Can one do calculus in finitist mathematics Can one "do calculus" in weak formal systems like primitive recursive arithmetic or elementary function arithmetic? Can one  at least explain the physics of classical mechanics in an elementary formal mathematical system like the ones described above ?
I admit this is  a vague question so I leave to you the task of giving this question  a reasonable rigorous interpretation (if any). I feel that with lot's of work , one can reproduce most of the results of calculus in the form of multivariable inequalities between some integers.
 A: Unsurprisingly, the answer is "yes, depending what you mean." Relevant topics include arithmetization, computable analysis, and reverse mathematics.

The first point of vagueness is exactly what "finitistic mathematics" means. There are various candidates for this one might propose - in the setting of first-order arithmetic we have $\mathsf{I\Sigma_1}$ and the substantially-weaker $\mathsf{EFA}$, and in the setting of second-order arithmetic we have $\mathsf{RCA_0}$ and its various weakenings.

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*Note that there's a horrible terminology issue here: both "first-order arithmetic" and "second-order arithmetic" here refer to languages in first-order logic. Terms like "one-sorted arithmetic" and "two-sorted arithmetic," or "arithmetic" and "analysis," respectively would be much better; unfortunately the above usage is entrenched. Boo I say.
Even beyond the specific choice of theory, though, there's a clear "super-vagueness:" what language are we even posing this theory in? Strength of theory and strength of language are a priori unrelated: there are very strong theories in very limited languages and very weak theories in very rich languages. For example, $\mathsf{PA}$ has much greater logical strength than $\mathsf{RCA_0}$ (as measured e.g. by the two systems respective provably total functions), but the language of $\mathsf{RCA_0}$ is richer than that of $\mathsf{PA}$ in a very important way. The linguistic aspect of finitization is more subtle in my experience, so let me focus on that first. For simplicity, I'm going to imagine that we only ever care about continuous functions defined on all of $\mathbb{R}$ going forward.
"Calculus results," broadly speaking, fall into two categories: results about specific continuous functions of interest, and general results about the class of all continuous functions as a whole. Now a priori a single real number lives at the same "type level" as a whole set of natural numbers, and so a function on real numbers like $\sin(x)$ lives at the same "type level" as a whole set of sets of natural numbers. Fortunately, however, continuity comes to the rescue:

Fix an appropriate bijection $b:\mathbb{Q}^4\rightarrow\mathbb{N}$. Given a continuous $f:\mathbb{R}\rightarrow\mathbb{R}$, let $$\mathsf{CODE}(f)=\{b(a,b,c,d)\in\mathbb{Q}^4:\forall x\in\mathbb{R}(\vert x-a\vert<b\implies \vert f(x)-c\vert<d)\}.$$ Then $\mathsf{CODE}(f)$ determines $f$ uniquely amongst all continuous functions $\mathbb{R}\rightarrow\mathbb{R}$, and basic operations on continuous functions correspond to arithmetically definable operations on their $\mathsf{CODE}$s.

This gives us two useful tricks. First, "naturally occurring" continuous functions - including $\sin(x)$ - have very simple (e.g. primitive recursive or better) $\mathsf{CODE}$s. Consequently results about specific such functions can be faithfully stated in the language of $\mathsf{EFA}$, and usually proved there too. Second, even when we look beyond specific functions of interest, we're still able to make things surprisingly concrete: via $\mathsf{CODE}$ing a continuous function on $\mathbb{R}$ is "morally equivalent" to a set of natural numbers, and so we can faithfully ask what general calculus theorems are provable in (say) $\mathsf{RCA_0}$. But we can't get all the way down to pure arithmetic, and this isn't a matter of strength: $\mathsf{PA}$ isn't an appropriate vehicle for this sort of thing either, despite being stronger than $\mathsf{RCA_0}$ in many senses.
