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

Correct answer: A. 6.25 ~m / s^2 aligned & Given, 102.5=u+ a 2 (2 n-1) & & 115=u+ a 2 (2 n+3) & 102.5=u+an- a 2 & & 115=u+an+ 3 a 2 & 12.5=2 a a=6.25 m / s^2 aligned

JEE Main 20244 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 4 April
Shift: Morning Shift
Subject: Physics
Chapter: Motion in One Dimension

JEE Main 2024 · 4 April · Morning Shift · Physics · Motion in One Dimension

A body travels 102.5 ~m in n^ th second and 115.0 ~m in ( n+ 2)^ th...

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 :

Options

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

Correct answer

A. $6.25 \mathrm{~m} / \mathrm{s}^2$

Step-by-step solution

\begin{aligned} & \text{ 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=\mathrm{u}+\mathrm{an}+\frac{3 \mathrm{a}}{2} \\ & 12.5=2 \mathrm{a} \Rightarrow \mathrm{a}=6.25 \mathrm{ m} / \mathrm{s}^2 \end{aligned}

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