Distance In Nth Second

The distance covered in the $n$th second of uniformly accelerated motion is given by the standard relation $s_n=u+\frac{a}{2}(2n-1)$, which is not a full second's average distance but the gap between total distances at the end of consecutive seconds. It is derived by subtracting the displacement up to time $(n-1)$ from the displacement up to time $n$. Here the particle starts from rest, so $u=0$, with $a=6{\text{ m/s}}^2$ and $n=4$. Substituting gives $s_4=0+\frac{6}{2}(2\times 4-1)=3\times 7=21\text{ m}$. The option 48 m is wrong because it computes the total displacement in 4 s using $\frac{1}{2}a{t}^2$, not the distance in the 4th second alone. The option 24 m wrongly uses $a\times n$. The option 18 m mistakenly takes $(2n-1)$ as 6 instead of 7. This is the standard NCERT result for displacement in a specified second. As a check, the distances in successive seconds form an arithmetic progression with common difference $a$, and 21 m fits the pattern 3, 9, 15, 21.