The displacement and the increase in the velocity of a moving particle in the time interval of $$t$$ to $$(t + 1)$$ s are $$125 \text{ m}$$ and $$50 \text{ m s}^{-1}$$, respectively. The distance travelled by the particle in $$(t + 2)^{th}$$ s is __________ m.

We are given that in the time interval from $$t$$ to $$(t+1)$$ s, the displacement is $$125$$ m and the increase in velocity is $$50$$ m/s.

Since the increase in velocity in 1 second equals the acceleration:

$$a = 50 \text{ m/s}^2$$

The displacement in the interval from $$t$$ to $$(t+1)$$ s can be written as:

$$s = v_t \cdot (1) + \frac{1}{2}a(1)^2 = v_t + \frac{a}{2}$$

where $$v_t$$ is the velocity at time $$t$$.

$$125 = v_t + \frac{50}{2} = v_t + 25$$

$$v_t = 100 \text{ m/s}$$

The velocity at time $$(t+1)$$:

$$v_{t+1} = v_t + a = 100 + 50 = 150 \text{ m/s}$$

The distance travelled in the $$(t+2)$$-th second (i.e., from $$(t+1)$$ to $$(t+2)$$ s) is:

$$s_{(t+2)\text{th}} = v_{t+1} + \frac{a}{2} = 150 + \frac{50}{2} = 150 + 25 = 175 \text{ m}$$

The correct answer is $$175$$ m.