jmc

The abrupt jump in the middle of the ant's trip makes the problem difficult to analyze geometrically. (A solution using calculus is possible, but the algebra becomes a little intense.) The speeds of $\sqrt{2}$ and 2 units per second are also suggestive, as is the fact that the teleportation begins on the $x$-axis and ends on the line $y=x$, which makes an angle of $45^{\circ }$ with the $x$-axis. Therefore we transform the entire latter portion of the ant's trip by rotating it clockwise by $45^{\circ }$ and scaling it down by $\sqrt{2}$. This has the effect of removing the teleportation all together, reduces the ant's speed to $\sqrt{2}$ for the second part of the journey, and moves the destination point to $(37,37)$.

In other words, an equivalent problem is to ask where the ant should cross the $x$-axis if it wishes to crawl from $(0,-63)$ to $(37,37)$ at a uniform rate of $\sqrt{2}$ units per second in the least amount of time. Of course, it is now clear that the ant should crawl in a straight line. The equation of this line is $y=\frac{100}{37}x-63$, and it crosses the $x$-axis when $y=0$, so $$0=\frac{100}{37}x-63\to x=\frac{37\cdot 63}{100}=\boxed{23.31}.$$