how to calculate a negative acceleration?

Question

I'm implementing the scrolling behaviour of a touch screen UI but I'm too tired in the moment to wrap my mind around some supposedly trivial piece of math:

y (distance/velocity)
|********
|        ******
|              ****
|                  ***
|                     ***
|                        **
|                          **
|                            *
|                             *
-------------------------------- x (time)

f(x)->?

The UI is supposed to allow the user to drag and "throw" a view in any direction and to keep it scrolling for a while even after he releases the finger from the screen. It's sort of having a momentum that depends on how fast the user was dragging before he took off the finger.

So I have a starting velocity (v0) and every 20ms I scroll by a amount relative to the current velocity. With every scrolling iteration I lower the velocity a bit until it falls below a threshold when I stop it. It just doesn't look right when I decrement it by a fixed amount (linear), so I need to model a negative acceleration but fail to come up with a decent simple formula how to calculate the amount by which I have to lower the velocity in every iteration.

Update:

Thank you for your responses so far but I still didn't manage to derive a satisfying function from the feedback yet. I probably didn't describe the desired solution good enough, so I'll try to give a real world example that should illustrate what kind of calculation I would like to do:

Assume there is a certain car driving on a certain street and the driver hits the brakes to a max until the car comes to a halt. The driver does this with the same car on the same street multiple times but begins to brake at varying velocities. While the car is slowing down I want to be able to calculate the velocity it will have exactly one second later solely based on it's current velocity. For this calculation it should not matter at which speed the car was driving when the driver began to break since all environmential factors remain the same. Of course there will be some constants in the formula but when the car is down to i.e. 30 m/s it will go the same distance in the next second, regardless whether it was driving 100 or 50 m/s when the driver began to break. So the time since hitting the breaks would also not be a parameter of the function. The deceleration at a certain velocity would always be the same.

How do you calculate the velocity one second later in such a situation assuming some arbitrary constants for deceleration, mass, friction or whatever and ignoring complicating influences like air resistance? I'm only after the kinetic energy and the it's dissipation due to the friction from breaking the car.

Update 2 I see now that the acceleration of the car would be liniear and this is actually not what I was looking for. I'll clean this up and try out the new suggestions tomorrow. Thank you for your input so far.

Answer 1

acceleration = (force / mass) 
velocity = (acceleration * time)
(force from user's finger) = acceleration / mass = velocity / time

1. give the view a mass (tweak it until things seem reasonable, and let the user tweak it)
2. identify some new force (drag)
3. give the new force (drag) a magnitude (tweak until reasonable, let the user tweak it)
4. apply the new force to the object and watch it slow down

Answer 2

I have tried this, which works (in Ruby). Not sure if the math is sound but the output looks right, meaning that you get faster as you get towards the center:

velocity=100;
(100.downto(0)).each { |distance_from_black_hole |  velocity=velocity+9.8/distance_from_black_hole; puts velocity; }

Answer 3

A bit of non programming discussion about the car example.

First, I'll assume that the driver cannot cause the brakes to lock at speed.

The first thing (or maybe the second or third thing) most new drivers learn is that the natural tendency when braking is to hold the brake pedal at a fixed position. The result is a sudden lurch forward as the car goes from moving slowly to stopping. This happens because the brakes are transitioning from dynamic friction, where braking force is proportional to brake pressure, to static friction, where braking force is restoring the forward momentum of the car. this sudden jump in acceleration is unpleasant, and the new driver learns to feather the pedal at the very end of deceleration to stop.

This behavior masks another pecularity, but this can be noticed during normal acceleration in a manual transmission car. when accelerating (or decelerating), if the driver suddenly pops the transmission out of gear, all of the passengers will suddenly lurch forward. What is actually happening, is the accelerating force that was pressing them into the backs of their seats is suddenly removed, and they spring back to a neutral sitting position. A more comfortable way to drive is to gradually feather the clutch so the motive force of the engine is removed gradually.

In both cases, the more aesthetic driving style involves smoothing the acceleration, removing sudden jumps. This is basically another way of talking about continious second derivative. Almost any motion with this property will seem natural.