QJEEM24Ph9b42: JEE Main 2024 Physics - Class 11 JEEM24Ph9b42 Solution: The displacement and the increase in the velocity of a moving…

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 $a = 50$ :

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