Halting problem within a finite time interval? Let the finite time halting problem to be one where the program is counted as "not halting" if the Turing machine takes more than a given time interval to run the program. 
Is this finite time halting problem undecidable?

This subject has annoying terminological redundancy - e.g. "decidable," "computable," and "recursive" all mean the same thing in this context. Below I'm using "computable" exclusively, since "recursive" is slightly old-fashioned and "decidable" does significant double-duty elsewhere in logic.
It is computable.
Precisely: the "general finite halting problem" $X$ of triples $(e,m,s)$ such that the $e$th Turing machine on input $m$ halts in at most $s$ steps is computable. Indeed, $X$ (or more precisely, the characteristic function of $X$) is even primitive recursive. (This is all related to Kleene's $T$ predicate.)
This is essentially a quick application of the universal Turing machine: to test whether $(e,m,s)\in X$ we want to run the $e$th Turing machine on input $m$ for $s$ steps and see what happens. And this is basically exactly what a universal Turing machine is for. It's a good exercise to write out the details, but there aren't any surprises.

Note that the halting problem $H=\{e:\Phi_e(e)\downarrow\}$ is a "projection" of $X$ in a sense: $$H=\{e:\exists s((e,e,s)\in X)\}.$$ In general, the image of a computable set under a computable function need not be computable but will always be computably enumerable.
