How Do You Calculate Acceleration from Position and Time Data?

[mixedtape_15]

Need Help Please! (Acceleration Problem)

I need help really badly with this problem. I've been trying to figure it out and I feel like a total idiot because I keep getting the same answer and the computer keeps telling me that I'm wrong.

The Question Is
An object is moving in a straight line with a constant acceleration. Its position is measured at three different times, as shown in the table below.

Time (s) Position, (m)
44.00 8.500
45.70 14.573
47.40 25.993

Calculate the magnitude of the acceleration at t=45.70 s.

And What I did was find the initial Velocity and the final velocity using those times and positions.
and for vi I ended up with 3.837 m/s and for vf i ended up with 6.717 m/s
then I took those and the times and put them into the equation Delta V / Delta T to get the acceleration and ended up with 1.694 m/s^2
But it keeps telling me that its wrong..and I'm completely confused..because I'm new to the whole physics thing. So yeah I need help really badly.
Thanks in advance.

 

[Astronuc]

What did one use for times to calculate the acceleration?

3.837 m/s and 6.717 m/s are the 'average' speeds during the first and second intervals, so using the intial and final time as the time interval would not be appropriate if one takes the difference in average velocities and divides by the difference in time.

Or one could solve 3 simultaneous equations for x₀, v₀ and a :

x₁ = x₀ + v₀*t₁ + 1/2a*t₁²

x₂ = x₀ + v₀*t₂ + 1/2a*t₂²

x₃ = x₀ + v₀*t₃ + 1/2a*t₃²

 

[mixedtape_15]

Okay this is exactly what I did
vi = (14.573m-8.500m)/(45.70s-44.00s) = 3.572m/s
vf = (25.993m-14.573m)/(47.40s-47.40s) = 6.717 m/s

a = (vf-vi)/delta t
a = (6.716m/s-3.572m/s)/(1.70s) = 1.850 m/s^2
I'm getting a different answer now than I did in the beginning..which is odd. And now its working out...okay..thats great. I must have screwed it up all those other times. Thanks for trying to help me...but I have the right answer now. But yeah thanks a bunch.