How to decelerate from velocity $v$ to stop time $t$ over distance $d$? I'd be grateful for some help with this problem I am trying to solve.
Let's say that I have an object travelling at a velocity $v$. I want that object to come to a halt in time $t$ AND travel exactly distance $d$ within that time.
So if we are at $t_0$ when we are at velocity $v$ and apply the brakes, the distance traveled since I applied the brakes should be $d$ and the time taken to cover $d$ should be $t$ and the velocity at that point be $0$.
How should I decelerate?
Some concrete numbers might help.
My initial velocity is $1498$ (let's say $\rm{m/s}$) with a distance left to go of $601$ (let's say metres) and $2.535$ seconds left.
If I concentrate on $v$ apply a constant deceleration, then
$$a = -\frac vt$$
and the distance I would travel would be
$$d = \frac{vt}{2} = 1898.7$$ this much higher than the $601$ I have.
It seems to me that I need some kind of decay-like curve.
I'm making a kinetic sculpture and am trying to drive a stepper motor so that the sculpture moves in a breathing-like motion. Inhalation is similar to an 'S'-curve (sigmoid curve). I have that working and all is fine.
Exhalation is similar to exponential decay. The problem with that is the acceleration is far too high for the system to cope. Instead, I decided to try accelerating as fast as I could and then at some point (that I would determine through experimentation) I would decelerate to complete movement required in the time given.
 A: Assume an acceleration of the form
$$a(t)=A+Bt.$$
Then
$$v(t)=V+A t+\frac{1}{2}B t^2$$
and
$$x(t)=Vt+\frac{1}{2}A t^2+\frac{1}{6}B t^3$$
where we assume that $v(0)=V$ and $x(0)=0$.
Now demand that $v(T)=0$ and $x(T)=d$. This gives two equations for $A$ and $B$,
$$0=V+A T +\frac{1}{2}B T^2$$
$$d=V T + \frac{1}{2}A T^2 + \frac{1}{6}B T^3$$
which can be solved to give
$$A=-\frac{2(2TV-3d)}{T^2}$$
$$B=\frac{6(TV-2d)}{T^3}.$$
This is not the only possible acceleration function with a specified stopping time and stopping distance, just a particularly simple one. You could assume other forms for $a(t)$ with two parameters and fit those parameters to have the specified stopping time $T$ and stopping distance $d$.
For $V=1498\,\text{m/s}$, $T=2.535\,\text{s}$, and $d=601\,\text{m}$, I find
$$a=-(1802.57\,\text{m/s}^2)+(955.931{m/s}^3)t.$$
This may not suit the OP's use case, because it starts as a deceleration but ends as an acceleration.
A: You are right that you cannot use constant acceleration. You are putting too many constraints on your system for a constant acceleration system.
If you specify an initial velocity $v(0)$ and a final velocity $v(T)$, then this determines a single acceleration $$a=\frac{v(T)-v(0)}{T}$$
now, according to the equation $$\Delta x=x(T)-x(0)=v(0)T+\frac12aT^2$$ this means that there is now only a single allowed stopping distance. In other words, you are not free to choose any stopping distance you want as well.
You can go the other way too. Let's say we specify a distance traveled $\Delta x$ along with some initial velocity $v(0)$. Then we can determine a unique acceleration:
$$a=\frac{2(\Delta x-v(0)T)}{T^2}$$
but then we know that the final velocity is given by
$$v(T)=v(0)+aT$$
which means there is only one allowed velocity after the time $T$. i.e. we are not free to choose any final velocity we want.
There is just no way to specify all that you want under constant acceleration motion. The only way you can do this is if your third constraint is consistent with the other two. Another way to see the impossibility is that you would need a function for $a$ that involves all of your specified variables $\Delta x$, $v(0)$, $v(T)$, and $T$. None exists for a constant acceleration system.
So what is the fix? Well you need to allow for time varying acceleration, but at this point there is no one way to do this to allow for a set stopping distance, a set initial velocity, and a set final velocity. You would have to specify how you want the acceleration to behave over time to go any further.
