delta t$ can be negative. The trick part calculating the coefficients and it seems people have a hard time doing that. If you assume $\Delta t$ can be negative then it is possible to calculate a position, velocity and acceleration before time 0 for that particular polynomial. Notice when $\Delta t$ is negative then the the sign of all the odd powers of $\Delta t$ are negative. The OP would be better served if he explained what he is trying to do in the real world.

What we want to do is derive the kinematic equation from first principles under this assumption, not assume a polynomial and figure out the coefficients. It's not about getting the result as much as the process of getting there.

 

[pnachtwey]

Which kinematic equation do you want to derive? There are lots of them! What I have seen above is that people don't even know how to integrate or differentiate.

Most of the time you know the initial position, velocity and acceleration and you want to move to another position, velocity and acceleration after time t.

For 3rd order polynomials the jerk is constant.
j
integrate to get the acceleration as function of time. The c after integration is the same as a(0).
a(t)=a(0)+j*t // sometime is just write a0 since it is a constant.
v(t) = v(0)+a(0)*t+(1/2)*j*t^2
x(s) = x(0)+v(0)*t+(1/2)*a(0)*t^2+(1/6)*j*t^3

Start with the highest constant derivative and start integrating. When you integrate you get a constant. That is the initial condition. The trick is being able to choose which derivative is going to be constant and what its value should be to end up at the desired position, velocity and acceleration.
I think this is clear in my wxMaxima post.

A simple applications is what motion people call the PVT command. Basically you know the initial position and velocity and want to end up at a final position and velocity at time t. This requires a third order polynomial. The initial position and velocity define the first to coefficients of the third order polynomial. Now you must solve two equations for the two higher order coefficients.

If this doesn't answer your question then give me an example with numbers. A real story problem :)

 

[bob012345]

Which kinematic equation do you want to derive? There are lots of them!

All the orientation you need is in post #1 and post #7.

What I have seen above is that people don't even know how to integrate or differentiate.

This is the third time you have complained that the people trying to help the OP in this thread don't know what they are doing. Please tell us which posts do you take issue with and why?

Most of the time you know the initial position, velocity and acceleration and you want to move to another position, velocity and acceleration after time t.

It is clearly stated in the OP that is not what is being asking here.

For 3rd order polynomials the jerk is constant.
j
integrate to get the acceleration as function of time. The c after integration is the same as a(0).
a(t)=a(0)+j*t // sometime is just write a0 since it is a constant.
v(t) = v(0)+a(0)*t+(1/2)*j*t^2
x(s) = x(0)+v(0)*t+(1/2)*a(0)*t^2+(1/6)*j*t^3

This result is above too by others. The OP wants the time in terms of an unknown start time $t$ given a known finish time $t_f$. In this case a(0) is zero. $$x_f = x_i+v_i*(t_f- t )+(1/6)*j*(t_f - t )^3$$

Edit: I'm beginning to doubt myself here. Usually acceleration is zero or a constant in these problems so only the $\Delta{t}$ matters but here it matters where one is compared to zero and it does not only matter what the $\Delta{t}$ is it matters what t is too.

 

[pnachtwey]

All the orientation you need is in post #1 and post #7.This is the third time you have complained that the people trying to help the OP in this thread don't know what they are doing. Please tell us which posts do you take issue with and why?

It is clearly stated in the OP that is not what is being asking here.

This result is above too by others. The OP wants the time in terms of an unknown start time $t$ given a known finish time $t_f$. In this case a(0) is zero. $$x_f = x_i+v_i*(t_f- t )+(1/6)*j*(t_f - t )^3$$

What confuses me, actually it is the rest of you that are confused, is that your $$x_f = x_i+v_i*(t_f- t )+(1/6)*j*(t_f - t )^3$$ equation shows that x_f and x_i are constants but then the times should be constants too such as (t_f-t_i) . (t_f-t_i) is a $\Delta t$. There should be no variable t.

I have provided the answer above when I used the formula with the $\Delta t$
$\Delta t$ can be t_f-ti, actually a t_f-t_i
What you just wrote looks OK but you forgot the initial acceleration term.
Even with the initial acceleration term, what do you do with this equation?

So what is x_i, v_i and a_i?
So how do you calculate t if you don't know x_i, v_i,and a_i?
If v_f and a_f are also known then there is some hope because then $\Delta t$ can be negative and x_i, v_i and a_i can be calculated.

The OP should provide an example of what he is trying to do..

 

[bob012345]

What confuses me, actually it is the rest of you that are confused, is that your $$x_f = x_i+v_i*(t_f- t )+(1/6)*j*(t_f - t )^3$$ equation shows that x_f and x_i are constants but then the times should be constants too such as (t_f-t_i) . (t_f-t_i) is a $\Delta t$. There should be no variable t.

I have provided the answer above when I used the formula with the $\Delta t$
$\Delta t$ can be t_f-ti, actually a t_f-t_i
What you just wrote looks OK but you forgot the initial acceleration term.
Even with the initial acceleration term, what do you do with this equation?

So what is x_i, v_i and a_i?
So how do you calculate t if you don't know x_i, v_i,and a_i?
If v_f and a_f are also known then there is some hope because then $\Delta t$ can be negative and x_i, v_i and a_i can be calculated.

The OP should provide an example of what he is trying to do..

I take the acceleration as simply alpha*t so in your notation a(0)= 0 and a = 0 + alpha *0 at t=0.

So, the unusual acceleration is confusing the usual derivation of the kinematic equations where as in my edited note above it does depend on actual times and not just time differences. But what the OP asked for is how to treat the initial start time as the unknown. Usually it is easy to start the problem at t=0 and integrate to the unknown time t. The OP wanted to assume one knows the end time but not the start time. As I see it in general if everything is labeled as constants initial and final that shows the relationship between initial and final positions, velocities and the acceleration but one can treat after the fact anyone of those as unknowns to be solved in terms of the other if they are all knowns.

For example for a constant acceleration we have

$$x_f = x_i + v_i(t_f - t_i) + a/2 ( t_f - t_i)^2$$

We might be given some but not others. Suppose we are given at the start the initial velocity and position are some values but we are not told what time that was. Given a there are five initial or final conditions. We might know any combination of 4 and compute the fifth. Just because something is labeled does not mean it can't be treated as an unknown and solved.

 

[annamal]

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.

If $t_f$ is the constant one and $t_i = t$ is variable,
with $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
If we integrate v(t) to derive x(t)
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$
Yes, I see how @bob012345 's answer
$x_f = x + v_i (t_f-t)+ \frac{\alpha (t_f-t) ^3 }{6}$ is correct

but I am confused how come my derivation didn't create the correct $x_f = x(t)$

 

[bob012345]

If $t_f$ is the constant one and $t_i = t$ is variable,
with $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$
If we integrate v(t) to derive x(t)
$x_f = x_i + v_i(t_f - t) + \frac{\alpha(t_f^2)}{2}(t_f - t) - \frac{\alpha t^3}{6}|_t^{t_f}$
Yes, I see how @bob012345 's answer
$x_f = x + v_i (t_f-t)+ \frac{\alpha (t_f-t) ^3 }{6}$ is correct

but I am confused how come my derivation didn't create the correct $x_f = x(t)$

Actually, I'm very sorry, I now believe my answer was not correct. Under the usual circumstances where acceleration is either zero or constant using a $\Delta{t}$ like that doesn't matter and works out, it only depends on the difference but when $a= \alpha t$ then something is being assumed which is a specific importance of the time $t=0$. In that case we have to integrate from zero to $t$. If we do that we get

$$x_f = x(t) = x_i + v_i t+ \frac{\alpha t ^3 }{6}$$ which should always work but $x_i$ and $v_i$ are the values at $t=0$. We can construct equations from this that relate two arbitrary times that we can call $t_f$ and $t_i$ assuming $t_f > t_i$.

There is also an issue that we cannot arbitrarily choose which time is going to be the constant because time integration assumes time goes forward and so it assumes the lower limit is the constant $t_i$. Basically we can't mess with the integration but we can choose which time will be considered the unknown to solve for given the other after the fact of integration. It is late and I will finish that in the morning.

How did you test to see if your derivation was correct or not?

 

[PeroK]

If $t_f$ is the constant one and $t_i = t$ is variable,
with $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$

If you plug $t = t_f$ into that equation you get $$v(t_f) = v_i$$which isn't right at all.

 

[PeroK]

If we want to generalise, then we ought to have some reference time $t_0$, with $x_0 = x(t_0)$ and $v_0 = v(t_0)$, where this is simply some point in the trajectory. It could be that $t_0 = 0$, but it doesn't have to be. Then, with acceleration as a function of time $a(t)$, we have:
$$v(t) = v(t_0) + \int_{t_0}^t a(t')dt' = v_0 + \int_{t_0}^t a(t')dt'$$Note that this is valid even when $t < t_0$, as in this case we have:
$$v(t) = v(t_0) - \int_{t}^{t_0} a(t')dt' = v(t_0) + \int_{t_0}^t a(t')dt'$$Using the properties of the definite integral.

Likewise:$$x(t) = x(t_0) + \int_{t_0}^t v(t')dt'= x_0 + v_0(t - t_0) + \int_{t_0}^t \int_{t_0}^{t'} a(t'')dt'' \ dt'$$

 

[Steve4Physics]

This may be an unnecessary post now, but since I’ve written it, here's an explanation of what (I think) went wrong with @annamal's logic/working.

$a = \alpha t$ gives $\frac {dv}{dt} = \alpha t$.
Integrating from $t=t_i$ to $t=t_f$ gives
$v_f - v_i = \frac{\alpha(t_f^2 – t_i^2)}{2}$

In the above equation, you can choose two quantities as (dependent and independent) variables - the other quantities are then treated as constants. You must then stick to these choices.

Assuming we are not interested in the effects of changing $\alpha$ our four options are:

a) $t_f$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_i + \frac{\alpha(t^2 – t_i^2)}{2}$
(This is the conventional choice.)

b) $t_i$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_f - \frac{\alpha(t_f^2 – t^2)}{2}$

c) $t_f$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_f - \frac{\alpha(t^2 – t_i^2)}{2}$

d) $t_i$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_i + \frac{\alpha(t_f^2 – t^2)}{2}$

Note that choice d) is @annamal's equation $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$.

The meaning of v(t) in the equation is the final velocity as a function of of the start time. It is not the velocity as a function of time during the motion, So it makes no sense to integrate it to find the displacement.

 

[vanhees71]

I don't know, what you guys really are after here, but are you maybe just discussing the Taylor expansion of a function in a very complicated way, which leads to confusion:
$$x(t)=x(t_0) + \dot{x}(t_0) (t-t_0) + \frac{1}{2} \ddot{x}(t_0) (t-t_0)^2 + \cdots = \sum_{k=0}^{\infty} \frac{1}{k!} x^{(k)}(t_0) (t-t_0)^k.$$
Here $x^{(k)}$ means the $k$-th derivative of the function $x(t)$.

 

[PeroK]

I don't know, what you guys really are after here, but are you maybe just discussing the Taylor expansion of a function in a very complicated way, which leads to confusion:
$$x(t)=x(t_0) + \dot{x}(t_0) (t-t_0) + \frac{1}{2} \ddot{x}(t_0) (t-t_0)^2 + \cdots = \sum_{k=0}^{\infty} \frac{1}{k!} x^{(k)}(t_0) (t-t_0)^k.$$
Here $x^{(k)}$ means the $k$-th derivative of the function $x(t)$.

That's a good way to avoid doing the integration for the OP's acceleration $a = \alpha t$:
$$x(t)=x_0 + v_0(t-t_0) + \frac 1 2 \alpha t_0(t-t_0)^2 + \frac 1 6 \alpha(t - t_0)^3$$

 

[vanhees71]

Hm, I get
$$v(t)=v_0+\int_{t_0}^t \mathrm{d} t' \alpha t' = v_0 + \frac{\alpha}{2} (t^2-t_0^2)$$
and then, integrating again and after some algebra,
$$x(t)=x_0 + v_0 (t-t_0) +\frac{\alpha}{6} (t-t_0)^2 (t+2 t_0) + v_0 (t-t_0)+x_0.$$
[EDIT:] Well, it's the same. So never mind ;-).

 

[PeroK]

Hm, I get
$$v(t)=v_0+\int_{t_0}^t \mathrm{d} t' \alpha t' = v_0 + \frac{\alpha}{2} (t^2-t_0^2)$$
and then, integrating again and after some algebra,
$$x(t)=x_0 + v_0 (t-t_0) +\frac{\alpha}{6} (t-t_0)^2 (t+2 t_0) + v_0 (t-t_0)+x_0.$$
[EDIT:] Well, it's the same. So never mind ;-).

Both methods simplify to:
$$x(t)=x_0 + v_0(t-t_0) + \frac \alpha 6 \big [t^3 - 3t_0^2 t + 2t_0^3 \big ]$$

 

[annamal]

This may be an unnecessary post now, but since I’ve written it, here's an explanation of what (I think) went wrong with @annamal's logic/working.

$a = \alpha t$ gives $\frac {dv}{dt} = \alpha t$.
Integrating from $t=t_i$ to $t=t_f$ gives
$v_f - v_i = \frac{\alpha(t_f^2 – t_i^2)}{2}$

In the above equation, you can choose two quantities as (dependent and independent) variables - the other quantities are then treated as constants. You must then stick to these choices.

Assuming we are not interested in the effects of changing $\alpha$ our four options are:

a) $t_f$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_i + \frac{\alpha(t^2 – t_i^2)}{2}$
(This is the conventional choice.)

b) $t_i$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_f - \frac{\alpha(t_f^2 – t^2)}{2}$

c) $t_f$ and $v_i$ are variables; so we can rename them $t$ and $v(t)$:
$v_f - v(t) = \frac{\alpha(t^2 – t_i^2)}{2}$
$v(t) = v_f - \frac{\alpha(t^2 – t_i^2)}{2}$

d) $t_i$ and $v_f$ are variables; so we can rename them $t$ and $v(t)$:
$v(t) - v_i = \frac{\alpha(t_f^2 – t^2)}{2}$
$v(t) = v_i + \frac{\alpha(t_f^2 – t^2)}{2}$

Note that choice d) is @annamal's equation $v(t) = v_i + \frac{\alpha(t_f^2 - t^2)}{2}$.

The meaning of v(t) in the equation is the final velocity as a function of of the start time. It is not the velocity as a function of time during the motion, So it makes no sense to integrate it to find the displacement.

Thanks. That was the solution. I derivated as d) instead of b)