Cop CAN catch the thief ... but only if he is MUCH faster - 20.5 times faster.
Lets invert numbers and say thief speed and set to 1, and
v is cop speed. Easier for me that way.
Vertices are labeled A1-D4, roads are labeled BC3 meaning road from B3 to C3 and B12 meaning road from B1 to B2. There are 16 vertices and 24 roads in total.
Let's start with vertex D4 where the cop starts. If the cop has speed 4, it is trivial to show that by going from C4 through D4 to D3 and back, thief can never reach D4 again. So, this patrol route at the cop speed 4 or greater is sufficient to make D4 safe forever. Now, we want to add D3 to the mix. Securing vertices is explained later, the first part is for patrol only - we want to know only how long it takes to ensure thief cannot ever reach D3 again. Securing means that the thief starts at D2, C3 or C4 (or further out) at any time of cop patrol (he might get closer during the patrol though). Required speed for cop's patrol is simple to calculate. Cop going the route 2xCD4, D34, 2xD23, 2xCD3, D34 leaves the cop back at D4 with full knowledge that if he manages to keep repeating that route with speed of at least 8 (the number of road travels), thief won't ever get to D3 or D4 again. Suppose thief started from D2 just behind cop's back - by the time thief would manage to reach D3, cop is already walking on D23 road (in case of speed 8 he just reached D3 to start on that route at the same time as thief).
OK, where now?
Is it better to add D2 or C4 to secured pieces? I don't know. We can try pushing thief up first, then left, or both at the same pace. But initial greedy algorithm might not be good. Let's assume we have secured the whole C and D, and are trying to prevent thief from ever going back there. It is fairly simple to extend the above movement to show speed of 14 is required to go 2xBC1, C12, 2xBC2 ... then return C34, C23, C12. There is no need to ever walk on roads CD or D, thief cannot be down there. Then we obviously add B4 to the stable of safe parts. Adding B4 among safe spots allows for patrol in same time, just going from C3 to B3 then B4 then up to A4 then back again. Same for adding other parts.
All in all,
This strategy of slowly expanding safe spaces guarantees full coverage of all the secured part with the speed of 14.
This is only for patrol. Cop needs some time allocated to secure some vertices. It is trivial to secure D1 and A4 - mere patrol covers that already. It is annoying to secure B2, B3, C2, C3. Those on the edge are in between these two, so the worst case is securing the ones in the center. Suppose we want to secure B3 now. We know that C3 and B4 are safe, B2 and A3 aren't.
How much more time we need for that?
While we know 2 of these vertices are safe, thief might still duck on that road briefly, only to return to the center. Say we just reached B3 from C3 during regular patrol. We first travel to B4. Then return to B3. Thief might have managed at most few steps towards C3 (at most 2/
v), so we go first towards C3 by the sufficient amount to catch the thief, then back to B3, then B4 etc, until we know thief cannot be found on the roads BC3 or B34.
We secure roads B23 and AB3. We first start with those tiny little steps in ALL 4 directions - go up towards A3 by infinitely small amount, then back to B3, then repeat for B2, B4, C3, increase the infinitely small amount a little bit etc. So, we know thief is not directly next to C3 now - our infinitely small steps would have caught him. We also know he isn't on the roads BC3 or B34, we checked those two roads before. So, we go on the road AB3 - just far enough to get back to B3 before thief if thief is hiding on the road B23. Then we go on the road B23, this time we can go further, etc, until we manage to reach B2 and A3 spots. When we reached those two, B3 is secured - thief cannot get to the B3 again.
How much time did that take?
Well ... the initial road securing took 2 extra travels on the road B34 (there and back; remember we reached B3 from C3 so that road is safe), then
4/v (2x 2/v) travels towards C3, then 2x 4/v^2 towards B4 etc, continue with the infinite sum ... As v is large (we know it has to be at least 16 - 14 + those 2 extra travels), this drops towards 0 rapidly - the whole infinite sum is less than 1/2.
Securing the other two roads is essentially 4 extra travels (2x B23 + 2x AB3) plus the same infinite sum (except we go from the center out, but the sum is the same), plus infinitely many steps with infinitely small time securing the immediate vicinity of C3. Seeing need for 20 speed without the infinite sum parts, term 4/v is mere 1/5 and the next term is 8/400 = 0.02, so we can safely claim cop speed of 20.5 is sufficient to catch the thief. 14 for patrol, 6.5 for securing.
Adding all together,
Strategy lets the cop catch the thief if his speed is approximately 20.5 times higher than the thief's. Note that this might not be the most optimal strategy, but I doubt any strategy where speed is below 20 is feasible.
That speed of 20.5 lets you ever increase amount of space covered. During any patrol cop's coverage increases by at least 1 square, therefore cop will take at most 20.5 * 16 roads to catch the thief. Most likely way fewer because it is easier to secure edges and corners; but I cannot be bothered to search for the minimum time needed.