JEE Physics 2024 Question with Solution

Given: The body starts from rest and moves with constant acceleration. It covers displacement $S_1$ in the first $(p-1)$ seconds and displacement $S_2$ in the first $p$ seconds.

Find: The time $t$ in which the displacement $S_1 + S_2$ is covered.

For motion from rest with constant acceleration,

$$S = \frac{1}{2}at^2$$

So,

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

and

$$S_2 = \frac{1}{2}ap^2$$

If the displacement $S_1 + S_2$ is covered in time $t$, then

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

Substituting the values of $S_1$ and $S_2$,

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

Cancelling $\frac{1}{2}a$ from both sides,

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

Now simplify,

$$\begin{aligned} t^2 &= p^2 - 2p + 1 + p^2 \\ &= 2p^2 - 2p + 1 \end{aligned}$$

Hence,

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

Therefore, the correct option is B.

The solution contains an inconsistency: one section states "The Correct Option is A", but the worked calculation concludes

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

which matches option B. Since the worked solution is the primary source, the answer is taken as B.