[asy][asy] draw((-5,-1)--(-5,5)); draw((-5,5)--(5,5)); draw((5,-1)--(5,5)); draw((-5,-1)--(5,-1)); draw((-5,-1)--(5,5)); draw((5,-1)--(-5,5)); label("A",(-5,5),N); label("B",(5,5),N); label("C",(5,-1),S); label("D",(-5,-1),S); label("O",(0,2),S); label("$60^\circ$",(0,2),W); label("$60^\circ$",(0,2),E); [/asy][/asy] The diagonals of a rectangle bisect each other, and because the diagonals are equal in length using the Pythagorean theorem, we find the part of the diagonals that are bisected are equal in length. Therefore, $\bigtriangleup{AOD}$ and $\bigtriangleup{BOC}$ are equilateral triangles because of the $3$ $60^\circ$ angles, and $\bigtriangleup{ACD}$ and $\bigtriangleup{ACB}$ are $90^\circ$ $60^\circ$ $30^\circ$ triangles. Therefore, traveling the route $A-C-B-D-A$ $10$ times has a length of $10\cdot{(2s+s+2s+s)}=10\cdot{6s}-=60s$, where $s$ is the length of $BC$. Traveling the route $A-D-A$ $15$ time has a length of $15\cdot{(s+s)}=15\cdot{2s}=30s$. Therefore, we must find $s$ such that $60s+30s=4.5$. Solving, we find $s=0.05$, and so $AC=2\cdot{0.05}=\boxed{0.1 km}$

[asy][asy] draw((-5,-1)--(-5,5)); draw((-5,5)--(5,5)); draw((5,-1)--(5,5)); draw((-5,-1)--(5,-1)); label("A",(-5,5),N); label("B",(5,5),N); label("C",(5,-1),S); label("D",(-5,-1),S); [/asy][/asy]

The diagonals form two equilateral triangles: $AED$ and $BCE$. Therefore, letting $AD$ be $s$, we have the equation $60s+30s=4.5\implies AD = 0.05 \implies \boxed{AC = 0.1}$