Get acceleration from accelerometer I'm currently developing an application. I'm practically done, but I have one big issue (for me). So I have 3-axis acceleration from my android device. Now I want to have the general acceleration at time t, because after that i want to calculate velocity. 
So, what I know: 
to get velocity there is this formula : v=u+at 
to get the general acceleration: √x^2+y^2+z^2
The problem: With this two formulas my acceleration is always positive so my velocity is always speeding up. 
Is this normal? How to I get negatives accelerations?
 A: The equation you gave, $\sqrt{x^2+y^2+z^2}$ gives you the magnitude of the acceleration.  The acceleration itself is a vector, $<x, y, z>$. Your velocities use the actual acceleration vector, not the magnitude.
If you want to think of magnitudes of acceleration and speeds, rather than accelerations and velocities, you'll have to remember what direction the acceleration is and change your equations to suit.  It's much easier, however, to simply use the correct vector equations.
And make sure you are aware that you can't just track velocity and position by repeatedly adding up accelerations.  You can get close, but there will be a drift which causes the position to slowly move off of the correct answer due to errors in the acceleration (no sensor is perfect).  This can surprise people if they aren't ready for it.
A: So, the SUVAT equations are useful when you are working with constant acceleration, which almost certainly will not be the case here.  I'm not sure how this problem is typically handled in phone apps - it is likely to be very specialised to work with all the quirks of using accelerometers, so I would suggest you seek that more specialised help.
However physically speaking, you would work with the vectors $\vec{a}$, and $\vec{v}$, where it has the components $\{v_x, v_y, v_z\}$. Then if we consider just a small period of time $\Delta t$, we can approximately treat $\vec{a}$ as being constant.  That would give us a change in velocity as:
$$\vec{v}(t+\Delta t) = \vec{v}(t) +\Delta t\times \vec{a}(t),$$
for some small instant of time.  You can recognise this as the vector form of your $v=u+at$ equation, however note I am using vectors, and only considering it on a small timescale.
In terms of a computer program, this could be say $\Delta t=20$ms, and you would just repeat that equation every $20$ ms, to update your velocity.
Vectors can point in all directions, and when you add negative acceleration, it will reduce your velocity.
A: There are two things you definitely need to get straight.
First - an accelerometer will usually measure (and display) the acceleration of gravity with any other acceleration; this means you need to know the magnitude and direction of local gravity and subtract it from your reading.
Second - acceleration is a vector; that means that you need both the magnitude (which you get from the formula you have), and the direction (which you get by dividing the acceleration vector by the magnitude).
Example. Imagine your accelerometer measures $a_x = -1, a_y = 10.2, a_z = 0$. If it is pointing such that the $y$ direction is pointing up, then you know that the 10.8 for $a_y$ is made up of 9.8 m/s$^2$ for gravity, and 1.0 m/s$^2$ for actual vertical acceleration. There is also an acceleration of -1.0 m/s$^2$ in the (negative) X direction. The magnitude of acceleration is 1.41 m/s$^2$, and it is pointing along the vector (-1 1 0)/$\sqrt{2}$.
As you can see, you can have "negative" acceleration...
A: There's a key challenge with regard to accelerometers: They do not sense gravitation. No local experiment can do so, and all that a non-local experiment can do is sense deviations from a uniform gravitational field (which remains undetectable).
The accelerometer in an Android is the quintessential example of a local experiment.
Think of it this way: Put your Android at rest on a table on the surface of the Earth. Its accelerometer will register an acceleration of about 1 g upward. Draw a free body diagram. The forces on your Android are gravitation, about 1 g times mass downward, and the normal force exerted by the table on your Android, which is about 1 g times mass upward. If your Android could sense gravitation, it would register a near zero acceleration. Since it can't sense gravitation, it registers the acceleration of what it can sense. In this case, that's the 1 g upward acceleration due to the normal force.
What this means is you need a model in your software of the local gravitational acceleration. This is a bit challenging for aircraft and spacecraft, but less so for a device that is occasionally at rest on the surface of the Earth. You can take advantage of those rest periods to measure the normal force as a surrogate for gravitation -- and you can also measure tilt. The latter is important because the rate gyros in your Android bite. Their angular random walk is atrociously bad.
