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JEE Main 2024 (5 Apr, Shift 1) · Equations of Kinematics

Equations of Kinematics question from JEE Main 2024 (5 Apr, Shift 1)

Q14JEE Main 2024 · 5 APR · SHIFT 1Medium

Equations of Kinematics

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 $\left(1 - \frac{2}{x}\right)$ for $n = 10$ . The value of $x$ is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 19

From rest, $S_n = \frac{a}{2}(2n - 1)$ and $S_{n-1} = \frac{a}{2}(2n - 3)$ .
$\frac{S_{n-1}}{S_n} = \frac{2n - 3}{2n - 1} = 1 - \frac{2}{2n - 1}$
For $n = 10$ : $x = 2n - 1 = 19$

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