Distance In Nth Second

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A particle moving from rest with uniform acceleration 6 m/s^2 travels along a straight line. How much distance does the particle cover specifically during the 4th second of its motion?

RankGuru · Accepted Answer

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 $a = 6	ext{ m/s}^2$ and $n = 4$. Substituting gives $s_4 = 0 + 	frac{6}{2}(2 	imes 4 - 1) = 3 	imes 7 = 21	ext{ m}$. The option 48 m is wrong because it computes the total displacement in 4 s using $	frac{1}{2}at^2$, not the distance in the 4th second alone. The option 24 m wrongly uses $a 	imes n$. 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.

Distance In Nth Second

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