A body moves on a frictionless plane starting from rest. If S n is distance moved between t = n -1 and t = n and S n -1 is distance moved between t = n -2 and t = n -1 , then the ratio frac S n -1 S n is (1-frac 2 x ) for n =10 . The value of x is

Physics · Motion in one Dimension

JEE Main 2024 — 5 April, Shift 1 — Question 54

A body moves on a frictionless plane starting from rest. If $\mathrm{S}_{\mathrm{n}}$ is distance moved between $\mathrm{t}=\mathrm{n}-1$ and t $=\mathrm{n}$ and $\mathrm{S}_{\mathrm{n}-1}$ is distance moved between $\mathrm{t}=\mathrm{n}-2$ and $\mathrm{t}=\mathrm{n}-1$ , then the ratio $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}$ is $\left(1-\frac{2}{\mathrm{x}}\right)$ for n $=10$ . The value of $x$ is $\qquad$

Answer: 19

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{S}_{\mathrm{n}}=\frac{1}{2} \mathrm{a}(2 \mathrm{n}-1)=\frac{19 \mathrm{a}}{2}$ $\mathrm{S}_{\mathrm{n}-1}=\frac{1}{2} \mathrm{a}(2 \mathrm{n}-3)=\frac{17 \mathrm{a}}{2}$ $\frac{\mathrm{S}_{\mathrm{n}-1}}{\mathrm{~S}_{\mathrm{n}}}=\frac{17}{19}=1-\frac{2}{\mathrm{x}} \Rightarrow \mathrm{x}=19$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 5 April, Shift 1
Subject: Physics
Chapter: Motion in one Dimension
Topic: Uniformly Accelerated Motion