A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $\mathrm{S}_2$ in first $p$ seconds. The displacement $\mathrm{S}_1+\mathrm{S}_2$ will be made in time :

<p>Let's denote the constant acceleration with which the body moves as $a$. We know that the displacement $S$ covered by a body starting from rest under a constant acceleration $a$ in time $t$ is given by the equation of motion: $S = \frac{1}{2} a t^2$.</p> <p>Considering the displacement $\mathrm{S}_1$ in first $(p-1)$ seconds, we apply the equation of motion:</p> $\mathrm{S}_1 = \frac{1}{2} a (p-1)^2$ <p>Similarly, for the displacement $\mathrm{S}_2$ in first $p$ seconds:</p> $\mathrm{S}_2 = \frac{1}{2} a p^2$ <p>To find out the total time it will take to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$, we first find the sum of these two displacements:</p> $\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a (p-1)^2 + \frac{1}{2} a p^2$ <p>Let's simplify this:</p> $\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left((p-1)^2 + p^2\right)$ $\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(p^2 - 2p + 1 + p^2\right)$ $\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a \left(2p^2 - 2p + 1\right)$ $\mathrm{S}_1+\mathrm{S}_2 = a \left(\frac{2p^2 - 2p + 1}{2}\right)$ <p>If we consider the total displacement $\mathrm{S}_1+\mathrm{S}_2$ is to be covered in a time $t$ seconds from rest, we should set this equal to the equation of motion:</p> $\mathrm{S}_1+\mathrm{S}_2 = \frac{1}{2} a t^2$ <p>Equating the two equations:</p> $a \left(\frac{2p^2 - 2p + 1}{2}\right) = \frac{1}{2} a t^2$ <p>Since $a \neq 0$, we can simplify by dividing both sides by $\frac{1}{2} a$:</p> $\frac{2p^2 - 2p + 1}{2} = \frac{t^2}{2}$ $2p^2 - 2p + 1 = t^2$ <p>To find $t$, we take the square root of both sides:</p> $t = \sqrt{2p^2 - 2p + 1}$ <p>Therefore, the time taken to cover the displacement $\mathrm{S}_1+\mathrm{S}_2$ will be:</p> $t = \sqrt{(2p^2 - 2p + 1)}\ s$ <p>Hence, the correct option would be:</p> Option D: $\sqrt{(2p^2 - 2p + 1)}\ s$