YES. Funnily enough, I invented the same problem not very long ago, in terms of a hound and a rabbit. Moreover, the only thing known is that the (maximal) speed of the rabbit is strictly less than the (maximal) speed of the hound (which makes the problem harder, because we don't know the exact ratio of these two speeds - in the case of the above $9/10$ - but only that it is subunitary). My proof follows. Let us first make some notations towards describing the steps $\Phi_0,\Phi_1,\ldots,\Phi_n,\ldots$ of a strategy $\Phi$, which within a finite amount of time will allow the hound to catch the rabbit. Take as origin of the straight line the point $o_0$ where the hound is at the moment in time $T_0 = 0$. Denote the speed of the hound by $v$ (so the maximal speed of the rabbit is bounded by a value strictly less than $v$, but unknown). We will now describe step $\Phi_n$. $\bullet$ Step $\Phi_n$ applies at the moment in time $T_n$. $\bullet$ Compute $p_n = (-1)^n \left (n + \left (v- \dfrac {v} {n+1} \right ) T_n \right )$, the farthest from the origin point that the rabbit can reach at the moment in time $T_n$, if its initial position belonged to the interval $[0, (-1)^nn]$, while its speed were at most $v- \dfrac {v} {n+1}$. $\bullet$ Compute $t_n = \dfrac {|p_n - o_n|} {v/(n+1)}$, the maximum time for the hound to run the above distance, hence catch the rabbit, under the above conditions. $\bullet$ Define now $T_{n+1} = T_n + t_n$, and $o_{n+1} = o_n + (-1)^nvt_n$ (the position of the hound at the end of step $\Phi_n$). All it is left to do is notice that the steps of the strategy are correctly defined, and that the number $n$ increases by $1$ with each step of the strategy, thus at some moment becoming large enough so that simultaneously to become larger than the absolute value of the initial position of the rabbit, and also larger than the ratio between the speed of the rabbit and the difference between the speed $v$ of the hound and that of the rabbit, hence the rabbit will be caught.

the following strategy is also available the cruiser go $1$ mile towards left,and then go $1+s$ miles towards right,then go $1+s+s^{2}$ miles towards left,and so on.here $s$ is extremely large

This strategy also works. Let the cruiser go $1!$ to left, $2!$ to right, $3!$ to left, $4!$ to right, $\dots$, $(2n+1)!$ to left, $(2n+2)!$ to right, $\dots$. If the thief went right, then we want that $$(2n)! - (2n-1)! + (2n-2)! - (2n-3)! + \dots + 2! - 1! \ge 0.9 ((2n)! + (2n-1)! + \dots + 2! + 1!) + N$$for some $N$. But this follows by taking $n$ to be arbitrarily large so that $(2n)! \ge 0.9 \cdot (2n)! + 3 \cdot (2n-1)! + N$; then the result follows from $(2n-k)! > (2n-k-1)!$ and $(2n-1)! = (2n-1) (2n-2)! > (2n-2)! + (2n-3)! + \dots + 1!.$ The cruiser also catches the thief if he goes left via a similar argument, so we are done. The police always prevail. This strategy also works if we replace $0.9$ with any number less than $1$.

I know this is a strange way to say, which can be definitely wrong: Let's say the police station, the road, and the police and the thief are all on a some planet, star, etc. They are all round. If the police when to the same way as the thief, the police will definitely catch the thief(obvious). If the police goes to the opposite way as the thief, the planet(let's say) is round, they will meet anyway. Sorry, It isn't mathematical.

See another related problem in the thread A car at the intersection of k+1 roads.

YS_PARKGONG wrote: I know this is a strange way to say, which can be definitely wrong: Let's say the police station, the road, and the police and the thief are all on a some planet, star, etc. They are all round. If the police when to the same way as the thief, the police will definitely catch the thief(obvious). If the police goes to the opposite way as the thief, the planet(let's say) is round, they will meet anyway. Sorry, It isn't mathematical. Same with my solution, praise for YS_PARKGONG