Does Kinetic Energy assume constant acceleration? Okay, I've always been someone who doesn't learn anything unless I can prove it. In this case, I set out to prove the Kinetic Energy formula and (wrongly?) found that it was based on constant acceleration. (I didn't use calculus, I proved the idea of calculus as I went, using geometry and basic algebra) I'm really not entirely sure how to ask this question, but I'll try to explain as much as I can. 
So a weight is connected to a spring, you pull the weight down $x$ distance, and you have to find the acceleration at that point along with the equation of motion. (This is for a controls class revolving around oscillations/frequency/settling time etc.) Now I didn't think this would be a problem, but since the answer involves equating kinetic energy of the spring and the system as well as equivalent masses and spring constants, I couldn't wrap my head around it. If Kinetic energy is based on constant acceleration and the spring force (and therefore, acceleration) is constantly changing, how can you equate it so simply? 
Thanks, and sorry if I made some problematic assumptions or there was a complete lack of clarity. 
 A: These equations use calculus but give a succinct proof that the force is always the change in kinetic energy with respect to distance, whether or not the acceleration in constant:
$$ F = ma = m \frac{dv}{dt} = mv \frac{dv}{dt} \frac{1}{v} = mv \frac{dv}{dt} \frac{1}{dx/dt} = mv \frac{dv}{dt} \frac{dt}{dx} = mv \frac{dv}{dx} = \frac{d}{dx} \left( \frac{1}{2} m v^2 \right) = \frac{d(KE)}{dx}.$$
Integrating with respect to $x$ gives
$$ \int_{x_1}^{x_2} F(x)\, dx = KE(x_2) - KE(x_1) = \Delta KE.$$
If the force is conservative then the LHS is the potential energy difference $U(x_1) - U(x_2)$, and rearranging gives
$$ KE(x_1) + U(x_1) = KE(x_2) + U(x_2),$$
which is the statement of conservation of energy.
A: This answer ignores the title question but I believe is relevant to what you are trying to understand:
You are right. The four kinematic equations are valid for constant or zero acceleration. They are also known as the constant acceleration equations. I'm not saying KE requires constant acceleration, just that those particular four equations require it. Not contradicting what Gert commented.
Edit:
What you have done right is discovering that you cannot use the four kinematic equations directly to calculate KE when a is varying. You need an equation which accounts for changing acceleration. Either one of the following:


*

*F(x), force as a function of displacement

*F(t), force as a function of time

*a(x), acceleration as a function of displacement

*a(t), acceleration as a function of time


The above are can be calculated from one another if you have the mass (Second Law). If you have F(x), this is easy-- the KE is simply the area under that curve. This curve will not be linear because acceleration is not constant. I'm using this example to hopefully show that your suggestion to multiply F into the area of the v(x) curve will not give you the correct KE quantity, because you are using F as a constant (due to changing acceleration, F used must be a function of something else).
A: Since i dont have enough reputation...I can comment so im posting an answer that is more like a comment.....anyways back to the question
When the force acting on a system is a variable then work done by the force will be equal to the integral of F.dx which will actually be equal to the change in kinetic energy of the system...Hence as u rightly mentioned in ur last comment....the area under force-displacement graph would have to be intregated as u cannot use the kinematic eqns. since they are valid for only constant acceleration. For getting the change in kinetic energy u would indeed need any one of the functions mentioned by gregsan in his answer...so that u can integrate it....
Well I hope ur doubt is cleared....
