Find the constant acceleration of an object that slows down to a standstill

[Guitz]

Hi all,

Let V be an initial velocity, D any distance, and T a time duration. which algorithm allows to obtain a constant acceleration A that make the object to brake from V to a zero speed over D during T?

Thanks

 

[kuruman]

None. Constant acceleration means that it is still ON when the object comes instantaneously to rest. This means that at the next moment, it will move in a direction opposite to its previous non-zero velocity. Throw a rock straight up in the air and see what happens. The acceleration is constant.

Note: The title says "standstill" which means that the object comes to rest and stays at rest.

 

[Guitz]

well i will fix that this way :
if my object is close enough to a certain position X (after covering the distance D during the time interval T), then i update both the velocity and acceleration to zero, and i move my object at this precise location X.

 

[Dale]

Hi all,

Let V be an initial velocity, D any distance, and T a time duration. which algorithm allows to obtain a constant acceleration A that make the object to brake from V to a zero speed over D during T?

Thanks

You have $$D=\frac{1}{2}aT^2 + V T$$ which is one equation in one unknown.

Edit: never mind there is also $$aT+V=0$$ which gives two equations in one unknown. The problem is over specified and cannot be solved. You can specify either the time it takes to reach $D$ or you can specify that you stop at $D$, but you cannot specify both that it takes a certain time and that you stop at $D$.

 

[Guitz]

Thanks for your replies.

No berkeman this is not for schoolwork. I'm developping a software plugin in which cars follow spline S. An array of struct (SpeedLimit, Acceleration, Position, Tangent,...) is used, among other things, to define the speed of the car along S. When the car crosses the penultimate element of the array I must update the acceleration so that the car stops neither before nor after the destination position vector.

 

[Dale]

When the car crosses the penultimate element of the array I must update the acceleration so that the car stops neither before nor after the destination position vector.

Then you cannot specify the time $T$ that is required

 

[Guitz]

Then you cannot specify the time $T$ that is required

I'll pick T = 2 seconds

 

[Dale]

I'll pick T = 2 seconds

You cannot do that. You must leave $t$ as an unknown to be solved.

 

[Guitz]

You cannot do that. You must leave $t$ as an unknown to be solved.

The unknown is A my friend

 

[pasmith]

You have four constraints:
- At $t = 0$, $s = 0$ and $\dot s = v$.
- At $t = T$, $s = d$ and $\dot s = 0$.

Using these, you can find $s$ as a cubic function of $t$ with $\dddot s = k$; the acceleration will not be constant, but vary linearly with $t$.

If you impose the condition that the acceleration is constant, then $d$ and $t$ must be related by $$d = \frac{vt}{2}$$ and under this constraint $$a = -\frac{v^2}{2d}.$$

 

[kuruman]

Since this is a software-controlled moving object, you know where the object is and how fast it is moving as you advance time by equal intervals $dt$. When it is at distance $D_0$ from where you where you want it to stop and is moving with speed $V_0$ then

1. Use the equation $$2AD_0=-V_0^2$$ to find the acceleration.
2. Let the algorithm move the car to position $D_k$ according to $$D_k=V_0(k~dt)+\frac{1}{2}A(k~dt)^2$$ but monitor its velocity using $$V_k=V_0+A(k~dt)$$ Here, $k = 1,2,3,\dots~$ is a running counter that marks the passage of time in increments $dt$.
3. At the $kth$ step test the velocity $V_{k+1}$ coming up at the next step. If it is negative, tell the algorithm to leave the object where it is and stop.

The perceived motion would be like throwing a rock straight up in the air and have someone catch it when it's almost at rest. How close is "almost" depends on the size of your $dt$.

 

[Dale]

The unknown is A my friend

You have two equations so you need two unknowns. Two equations in two unknowns is solvable. Two equations in one unknown is not solvable. You need both $a$ and $t$ as unknowns.

 

[Guitz]

thanks again for your help, it works like a charm

 

[kuruman]

thanks again for your help, it works like a charm

If at all possible can you post a video? I am curious to see how it turned out. Thanks.

 

[Guitz]

If at all possible can you post a video? I am curious to see how it turned out. Thanks.

Sure i post it tomorrow my friend.

 

[Guitz]

Hi kuruman,

there she is :