$$ \begin{array}{c|c|c|c|c|c|c|c} t & 0 & 0.25 & 0.5 & 0.75 & 1 & 1.25 & 1.5\\ \hline x & 0 & 4.3 & 10.2 & 17.2 & 26.2 & 33.1& 39.1\\ \end{array} $$
$t$ is time in second and $x$ is distance in meter. I wanted to find the acceleration when $t = 0.25$
I constructed this $$f''(0.25) = \frac{f(0.25+0.25) - 2f(0.25) + f(0.25-0.25)}{(0.25)^2}$$
The result i get is $25.6~m/s^2$
but the actual result I am given is $0.4~m/s^2$
What am I doing wrong?
You are using 2nd order central difference method to find the second derivative. :
$$f''(t) \approx \frac{f(t+h)-2f(t)+f(t-h)}{h^2}$$
with $t = 0.25$ and $h = 0.25$.
Your calculations are true according to the given data.