When it's better choose an specific Dx than calculate the Derivative?

Question

I was thinking about this problem: a car that moves as x(t) = t^3 in t=0 if i calculate the derivative in that point is 0. But in the reality the car near the time 0 its accelerating. If i choose certain dt like 0,01 and i solve ((t + dt)^3 - t^3)/dt when t=0, the rate of change it's 0,001. Little but not cero. Therefore Can i say its a better aproximation than the derivative? It exits some clue about when its better choose a specific dt than make dt aproaches to 0 (derivative)?

Answer 1

The velocity at $t=0$ is indeed zero. What you've calculated is the average velocity between $t=0$ and $t=0.01$, which is a different, albeit useful, value. In general, average velocity & instantaneous velocity (i.e. the usual velocity) are equal in case of constant-speed motion, which will have an equation of the form $x(t)=at+b$ (here, both velocities will be equal to $a$).