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$$ \begin{array}{c|c|c|c|c|c|c|c} t & 0 & 0.25 & 0.5 & 0.75 & 1 & 1.25 & 1.5\\ \hline x & 0 & 4.3 & 10.2 & 17.2 & 26.2 & 33.1& 39.1\\ \end{array} $$

$t$ 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?

Ahmed
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Thunfische
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1 Answers1

2

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.

Ahmed
  • 1,338