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

<p>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: <p>Given that the acceleration is constant, we start with: </p> <p>$v = u + at$</p> <p>When the velocity has increased by $50 \mathrm{~m}/\mathrm{s}$, the equation becomes:</p> <p>$u + 50 = u + a \quad \Rightarrow \quad a = 50 \mathrm{~m}/\mathrm{s}^2$</p> <p>Next, we consider the displacement: </p> <p>$125 = u t + \frac{1}{2} a t^2$</p> <p>Since this is given over a unit time interval (from $t$ to $(t + 1)$), we use: </p> <p>$125 = u + \frac{a}{2}$</p> <p>Substituting $a = 50 \mathrm{~m}/\mathrm{s}^2$: </p> <p>$$ 125 = u + \frac{50}{2} \quad \Rightarrow \quad 125 = u + 25 \quad \Rightarrow \quad u = 100 \mathrm{~m}/\mathrm{s} $$</p> <p>To find the distance traveled by the particle in $(t + 2)^\text{th}$ second, we use: </p> <p>$S_n = u + \frac{a}{2} [2n - 1]$</p> <p>For $n = t + 2$ (i.e., the (t+2)th second): </p> <p>$S_{(t+2)} = u + \frac{a}{2} [2(t+2) - 1]$</p> <p>With $u = 100$ and $a = 50$: <p> <p>$$ S_{(t+2)} = 100 + \frac{50}{2} [2(t+2) - 1] = 100 + 25 \times [2(t+2) - 1] = 100 + 25 \times (2t + 4 - 1) = 100 + 25 \times (2t + 3) = 100 + 25 \times 5 = 100 + 125 = 225 \mathrm{~m} $$</p>