The displacement and the increase in the velocity of a moving particle from time $$t$$ to $$(t + 1) \mathrm{s}$$ are $$125 \mathrm{~m}$$ and $$50 \mathrm{~m} / \mathrm{s}$$, respectively. The distance traveled by the particle in $$(\mathrm{t} + 2)^{\mathrm{th}} \mathrm{s}$$ is calculated as follows:

Given that the acceleration is constant, we start with:

$$ v = u + at $$

When the velocity has increased by $$50 \mathrm{~m}/\mathrm{s}$$, the equation becomes:

$$ u + 50 = u + a \quad \Rightarrow \quad a = 50 \mathrm{~m}/\mathrm{s}^2 $$

Next, we consider the displacement:

$$ 125 = u t + \frac{1}{2} a t^2 $$

Since this is given over a unit time interval (from $$t$$ to $$(t + 1)$$), we use:

$$ 125 = u + \frac{a}{2} $$

Substituting $$a = 50 \mathrm{~m}/\mathrm{s}^2$$:

$$ 125 = u + \frac{50}{2} \quad \Rightarrow \quad 125 = u + 25 \quad \Rightarrow \quad u = 100 \mathrm{~m}/\mathrm{s} $$

To find the distance traveled by the particle in $$(t + 2)^\text{th}$$ second, we use:

$$ S_n = u + \frac{a}{2} [2n - 1] $$

For $$n = t + 2$$ (i.e., the (t+2)th second):

$$ S_{(t+2)} = u + \frac{a}{2} [2(t+2) - 1] $$

With $$u = 100$$ and $$a = 50$$:

$$ S_{(t+2)} = 100 + \frac{50}{2} [2(t+2) - 1] = 100 + 25 \times [2(t+2) - 1] = 100 + 25 \times (2t + 4 - 1) = 100 + 25 \times (2t + 3) = 100 + 25 \times 5 = 100 + 125 = 225 \mathrm{~m} $$