Calculate initial velocity to move a set distance with inertia

Question

I want to move something a set distance. However in my system there is inertia/drag/negative accelaration. I\'m using a simple calculation like this for it:

Answer 1

Distance travelled is just the integral of velocity with respect to time. You need to integrate your expression with respect to time with limits [v, 0] and this will give you an expression for distance in terms of v (initial velocity).

Answer 2

If you want to move a set distance, use the following:

Answer 3

This is a simple kinematics problem.

At some time t, the velocity (v) of an object under constant acceleration is described by:

v = v0 + at

Where v0 is the initial velocity and a is the acceleration. In your case, the final velocity is zero (the object is stopped) so we can solve for t:

t = -v0/a

To find the total difference traveled, we take the integral of the velocity (the first equation) over time. I haven't done an integral in years, but I'm pretty sure this one works out to:

d = v0t + 1/2 * at^2

We can substitute in the equation for t we developed ealier:

d = v0^2/a + 1/2 * v0^2 / a

And the solve for v0:

v0 = sqrt(-2ad)

Or, in a more programming-language format:

initialVelocity = sqrt( -2 * acceleration * distance );

The acceleration in this case is negative (the object is slowing down), and I'm assuming that it's constant, otherwise this gets more complicated.

If you want to use this inside a loop with a finite number of steps, you'll need to be a little careful. Each iteration of the loop represents a period of time. The object will move an amount equal to the average velocity times the length of time. A sample loop with the length of time of an iteration equal to 1 would look something like this:

position = 0;
currentVelocity = initialVelocity;
while( currentVelocity > 0 )
{
    averageVelocity = currentVelocity + (acceleration / 2);
    position = position + averageVelocity;
    currentVelocity += acceleration;
}