Cart on Ramp with Varying Acceleration For a lab that I have to do, the teacher has given me a data table of the displacement of a cart on a ramp after successive intervals of time. I have used this data to derive the velocities and accelerations of the cart.
This is all fine, the problem is that I am now being asked to find the angle of the ramp. I don't know how to figure this out because the acceleration is not constant. After every half a second the acceleration appears to switch between -1.68 m/s2 and -1.72 m/s2. There is no indication that any other forces are involved aside from gravity. How can this be possible if the acceleration of gravity is constant?
My initial thought was that it might be a circular or spiral ramp of some sort, but the displacement never decreases as I would expect after the cart would reach a displacement equal to the diameter of the circle. What else could it be and how could I find the angle from it?
Here is the data that I am given:
Time (s) | Displacement (m)
0.00     | -10.00
0.50     | -5.21
1.00     | -0.85
1.50     | 3.09
2.00     | 6.60
2.50     | 9.69
3.00     | 12.35
3.50     | 14.59
4.00     | 16.40

And here is the data that I have calculated:
Time (s) | Velocity (m/s)
0.50     | 9.6
1.00     | 8.73
1.50     | 7.88
2.00     | 7.02
2.50     | 6.18
3.00     | 5.32
3.50     | 4.48
4.00     | 3.62


Time (s) | Acceleration (m/s^2)
0.75     | -1.7
1.25     | -1.68
1.75     | -1.72
2.25     | -1.68
2.75     | -1.72
3.25     | -1.68
3.75     | -1.72

 A: I agree with kevinsa5 that the variation in $a$ is due to rounding errors, but I'd like to suggest a better way to analyse the data.
Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is constant the distance is given by the SUVAT equation:
$$ s = ut + \tfrac{1}{2}a t^2 $$
where $u$ is the initial velocity and $a$ is the acceleration. The graph of $s(t)$ isn't a straight line, but we can convert it to a straight line by dividing through by $t$ to get:
$$ \frac{s}{t} = u + \tfrac{1}{2}a t $$
So a graph of $s/t$ against $t$ will be a straight line with gradient $\tfrac{1}{2}a$ and intercept $u$. There's a minor wrinkle that in your data when $t = 0$ $s = -10$, however we can add $10$ to all the distances to make $s(0) = 0$ as long as we remember to subtract if off again at the end.
If we do this to your data we get:
 t    s (+10)  s/t
0.0    0.00   
0.5    4.79   9.58
1.0    9.15   9.15
1.5   13.09   8.73
2.0   16.60   8.30
2.5   19.69   7.88
3.0   22.35   7.45
3.5   24.59   7.03
4.0   26.40   6.60

and a graph of $s/t$ against $t$ looks like:

The blue squares are your data and the red line is a linear regression. The data obviously lies on the straight line, and the linear regression gives:
Gradient   -0.8508
Intercept   10.003

So the initial velocity is $u = 10.00$ and the acceleration is $a = -1.70$ (rounding everything to 2 decimal places).
A: The best thing to do is draw a graph of velocity versus time. Knowing the acceleration is constant, you draw the best straight line. The inclination will give you the acceleration. 
You can even draw the maximum inclined and least inclined line to determine your uncertainty. 
