A body travels 102.5 m \; in \; n th second and 115.0 m in ( n +2) th second. The acceleration is :

Physics · Motion in one Dimension

JEE Main 2024 — 4 April, Shift 1 — Question 43

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

Option A:
$9 \mathrm{~m} / \mathrm{s}^{2}$
Option B:
$6.25 \mathrm{~m} / \mathrm{s}^{2}$
Correct
Option C:
$12.5 \mathrm{~m} / \mathrm{s}^{2}$
Option D:
$5 \mathrm{~m} / \mathrm{s}^{2}$

Answer: B

Step-by-step solution

Sol. Given, $102.5=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)$ & $115=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}+3)$

$\Rightarrow 102.5=\mathrm{u}+\mathrm{an}-\frac{\mathrm{a}}{2} \&$ $115=u+a n+\frac{3 a}{2}$ $12.5=2 \mathrm{a} \Rightarrow \mathrm{a}=6.25 \mathrm{~m} / \mathrm{s}^{2}$

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion

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