Obtaining Acceleration from Position vs Time graph

[aleronLolli]

I have collected data on a car accelerating down a straight track. I graphed it Position vs Time. Then I found the tangent slope at 5 points along the curve and plotted them as velocity. But then I linearized the position vs Time graph (position vs Time^2.) However, the slope of the velocity vs Time graph i got from taking the slope of the tangents was about double of what I got from linearization. This doesn't make sense to me...

Equations:
Position vs Time:
x = 9.396x^2-1.188x+6.014
Velocity vs Time: (from tangent slopes)
18.7798
Position vs Time^2: (from linearization)
9.188

 

[FactChecker]

Both of those results are reasonably close to what they should be. If position = 9.396*time^2-1.188*time+6.014, then velocity = 2*9.396*time-1.188. Remember that the position at a later time (when the velocity is high) has the accumulated effect of all the proceeding time (when velocity was low). So you should not expect the same multipliers. Velocity's multiplier of time should be twice as high as position's multiplier of time.

Velocity is the derivative of position.

 

[aleronLolli]

So what would the acceleration of the car be? 9.188 or 18.7798?

 

[FactChecker]

Is this a homework problem? If so, there is a format for those questions and I will not help more except to say that acceleration is the derivative of velocity.