I have this function here
def gravitationalForce(p1,p2):
G = 1=
rVector = p1.pos - p2.pos
# if vp.mag(rVector) <.01:
# planet1.pos =- planet1.pos
# star.pos = - star.pos
rMagnitude = vp.mag(rVector)
rHat = rVector / rMagnitude
F = - rHat * G * p1.mass * p2.mass /rMagnitude**2
return F
and what I'm trying to do is make a single object/sphere that starts with a momentum of zero, some distance from a massive object. Over time the object pulls it in, and then it flies through to the other side, slowing down, and osculating back and forth. My issue is that while I am accurately simulating gravity between the two points, the moment the magnitude of the radius hits 0, the force seems to go to infinity, shooting the particle out at very high speeds in the following time steps, before it has a chance to be slowed down my the force of gravity on the other side. I tried to skip over the center when it was some small radius by implementing the conditional
# if vp.mag(rVector) <.01:
# planet1.pos =- planet1.pos
# star.pos = - star.pos
but this made no change and the object still shoots out.
Here are the given objects I generate
star = vp.sphere(pos=vp.vector(0,0,0), radius=0.2, color=vp.color.yellow,
mass = 2.0*10000, momentum=vp.vector(0,0,0), make_trail=True)
planet1 = vp.sphere(pos=vp.vector(-1,0,0), radius=0.05, color=vp.color.green,
mass = 1, momentum=vp.vector(10,0,0), make_trail=True)
You need to check to see whether the next step would, if carried out, put the planet inside the star. If so, you might move to the near side of the star, then move at (say) a constant speed to the other side of the star and turn on gravity again. The alternative would be to take very short steps while inside the star. This isn't quite physical, because you really need to use Gauss's law to calculate (in the approximation of a constant density for the star) the actual gravitational force.
Related
I'm trying to create a solar system simulation, and I'm having problems trying to figure out initial velocity vectors for random objects I've placed into the simulation.
Assume:
- I'm using Gaussian grav constant, so all my units are AU/Solar Masses/Day
- Using x,y,z for coordinates
- One star, which is fixed at 0,0,0. Quasi-random mass is determined for it
- I place a planet, at a random x,y,z coordinate, and its own quasi-random mass determined.
Before I start the nbody loop (using RK4), I would like the initial velocity of the planet to be such that it has a circular orbit around the star. Other placed planets will, of course, pull on it once the simulation starts, but I want to give it the chance to have a stable orbit...
So, in the end, I need to have an initial velocity vector (x,y,z) for the planet that means it would have a circular orbit around the star after 1 timestep.
Help? I've been beating my head against this for weeks and I don't believe I have any reasonable solution yet...
It is quite simple if you assume that the mass of the star M is much bigger than the total mass of all planets sum(m[i]). This simplifies the problem as it allows you to pin the star to the centre of the coordinate system. Also it is much easier to assume that the motion of all planets is coplanar, which further reduces the dimensionality of the problem to 2D.
First determine the magnitude of the circular orbit velocity given the magnitude of the radius vector r[i] (the radius of the orbit). It only depends on the mass of the star, because of the above mentioned assumption: v[i] = sqrt(mu / r[i]), where mu is the standard gravitational parameter of the star, mu = G * M.
Pick a random orbital phase parameter phi[i] by sampling uniformly from [0, 2*pi). Then the initial position of the planet in Cartesian coordinates is:x[i] = r[i] * cos(phi[i]) y[i] = r[i] * sin(phi[i])
With circular orbits the velocity vector is always perpendicular to the radial vector, i.e. its direction is phi[i] +/- pi/2 (+pi/2 for counter-clockwise (CCW) rotation and -pi/2 for clockwise rotation). Let's take CCW rotation as an example. The Cartesian coordinates of the planet's velocity are:vx[i] = v[i] * cos(phi[i] + pi/2) = -v[i] * sin(phi[i])vy[i] = v[i] * sin(phi[i] + pi/2) = v[i] * cos(phi[i])
This easily extends to coplanar 3D motion by adding z[i] = 0 and vz[i] = 0, but it makes no sense, since there are no forces in the Z direction and hence z[i] and vz[i] would forever stay equal to 0 (i.e. you will be solving for a 2D subspace problem of the full 3D space).
With full 3D simulation where each planet moves in a randomly inclined initial orbit, one can work that way:
This step is equal to step 1 from the 2D case.
You need to pick an initial position on the surface of the unit sphere. See here for examples on how to do that in a uniformly random fashion. Then scale the unit sphere coordinates by the magnitude of r[i].
In the 3D case, instead of two possible perpendicular vectors, there is a whole tangential plane where the planet velocity lies. The tangential plane has its normal vector collinear to the radius vector and dot(r[i], v[i]) = 0 = x[i]*vx[i] + y[i]*vy[i] + z[i]*vz[i]. One could pick any vector that is perpendicular to r[i], for example e1[i] = (-y[i], x[i], 0). This results in a null vector at the poles, so there one could pick e1[i] = (0, -z[i], y[i]) instead. Then another perpendicular vector can be found by taking the cross product of r[i] and e1[i]:e2[i] = r[i] x e1[i] = (r[2]*e1[3]-r[3]*e1[2], r[3]*e1[1]-r[1]*e1[3], r[1]*e1[2]-r[2]*e1[1]). Now e1[i] and e2[i] can be normalised by dividing them by their norms:n1[i] = e1[i] / ||e1[i]||n2[i] = e2[i] / ||e2[i]||where ||a|| = sqrt(dot(a, a)) = sqrt(a.x^2 + a.y^2 + a.z^2). Now that you have an orthogonal vector basis in the tangential plane, you can pick one random angle omega in [0, 2*pi) and compute the velocity vector as v[i] = cos(omega) * n1[i] + sin(omega) * n2[i], or as Cartesian components:vx[i] = cos(omega) * n1[i].x + sin(omega) * n2[i].xvy[i] = cos(omega) * n1[i].y + sin(omega) * n2[i].yvz[i] = cos(omega) * n1[i].z + sin(omega) * n2[i].z.
Note that by construction the basis in step 3 depends on the radius vector, but this does not matter since a random direction (omega) is added.
As to the choice of units, in simulation science we always tend to keep things in natural units, i.e. units where all computed quantities are dimensionless and kept in [0, 1] or at least within 1-2 orders of magnitude and so the full resolution of the limited floating-point representation could be used. If you take the star mass to be in units of Solar mass, distances to be in AUs and time to be in years, then for an Earth-like planet at 1 AU around a Sun-like star, the magnitude of the orbital velocity would be 2*pi (AU/yr) and the magnitude of the radius vector would be 1 (AU).
Just let centripetal acceleration equal gravitational acceleration.
m1v2 / r = G m1m2 / r2
v = sqrt( G m2 / r )
Of course the star mass m2 must be much greater than the planet mass m1 or you don't really have a one-body problem.
Units are a pain in the butt when setting up physics problems. I've spent days resolving errors in seconds vs timestep units. Your choice of AU/Solar Masses/Day is utterly insane. Fix that before anything else.
And, keep in mind that computers have inherently limited precision. An nbody simulation accumulates integration error, so after a million or a billion steps you will certainly not have a circle, regardless of the step duration. I don't know much about that math, but I think stable n-body systems keep themselves stable by resonances which absorb minor variations, whether introduced by nearby stars or by the FPU. So the setup might work fine for a stable, 5-body problem but still fail for a 1-body problem.
As Ed suggested, I would use the mks units, rather than some other set of units.
For the initial velocity, I would agree with part of what Ed said, but I would use the vector form of the centripetal acceleration:
m1v2/r r(hat) = G m1 m2 / r2 r(hat)
Set z to 0, and convert from polar coordinates to cartesian coordinates (x,y). Then, you can assign either y or x an initial velocity, and compute what the other variable is to satisfy the circular orbit criteria. This should give you an initial (Vx,Vy) that you can start your nbody problem from. There should also be quite a bit of literature out there on numerical recipes for nbody central force problems.
I am trying to calculate the forces that will act on circular objects in the event of a collision. Unfortunately, my mechanics is slightly rusty so i'm having a bit of trouble.
I have an agent class with members
vector position // (x,y)
vector velocity // (x,y)
vector forward // (x,y)
float radius // radius of the agent (all circles)
float mass
So if we have A,B:Agent, and in the next time step the velocity is going to change the position. If a collision is going to occur I want to work out the force that will act on the objects.
I know Line1 = (B.position-A.position) is needed to work out the angle of the resultant force but how to calculate it is baffling me when I have to take into account current velocity of the vehicle along with the angle of collision.
arctan(L1.y,L1.x) is am angle for the force (direction can be determined)
sin/cos are height/width of the components
Also I know to calculate the rotated axis I need to use
x = cos(T)*vel.x + sin(T)*vel.y
y = cos(T)*vel.y + sin(T)*vel.x
This is where my brain can't cope anymore.. Any help would be appreciated.
As I say, the aim is to work out the vector force applied to the objects as I have already taken into account basic physics.
Added a little psudocode to show where I was starting to go with it..
A,B:Agent
Agent {
vector position, velocity, front;
float radius,mass;
}
vector dist = B.position - A.position;
float distMag = dist.magnitude();
if (distMag < A.radius + B.radius) { // collision
float theta = arctan(dist.y,dist.x);
flost sine = sin(theta);
float cosine = cos(theta);
vector newAxis = new vector;
newAxis.x = cosine * dist .x + sine * dist .y;
newAxis.y = cosine * dist .y - sine * dist .x;
// Converted velocities
vector[] vTemp = {
new vector(), new vector() };
vTemp[0].x = cosine * agent.velocity.x + sine * agent.velocity.y;
vTemp[0].y = cosine * agent.velocity.y - sine * agent.velocity.x;
vTemp[1].x = cosine * current.velocity.x + sine * current.velocity.y;
vTemp[1].y = cosine * current.velocity.y - sine * current.velocity.x;
Here's to hoping there's a curious maths geek on stack..
Let us assume, without loss of generality, that we are in the second object's reference frame before the collision.
Conservation of momentum:
m1*vx1 = m1*vx1' + m2*vx2'
m1*vy1 = m1*vy1' + m2*vy2'
Solving for vx1', vy1':
vx1' = vx1 - (m2/m1)*vx2'
vy1' = vy1 - (m2/m1)*vy2'
Secretly, I will remember the fact that vx1'*vx1' + vy1'*vy1' = v1'*v1'.
Conservation of energy (one of the things elastic collisions give us is that angle of incidence is angle of reflection):
m1*v1*v1 = m1*v1'*v1' + m2*v2'+v2'
Solving for v1' squared:
v1'*v1' = v1*v1 - (m2/m1)v2'*v2'
Combine to eliminate v1':
(1-m2/m1)*v2'*v2' = 2*(vx2'*vx1+vy2'*vy1)
Now, if you've ever seen a stationary poolball hit, you know that it flies off in the direction of the contact normal (this is the same as your theta).
v2x' = v2'cos(theta)
v2y' = v2'sin(theta)
Therefore:
v2' = 2/(1-m2/m1)*(vx1*sin(theta)+vy1*cos(theta))
Now you can solve for v1' (either use v1'=sqrt(v1*v1-(m2/m1)*v2'*v2') or solve the whole thing in terms of the input variables).
Let's call phi = arctan(vy1/vx1). The angle of incidence relative to the tangent line to the circle at the point of intersection is 90-phi-theta (pi/2-phi-theta if you prefer). Add that again for the reflection, then convert back to an angle relative to the horizontal. Let's call the angle of incidence psi = 180-phi-2*theta (pi-phi-2*theta). Or,
psi = (180 or pi) - (arctan(vy1/vx1))-2*(arctan(dy/dx))
So:
vx1' = v1'sin(psi)
vy1' = v1'cos(psi)
Consider: if these circles are supposed to be solid 3D spheres, then use a mass proportional to radius-cubed for each one (note that the proportionality constant cancels out). If they are supposed to be disklike, use mass proportional to radius-squared. If they are rings, just use radius.
Next point to consider: Since the computer updates at discrete time events, you actually have overlapping objects. You should back out the objects so that they don't overlap before computing the new location of each object. For extra credit, figure out the time that they should have intersected, then move them in the new direction for that amount of time. Note that this time is just the overlap / old velocity. The reason that this is important is that you might imagine a collision that is computed that causes the objects to still overlap (causing them to collide again).
Next point to consider: to translate the original problem into this problem, just subtract object 2's velocity from object 1 (component-wise). After the computation, remember to add it back.
Final point to consider: I probably made an algebra error somewhere along the line. You should seriously consider checking my work.
I want to simulate a free fall and a collision with the ground (for example a bouncing ball). The object will fall in a vacuum - an air resistance can be omitted. A collision with the ground should causes some energy loss so finally the object will stop moving. I use JOGL to render a point which is my falling object. A gravity is constant (-9.8 m/s^2).
I found an euler method to calculate a new position of the point:
deltaTime = currentTime - previousTime;
vel += acc * deltaTime;
pos += vel * deltaTime;
but I'm doing something wrong. The point bounces a few times and then it's moving down (very slow).
Here is a pseudocode (initial pos = (0.0f, 2.0f, 0.0f), initial vel(0.0f, 0.0f, 0.0f), gravity = -9.8f):
display()
{
calculateDeltaTime();
velocity.y += gravity * deltaTime;
pos.y += velocity.y * deltaTime;
if(pos.y < -2.0f) //a collision with the ground
{
velocity.y = velocity.y * energyLoss * -1.0f;
}
}
What is the best way to achieve a realistic effect ? How the euler method refer to the constant acceleration equations ?
Because floating points dont round-up nicely, you'll never get at a velocity that's actually 0. You'd probably get something like -0.00000000000001 or something.
you need to to make it 0.0 when it's close enough. (define some delta.)
To expand upon my comment above, and to answer Tobias, I'll add a complete answer here.
Upon initial inspection, I determined that you were bleeding off velocity to fast. Simply put, the relationship between kinetic energy and velocity is E = m v^2 /2, so after taking the derivative with respect to velocity you get
delta_E = m v delta_v
Then, depending on how energyloss is defined, you can establish the relationship between delta_E and energyloss. For instance, in most cases energyloss = delta_E/E_initial, then the above relationship can be simplified as
delta_v = energyloss*v_initial / 2
This is assuming that the time interval is small allowing you to replace v in the first equation with v_initial, so you should be able to get away with it for what your doing. To be clear, delta_v is subtracted from velocity.y inside your collision block instead of what you have.
As to the question of adding air-resistance or not, the answer is it depends. For small initial drop heights, it won't matter, but it can start to matter with smaller energy losses due to bounce and higher drop points. For a 1 gram, 1 inch (2.54 cm) diameter, smooth sphere, I plotted time difference between with and without air friction vs. drop height:
For low energy loss materials (80 - 90+ % energy retained), I'd consider adding it in for 10 meter, and higher, drop heights. But, if the drops are under 2 - 3 meters, I wouldn't bother.
If anyone wants the calculations, I'll share them.
I wanted to simulate the following through animation :
A ball starts with a certain velocity at the bottom most point of
a vertical circular loop and keeps rolling in it until its velocity permits.
For this, I wanted to find velocity/x/y vs. time equation.
For e.g. if the ball had mass : 5Kg, radius of the circular loop = 10m,
and initial velocity of the ball is 200 m/s, what will its velocity and (x,y) position
be after 5 seconds?
thanks.
Sliding, frictionless case with a point-particle ball
In this case we aren't worrying about rotational energy and are assuming that the ball is actually a point particle. Then, in order for the ball to stay on at the top, the centripetal force condition has to be satisfied:
m * v_top^2 / r = m * g
so
v_top = sqrt(r * g)
So the minimum initial velocity is determined by:
1 / 2 * m * v0^2 >= 1 / 2 * m * v_top^2 + m * g * 2 * r
v0 >= sqrt(5 * r * g)
This is similar to what Pete said, except that he forgot the centripetal force condition to stay on at the top.
Next, the acceleration tangential to the track is given by:
a = - g * sin(theta)
but a = r * alpha = r * d^2(theta)/dt^2 where alpha is the rotational acceleration. Thus, we get
r * d^2(theta)/dt^2 = g * sin(theta)
However, I don't know of an analytical solution to this differential equation and Mathematica was stumbling with finding one too. You can't just move the dts to the other side and integrate because theta is a function of t. I would recommend solving it by numerical means such as a Runga-Kutte or maybe the Verlet method. I solved it using Mathematica for the parameters you gave, but with the ball moving so quickly, it doesn't really slow down much in going around. When I lowered the initial velocity though, I was able to see the speeding up and slowing down by plotting theta as a function of time.
Adding in other things like a finite ball radius, rotational energy and friction are certainly doable, but I would worry about being able to solve this first case before moving on because it only gets more complicated from here. By the way, with the friction you will have to choose some kinetic coefficient of friction for your given materials which will of course be proportional to the normal force exerted on the ball by the track which can be solved for by summing the force components along the radius of the circle and don't forget to include the centripetal force condition.
If you haven't done this sort of physics before, I definitely recommend getting a introductory good book on physics (with calculus) and working through it. You only need to bother with the sections that apply to mechanics though that is a very large section of the book probably. There might be better routes to pursue though like some of the resources in this question.
If there are no acceleration (x,y) =(xstart+ vx*time ,ystart + vy*time) and speed remain the same, and it is not related to the radius
Since the velocity is constant you will have an angular velocity of omega = vel / radius. You will obtain how many radians you ball will move per second over its circular path.
To get the position at time t you just have to exploit polar coordinates:
x = x_center + sin( 3/2*PI + omega*t)*radius
y = y_center + cos( 3/2*PI + omega*t)*radius
This because you start from bottom point of the circle (so its 3/2*PI) plus how many radiants you move every second (we obtained it from tangential velocity). All multiplied for the radius, otherwise you will consider a unity circle.
EDIT: Since you wonder how to find a position of an object that is subject to many different forces I can tell you that usually a physical engine doesn't care about finding equations of moving objects. It just applies forces to objects considering their intended motions (like your circular one) or environmental factors (like gravity or friction) and calculates coordinates step by step by applying forces and using an integrator to see the results.
Ignoring friction, the forces on the ball are gravity and the track.
First, there are two main cases - is the velocity enough for the ball to loop-the-loop or not:
initial energy = 1/2 m v² = 0.5 * 5 * 200 * 200
potential energy = m g h = 5 * 9.8 * 20
so it will go round the whole loop.
Initially the ball is at the bottom of the loop, theta = 0
The acceleration on the ball is the component of g along the track
a = g⋅sin theta
The distance travelled is theta * radius. It is also the double integral of acceleration against time.
theta ⋅ radius = double integral of acceleration against time
Integrating acceleration once gives velocity, integrating velocity gives distance.
so solve this for t:
theta ⋅ r = ∫(∫ g⋅sin theta.dt).dt
then your x and y are trivial functions of theta.
Whether you solve it analytically or numerically is up to you.
With dynamic friction, friction is usually proportional to the normal force on the bodies. So this will equal the centripetal force - proportional to the square of the angular velocity, and the component of gravity normal to the track (g sin theta)
You didn't tell anything about how you want your velocity to change. Do you have any friction model? If there is no friction, then the formulas are simple:
length = velocity*t
x = sin(length)*radius
y = -cos(length)*radius
If the velocity is changing, then you have to change length to something like
length = integral over dt[0..t] (velocity dt)
The only thing I wanted to add is the if this is real ball (sphere) with mass 5kg then it must have a diameter dia=(6*m/(PI*rho))^(1/3) where rho is the density of the material. For steel (rho=7680) the diameter is dia=0.1075 meters. Therefore the pitch radius (radius at which the center of gravity of the ball rides on) is equal to R=10-(dia/2) or R=9.9466 meters.
The problem gets a little more complex when friction is included. For one you have to consider the direction of friction (assuming dry friction theory). That depends on the amount the ball rotates in its axis and that depends on moment of inertia of the ball.
When you do the simulation you might want to monitor the total kinetic energy + the total potential energy and make sure your are not adding energy to the system (or taking away). [Don't forget to include the rotational component for the kinetic energy]
Get a standard book on dynamics, and I am sure a similar problem is already described in the book.I would recommend "Vector Mechanic for Engineers - Dynamics".
I've tried the typical physics equations for this but none of them really work because the equations deal with constant acceleration and mine will need to change to work correctly. Basically I have a car that can be going at a large range of speeds and needs to slow down and stop over a given distance and time as it reaches the end of its path.
So, I have:
V0, or the current speed
Vf, or the speed I want to reach (typically 0)
t, or the amount of time I want to take to reach the end of my path
d, or the distance I want to go as I change from V0 to Vf
I want to calculate
a, or the acceleration needed to go from V0 to Vf
The reason this becomes a programming-specific question is because a needs to be recalculated every single timestep as the car keeps stopping. So, V0 constantly is changed to be V0 from last timestep plus the a that was calculated last timestep. So essentially it will start stopping slowly then will eventually stop more abruptly, sort of like a car in real life.
EDITS:
All right, thanks for the great responses. A lot of what I needed was just some help thinking about this. Let me be more specific now that I've got some more ideas from you all:
I have a car c that is 64 pixels from its destination, so d=64. It is driving at 2 pixels per timestep, where a timestep is 1/60 of a second. I want to find the acceleration a that will bring it to a speed of 0.2 pixels per timestep by the time it has traveled d.
d = 64 //distance
V0 = 2 //initial velocity (in ppt)
Vf = 0.2 //final velocity (in ppt)
Also because this happens in a game loop, a variable delta is passed through to each action, which is the multiple of 1/60s that the last timestep took. In other words, if it took 1/60s, then delta is 1.0, if it took 1/30s, then delta is 0.5. Before acceleration is actually applied, it is multiplied by this delta value. Similarly, before the car moves again its velocity is multiplied by the delta value. This is pretty standard stuff, but it might be what is causing problems with my calculations.
Linear acceleration a for a distance d going from a starting speed Vi to a final speed Vf:
a = (Vf*Vf - Vi*Vi)/(2 * d)
EDIT:
After your edit, let me try and gauge what you need...
If you take this formula and insert your numbers, you get a constant acceleration of -0,0309375. Now, let's keep calling this result 'a'.
What you need between timestamps (frames?) is not actually the acceleration, but new location of the vehicle, right? So you use the following formula:
Sd = Vi * t + 0.5 * t * t * a
where Sd is the current distance from the start position at current frame/moment/sum_of_deltas, Vi is the starting speed, and t is the time since the start.
With this, your decceleration is constant, but even if it is linear, your speed will accomodate to your constraints.
If you want a non-linear decceleration, you could find some non-linear interpolation method, and interpolate not acceleration, but simply position between two points.
location = non_linear_function(time);
The four constraints you give are one too many for a linear system (one with constant acceleration), where any three of the variables would suffice to compute the acceleration and thereby determine the fourth variables. However, the system is way under-specified for a completely general nonlinear system -- there may be uncountably infinite ways to change acceleration over time while satisfying all the constraints as given. Can you perhaps specify better along what kind of curve acceleration should change over time?
Using 0 index to mean "at the start", 1 to mean "at the end", and D for Delta to mean "variation", given a linearly changing acceleration
a(t) = a0 + t * (a1-a0)/Dt
where a0 and a1 are the two parameters we want to compute to satisfy all the various constraints, I compute (if there's been no misstep, as I did it all by hand):
DV = Dt * (a0+a1)/2
Ds = Dt * (V0 + ((a1-a0)/6 + a0/2) * Dt)
Given DV, Dt and Ds are all given, this leaves 2 linear equations in the unknowns a0 and a1 so you can solve for these (but I'm leaving things in this form to make it easier to double check on my derivations!!!).
If you're applying the proper formulas at every step to compute changes in space and velocity, it should make no difference whether you compute a0 and a1 once and for all or recompute them at every step based on the remaining Dt, Ds and DV.
If you're trying to simulate a time-dependent acceleration in your equations, it just means that you should assume that. You have to integrate F = ma along with the acceleration equations, that's all. If acceleration isn't constant, you just have to solve a system of equations instead of just one.
So now it's really three vector equations that you have to integrate simultaneously: one for each component of displacement, velocity, and acceleration, or nine equations in total. The force as a function of time will be an input for your problem.
If you're assuming 1D motion you're down to three simultaneous equations. The ones for velocity and displacement are both pretty easy.
In real life, a car's stopping ability depends on the pressure on the brake pedal, any engine braking that's going on, surface conditions, and such: also, there's that "grab" at the end when the car really stops. Modeling that is complicated, and you're unlikely to find good answers on a programming website. Find some automotive engineers.
Aside from that, I don't know what you're asking for. Are you trying to determine a braking schedule? As in there's a certain amount of deceleration while coasting, and then applying the brake? In real driving, the time is not usually considered in these maneuvers, but rather the distance.
As far as I can tell, your problem is that you aren't asking for anything specific, which suggests that you really haven't figured out what you actually want. If you'd provide a sample use for this, we could probably help you. As it is, you've provided the bare bones of a problem that is either overdetermined or way underconstrained, and there's really nothing we can do with that.
if you need to go from 10m/s to 0m/s in 1m with linear acceleration you need 2 equations.
first find the time (t) it takes to stop.
v0 = initial velocity
vf = final velocity
x0 = initial displacement
xf = final displacement
a = constant linear acceleration
(xf-x0)=.5*(v0-vf)*t
t=2*(xf-x0)/(v0-vf)
t=2*(1m-0m)/(10m/s-0m/s)
t=.2seconds
next to calculate the linear acceleration between x0 & xf
(xf-x0)=(v0-vf)*t+.5*a*t^2
(1m-0m)=(10m/s-0m/s)*(.2s)+.5*a*((.2s)^2)
1m=(10m/s)*(.2s)+.5*a*(.04s^2)
1m=2m+a*(.02s^2)
-1m=a*(.02s^2)
a=-1m/(.02s^2)
a=-50m/s^2
in terms of gravity (g's)
a=(-50m/s^2)/(9.8m/s^2)
a=5.1g over the .2 seconds from 0m to 10m
Problem is either overconstrained or underconstrained (a is not constant? is there a maximum a?) or ambiguous.
Simplest formula would be a=(Vf-V0)/t
Edit: if time is not constrained, and distance s is constrained, and acceleration is constant, then the relevant formulae are s = (Vf+V0)/2 * t, t=(Vf-V0)/a which simplifies to a = (Vf2 - V02) / (2s).