Consider the kinematic equation of motion for moving particle: $f(t)=t^2$.
$df(t)/dt=2t$ and $df(t)=2tdt$
For $t=0$ we have $df(t)=0dt$.
Also we have $f(0+dt)-f(0)=0$. It proves that moving particle position is $0$ from $0$ to $dt$. (moving particle rest $dt$ time)
My question is: if moving particle rest $dt$ time then how is it possible that position of particle at $dt$ is equal to $f(dt)=dt^2$
How is it possible?
Thanks.
You are confusing instantaneous velocity at t=0 and average velocity over the infinitesimal interval between 0 and dt. The latter is infinitesimal. The former, being the standard part of the latter, is 0. The average velocity is the change in position divided by the length of the interval; namely $\frac{dt^2}{dt}$.
Edit : updated answer after (extensive) discussion in the comments.
One may start from the Taylor series of $f$ in order to write $$ f(t+\mathrm{d}t) = \sum_{n=0}^\infty \frac{f^{(n)}(t)}{n!}\mathrm{d}t^n, $$ with the derivatives being defined from $$ f'(t) = \operatorname{st}\left[\frac{f(t+\mathrm{d}t)-f(t)}{\mathrm{d}t}\right], $$ where the standard part $\operatorname{st}[\,\cdot\,]$ "kills" all infinitesimal terms by rounding to the closest real number below its argument. As a consequence, the expression $f'(t)\mathrm{d}t$ is not equivalent to $f(t+\mathrm{d}t) - f(t)$ anymore; indeed, one has $$ f'(t)\mathrm{d}t = \operatorname{st}\left[\frac{f(t+\mathrm{d}t)-f(t)}{\mathrm{d}t}\right]\mathrm{d}t $$ but $$ f(t+\mathrm{d}t)-f(t) = \sum_{n=\color{red}{1}}^\infty \frac{f^{(n)}(t)}{n!}\mathrm{d}t^n. $$ Loosely speaking, you can see the left-hand side as a "regularized" derivative and the right-hand side as a kind of "raw" derivative, which cannot be the same unless $f$ is (locally) affine $-$ or under an assumption such as $\mathrm{d}t^2 = 0$, but it doesn't correspond to the standard setup of nonstandard analysis (no pun intended).
With your concrete example, namely $f(t) = t^2$, the LHS is equal to $2t\mathrm{d}t$, which vanishes at $t = 0$, while the RHS is given by $(t+\mathrm{d}t)^2 - t^2 = 2t\mathrm{d}t + \mathrm{d}t^2$, which is reduced to $\mathrm{d}t^2 \neq 0$ when $t = 0$.
I’m assuming the function $f(t)=t^2$ represents the position of the particle over time. If $dt=0$, then there is no change between $f(0)$ and $f(dt)$. If $dt>0$, then the change in position is $f(dt)-f(0)$, or simply $dt^2$. If $dt<0$, then the change in position is $f(0)-f(dt)$, or simply $-dt^2$.
