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JEE PYQ: Motion in a Straight Line - Question ID 225ca3a05756 (JEE Main 2024)

ID: 225ca3a05756JEE Main 2024Single Correct MCQ

A body travels $102.5 \mathrm{~m}$ in $\mathrm{n}^{\text {th }}$ second and $115.0 \mathrm{~m}$ in $(\mathrm{n}+2)^{\text {th }}$ second. The acceleration is :

Select Option

Step-by-step Explanation

Core Formula & Concept:

When a body moves with constant acceleration $a$ , its displacement in the $k^{\text{th}}$ second is given by the formula $s_k = u + \frac{a}{2}\bigl(2k - 1\bigr),$ where

$u$ is the initial velocity at the start of the motion,
$a$ is the constant acceleration,
$k$ is the second number (1st, 2nd, …).

This formula comes from subtracting the displacement up to

(k-1)

seconds from the displacement up to

k

seconds:

s_k = \Bigl[u\,k + \tfrac12\,a\,k^2\Bigr] - \Bigl[u\,(k-1) + \tfrac12\,a\,(k-1)^2\Bigr].

Step-by-Step Derivation:

Let the body have initial velocity $u$ and constant acceleration $a$ .

1. Write the given displacements:

s_n = 102.5 = u + \frac{a}{2}\bigl(2n - 1\bigr),

s_{n+2} = 115.0 = u + \frac{a}{2}\bigl(2(n+2) - 1\bigr) = u + \frac{a}{2}\bigl(2n + 3\bigr).

2. Subtract the two equations to eliminate $u$ :

s_{n+2} - s_n = \Bigl[u + \tfrac{a}{2}(2n+3)\Bigr] - \Bigl[u + \tfrac{a}{2}(2n-1)\Bigr] = \tfrac{a}{2}\bigl[(2n+3)-(2n-1)\bigr] = \tfrac{a}{2}\cdot4 = 2a.

Numerically,

115.0 - 102.5 = 12.5 = 2a \quad\Longrightarrow\quad a = \frac{12.5}{2} = 6.25\ \mathrm{m/s^2}.

3. Conclusion: The acceleration is

6.25\ \mathrm{m/s^2}

, which corresponds to option A. Common Traps & Exam Tip:

Misapplying the displacement formula: Students sometimes use the total displacement formula $s = ut + \tfrac12 a t^2$ for the $n^{\text{th}}$ second instead of the correct $s_n = u + \tfrac{a}{2}(2n-1)$ . This leads to an incorrect equation.
Sign errors in subtraction: When subtracting $s_{n+2}-s_n$ , one must carefully expand $(2n+3)-(2n-1)=4$ . A common mistake is to get $2$ instead of $4$ , yielding $a=6.25$ incorrectly halved.
Forgetting to eliminate $u$ : Trying to solve for $u$ first is unnecessary and complicates the algebra. Subtracting the two equations directly gives $a$ without needing $u$ .

Exam Tip: Always write down the two displacement‐in‐the‐

n^{\text{th}}

‐second equations and subtract them immediately. This shortcut avoids dealing with the initial velocity and simplifies the problem to a single step.

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