There are a number of different approaches that can be used to solve this problem.
Rotation
Consider rotating the square $60^\circ$ clockwise about $S$.
Since $PQRS$ is a square, and $PST$ and $RSU$ are equilateral, $SR = SU = SP = ST = 1$.
As $\angle PST = 60^\circ$ (as the interior angle of an equilateral triangle), $P$ rotates onto $T$.
As $\angle RSU = 60^\circ$ (as the interior angle of an equilateral triangle), $R$ rotates onto $U$.
This means that the distance $UT$ is the same as $PR$, since distances are not changed by rotation.
Then, as $\angle PSR = 90^\circ$, Pythagoras' theorem can be used on triangle $PSR$ to find that:
$PR = \sqrt{1^2+1^2} = \sqrt{2}$.
Hence, $UT= PR = \sqrt{2}$.
Right Angle
$\angle PSR = 90^\circ$, as it is in a square.
$\angle USR = \angle TSP = 60^\circ$, as they are in equilateral triangles.
An equilateral triangle of side length $1$ has height $\frac{\sqrt{3}}{2}$.
Let $R$ be the origin.
The coordinates of $U$ are $\left(\frac{1}{2},\frac{\sqrt{3}}{2} \right)$.
The coordinates of $T$ are $\left( 1 + \frac{\sqrt{3}}{2},\frac{1}{2} \right)$.