How to solve the loss of values while using the diff function ?

Hey guys , So i was given the altitude and time data for a certain incident and i was required to find velocity and acceleration . This is what a part of my code looks like :

t=0:10:250;
tv=(t(1)+t(2))/2:10:t(end);
vel=diff(c)./diff(t);
acc=diff(vel)./diff(tv);

But what do i define my time for acceleration as? is it ta = (tv(1)+tv(2))/2:10:t(end); because that did not seems to work as it would have 25 elements while i only need 24 as the acceleration variable has only 24 elements . Please help ! thanks in advance!

Answer 1

[Stephen23]

Use gradient instead.

[Sean de Wolski]

Slick. I've always zero-padded or used one of the FEX entries but this is better.

Answer 2

[KSSV]

Why you are worried about size of time? You have to use time step (dt) which remains constant.

t=0:10:250;
dt = min(diff(t)) ;
% tv=(t(1)+t(2))/2:10:t(end);
vel=diff(c)./dt;
acc=diff(vel)./dt;