Calculating acceleration based on a starting and ending distance I have been trying to figure this problem out and just can't.  
The answer is supposed to be (A): 1.0 m/s/s.  
Can anyone tell me how they got that answer?  I thought this problem was based on the formula: Acceleration $\displaystyle = \frac {(v2 - v1)}{t}$, but I can't get that answer.  What am I doing wrong? Or is the correct answer not listed?  Thanks for any help.

Problem:
  A Pinewood Derby car rolls down a track.   The total distance travel is measured each second.   Use the chart to calculate the acceleration of the car.
Time(s) |  Dist(m)

0       |     0
1       |     1
2       |     3
3       |     6
4       |    10

A) 1.0  meters/sec/sec
B) 1.5  meters/sec/sec
C) 2.0 meters/sec/sec
D) 2.5 meters/sec/sec
 A: Your equation acceleration = (v2 - v1)/t is correct (if the acceleration is constant), but you don't have measurements of velocity.  You should also have one that says distance=acceleration*t^2/2 (also for constant acceleration and a standing start).  You do have measures of distance, and any of the measurements of distance will work.  You also have to note that the measurements you have are distance in the last second.  If you want to use the data from 4 seconds, you need the total distance traveled.
A: If $x(t)$ denotes the total distance travelled at time $t$, then you can check (according to your data) that
$x(t) = \dfrac{t^2+t}2$
where $x$ is in m, $t$ in s. The acceleration is given by $a = \ddot x$, hence $a = 1\,\text m/\text s/\text s$.
A: Remark: Since the data is discrete, we are going to assume mean velocities in each time intervals, otherwise the movement's law would have to be given explicitly.

Another approach is to compute the velocity first and then the acceleration, without using equations explicitly.
Since you are given five pairs $(t,d)$, where $t$ is the time and $d$ is the
distance, we can evaluate the average velocity $v=\frac{\Delta d}{\Delta t}$ over each period $\left[t_{i},t_{i+1}\right] $, starting at $t=0$ s, and $\Delta t_{i}=d_{i+1}-d_{i}$. The measurements are equally spaced, with $\Delta t=1$ s. We get the following average velocities: 


*

*$i=0\quad t\in \left[ 0,1\right] $ s,$\quad v=\frac{1-0}{1}=1$ ms$^{-1}=1 \text{m/s}$,

*$i=1\quad t\in \left[ 1,2\right] $ s,$\quad v=\frac{3-1}{1}=2$ ms$^{-1}$,

*$i=2\quad t\in \left[ 2,3\right] $ s,$\quad v=\frac{6-3}{1}=3$ ms$^{-1}$,

*$i=3\quad t\in \left[ 3,4\right] $ s,$\quad v=\frac{10-6}{1}=4$ ms$^{-1}$.


Now we compute the average acceleration from one period to the next $a=\dfrac{\Delta v}{%
\Delta t}$: 


*

*from $\left[ 0,1\right] $ to $\left[ 1,2\right] $ s,$\quad a=%
\frac{2-1}{1}=1$ ms$^{-2},$

*from $\left[ 1,2\right] $ to $\left[ 2,3\right] $ s,$\quad a=%
\frac{3-2}{1}=1$ ms$^{-2},$

*from $\left[ 2,3\right] $ to $\left[ 3,4\right] $ s,$\quad a=%
\frac{4-3}{1}=1$ ms$^{-2}.$


Thus the average acceleration $a$ is constant: $a=1$ $\text{ms}^{-2}=1\text{m/s}^{2}$. 
In summary we get the following table with data and computations: 
$$\begin{array}{ccccccccc}
i & t_{i}\text{ (s)} &  & d_{i}\text{ (m)} &  & v_{i}\text{ (ms}^{-1}%
\text{)} &  & a_{i}\text{ (ms}^{-2}\text{)} &  \\ 
&  &  &  &  &  &  &  &  \\ 
& 0 &  & 0 &  &  &  &  &  \\ 
0 & 1 &  & 1 &  & 1 &  &  &  \\ 
1 & 2 &  & 3 &  & 2 &  & 1 &  \\ 
2 & 3 &  & 6 &  & 3 &  & 1 &  \\ 
3 & 4 &  & 10 &  & 4 &  & 1 & 
\end{array}.$$
Notes: 


*

*Since the measurements are discrete, the velocity and accelerate
values are mean values and not instant one.

*The acceleration unit ms$^{-2}$ (or $\text{m}/\text{s}^2$) is the same as meters/sec/sec in the question, tought this is an abuse of notation, since the meaning is (meters/second)/second. 

A: OK, I actually think that Ross's answer makes more sense. His assumption is that the last column of your table is the distance travelled between the last two measurements. Also, he assumed (as I did) constant velocity $a$, in which case the distance travelled at time $t$ is
$x = \dfrac{at^2}2$
Now, the last column actually corresponds to
$\Delta x(t) = x(t)-x(t-\Delta t) = a\Delta t(t-\dfrac{\Delta t}2)$,
with $\Delta t$ (time between two successive measurements) is constantly $1\,\text s$, and $\Delta x$ is the distance travelled in this interval of time (in other words: the second column of the table). Then I think you do find that $a = 2\,\text m.\text s^{-2}$.
