How to decelerate from velocity v to stop time t over distance d?

Question

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.

Answer 1

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 $\frac va$

Answer 2

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 $a=-1866.86\text{m/s}^2$.