Q 11-02-059JEE MainJEE Main 2024 (5 Apr, Shift 1)Medium
A body moves on a frictionless plane starting from rest. If $S_n$ is the distance moved between $t = n - 1$ and $t = n$, and $S_{n-1}$ is the distance moved between $t = n - 2$ and $t = n - 1$, then the ratio $\dfrac{S_{n-1}}{S_n}$ is $\left(1 - \dfrac{2}{x}\right)$ for $n = 10$. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 19
Starting from rest with uniform acceleration, the distance in the $n^{\text{th}}$ second is $\dfrac{a}{2}(2n - 1)$:
$$\frac{S_{n-1}}{S_n} = \frac{2n - 3}{2n - 1} = \frac{17}{19} = 1 - \frac{2}{19} \Rightarrow x = 19$$
Solution by Sreeraj P, M.Sc Physics