1. Assume uniform acceleration

Since the question gives displacement in one-second intervals and increase in velocity over one second, we take the motion to be with constant acceleration.

Let acceleration be $a$ and velocity at time $t$ be $v_t$.

Given that the increase in velocity from $t$ to $(t+1)$ s is $50\,\text{m/s}$.

Over $1$ second, $a = 50\,\text{m/s}^2$

2. Use displacement from $t$ to $(t+1)$ s

Displacement in this 1-second interval is: $s = v_t(1) + \frac{1}{2}a(1)^2$

Given $s = 125$ m and $a=50$: $125 = v_t + \frac{1}{2}(50)$ $125 = v_t + 25$ $v_t = 100\,\text{m/s}$

3. Find velocity at time $(t+2)$

Velocity increases by $50\,\text{m/s}$ every second.

So, $v_{t+1} = 100 + 50 = 150\,\text{m/s}$ $v_{t+2} = 150 + 50 = 200\,\text{m/s}$

4. Find distance travelled in the $(t+2)^{\text{th}}$ second

The $(t+2)^{\text{th}}$ second means the interval from $(t+1)$ to $(t+2)$.

Distance in this 1-second interval: $s_{(t+2)^{\text{th}}} = v_{t+1}(1) + \frac{1}{2}a(1)^2$ $= 150 + 25$ $= 175\,\text{m}$

5. Final answer

$\boxed{175}$