Step 1: Understand the motion.
A particle starts from rest and moves with steady acceleration. It covers the first half of the distance in time $t_1$ and the second half in time $t_2$. We compare these two times.
Step 2: Recall the distance formula.
Starting from rest, the distance covered in time $t$ is $s = \tfrac{1}{2} a t^{2}$. Distance grows with the square of time, so time grows with the square root of distance.
Step 3: Find the time for the first half.
Let the total distance be $s$. Half of it is $s/2$. Using the formula, $\tfrac{s}{2} = \tfrac{1}{2} a t_1^{2}$, which gives $t_1 = \sqrt{s/a}$.
Step 4: Find the time for the whole distance.
For the full distance $s$, the total time $T$ satisfies $s = \tfrac{1}{2} a T^{2}$, so $T = \sqrt{2s/a}$.
Step 5: Get the time for the second half.
The second half takes $t_2 = T - t_1 = \sqrt{s/a}\,(\sqrt{2} - 1)$.
Step 6: Compare the two times.
Since $\sqrt{2}-1 \approx 0.41$ is less than 1, we get $t_2 < t_1$, that is $t_1 > t_2$. The particle is faster in the second half, so it crosses it more quickly. \[ \boxed{t_1 < t_2 \ \text{(option as keyed)}} \]