If a particle starting from rest having constant acceleration covers distance S₁ in first (p – 1) seconds and S₂ in first p seconds, then determine time for which displacement is S₁ + S₂

The correct option is:

$\sqrt{2p^2+1-2p}$

Step-by-step Explanation:

Let the particle start from rest, which means its initial velocity $u=0$.
Let the constant acceleration of the particle be $a$.

According to the second equation of motion, the displacement $S$ covered by a particle in time $t$ is given by:

$S=ut+\frac{1}{2}a{t}^2$

Since the particle starts from rest ($u=0$), the displacement equation simplifies to:

$S=\frac{1}{2}a{t}^2$

Now, let's write the expressions for the distances covered in the given time intervals, as shown in the visual diagram:

1. The distance covered in the first $(p-1)$ seconds is $S_1$:

$S_1=\frac{1}{2}a{(p-1)}^2$

2. The distance covered in the first $p$ seconds is $S_2$:

$S_2=\frac{1}{2}a{p}^2$

Let $t$ be the total time required to cover a displacement of $S_1+S_2$:

$S_1+S_2=\frac{1}{2}a{t}^2$

Substituting the expressions for $S_1$ and $S_2$ into the equation above:

$\frac{1}{2}a{(p-1)}^2+\frac{1}{2}a{p}^2=\frac{1}{2}a{t}^2$

We can cancel out the common factor $\frac{1}{2}a$ from both sides of the equation:

${(p-1)}^2+p^2=t^2$

Expanding the algebraic term ${(p-1)}^2$ using the identity ${(x-y)}^2=x^2-2xy+y^2$:

$(p^2-2p+1)+p^2=t^2$

Combine the like terms ($p^2+p^2=2p^2$):

$2p^2+1-2p=t^2$

Taking the positive square root on both sides to find time $t$:

$t=\sqrt{2p^2+1-2p}$