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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 t0 when we are at velocity v and apply the brakes, the distance travelled since I applied the brakes should be d and the time taken to cover d should be t and the velocity at that point, 0.

How should I decelerate?

The specific details of my problem are that I have an object travelling at 1498 (let's say 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 = -v/t

and the distance I would travel would be

d = vt/2 = 1898.7, much higher than the 601 I have.

It seems to me that I need some kind of sigmoid-like curve.

Edit: based on others' questions / comments, perhaps some background might help. I am an artist who is trying to make a kinetic sculpture that mimics the motion of breathing. According to research I have come across the motion of inhaling is similar to a sigmoid curve, whilst exhaling is like an exponential decay. However, a true exponential decay requires massive acceleration, so I thought of synthesising this by modelling the initial phase with the maximum acceleration that the mechanics of my system will allow. I then chose an arbitrary point that I would refine by experimentation and the decelerate from there to reach the end point (lungs empty) within time t, before then beginning to 'inhale'. I thought that this pragmatic approach might be simpler. Ha!

The numbers I have quoted are based on my simulations in Excel, but I've changed the units to SI for simplicity of explanation.

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  • $\begingroup$ Is there a limit to the acceleration? One approach would be to continue at your speed to the destination, decelerate instantly to zero, and wait for the allotted time to arrive. If you don't like infinite acceleration you can do something very similar by getting close to the target, slowing very rapidly to low speed, then coasting in to the target at low speed. The same approach I give in my answer will work. $\endgroup$ – Ross Millikan Apr 7 at 22:19
  • $\begingroup$ @RossMillikan, I've now edited my original post which I hope answers your question. I need a smooth transition; something like an exponential decay. The answer certainly isn't linear. $\endgroup$ – the_ether Apr 8 at 1:58
  • $\begingroup$ You should have said that before. The original question really just asked for a function with the proper first two integrals. The first integral would make the velocity $0$ at the proper time and the second would make the position correct as well. There are infinitely many functions that satisfy those constraints. I picked piecewise linear because it is easy to integrate, so easy to make it satisfy the constraints. Demanding smoothness doesn't restrict us much, either. It is easy to round off the corners in a way that you get infinite differentiability. $\endgroup$ – Ross Millikan Apr 8 at 2:04
  • $\begingroup$ If you have a maximum acceleration that is a very important constraint. It sounds like you want the acceleration to be monotonic as well. That is also very important. You need to define the range of acceptable functions. $\endgroup$ – Ross Millikan Apr 8 at 2:06
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You have two constraints: the time of the trajectory and the change in velocity over the trajectory. You need two knobs to turn to satisfy these. One way is to imagine you wait time $t_1$ before you decelerate, then decelerate at $a$ m/s$^2$, coming to rest at the time and position you desire.

The time of deceleration is $\frac va$ and the distance covered in that time is $\frac 12a\left(\frac va\right)^2=\frac {v^2}{2a}$
Before you decelerate you cover $vt_1$ in time $t_1$.

Now from the time requirement we need $\frac va+t_1=t$
From distance covered we need $vt_1+\frac {v^2}{2a}=d$

That is two equations in two unknowns, giving you $a,t_1$ when you solve them. If that fails by giving an imaginary solution you need to invert it. You decelerate at some rate that covers $d$ during the deceleration and gets to the end point early with zero velocity. You can then just wait for the required time.

The acceleration needed to be at zero velocity after distance $d$ is $a=\frac {v^2}{2d}$. The time taken is $\frac va$

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  • $\begingroup$ Thanks @Ross Millikan but that doesn't appear to be correct. According to you, a = 1498^2 / (2 x 601) = 1866.89. If we use that as a deceleration, then the time to come to a halt would be 1498 / 1866.89 = 0.8s, but in my problem I stated that I should come to a halt in 2.535 seconds. $\endgroup$ – the_ether Apr 7 at 17:31
  • $\begingroup$ If you come to a halt too soon, but at the right place, you can just sit there and wait. In my first approach you need to wait before you decelerate so you can arrive at the right place at zero velocity. In the second you may need to wait at the end. Nothing in the question prohibits this. $\endgroup$ – Ross Millikan Apr 7 at 22:17
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To solve this type of problem we need to do two steps. First we find the time it takes for the object to come to a stop using the equation

$$v=v_0+at$$

where $v_0$ is the initial velocity. Solving this equation for $v=0$ results in

$$t=-\frac{v_0}{a}$$

The position of an object is given by the kinematic equation

$$x=x_0+v_0t+\frac{1}{2}at^2$$

where $x_0$ is the initial position, $v_0$ is the initial velocity, and $a$ is the acceleration. Subtracting $x_0$ from both sides and defining the displacement as $d=x-x_0$. Thus the equation for finding the displacement is

$$d=v_0t+\frac{1}{2}at^2$$

If we now plug the equation for time that we found in the first part, into the equation for displacement we get

$$d=-v_0\frac{v_0}{a}+\frac{1}{2}a\left(\frac{v_0}{a}\right)^2$$

Simplifying this equation and solving for the acceleration $a$ gives

$$a=-\frac{v_0^2}{2d}$$

If we plug in the values given in the problem $d=601$ m, and an initial velocity $v=1498$ m/s, this means that the object must accelerate at a rate of $a=-1866.86\text{m/s}^2$.

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  • $\begingroup$ Thanks v much @cphys, but your value of a doesn't seem to be correct. As you've stated, t = -vo / a So, with this acceleration, the object would halt in t = -1498 / -1866.86 = 0.8 seconds But my object isn't supposed to come to a halt until 2.533 seconds $\endgroup$ – the_ether Apr 7 at 17:23

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