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Distance In Nth Second

Hardphysics

A body starting from rest moves with uniform acceleration of 6 m/s² along a straight road, what distance does it travel during the third second of its motion?

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About This Question

Subject
physics
Chapter
kinematics
Topic
distance in nth second
Difficulty
Hard
Year
2025
Tags
distance in nth seconduniform accelerationinterval distancekinematic interval formularest start

Solution

Correct Answer:

15 m

As stated in NCERT Class 11, Chapter 3 (Motion in a Straight Line), the distance covered in the nth second of uniformly accelerated motion is given by s_n = u + (a/2)(2n − 1), which is not the same as total distance but the distance in that particular one-second interval. Starting from rest, u = 0, so s_n = (a/2)(2n − 1). For the third second, n = 3, giving s_3 = (6/2)(2×3 − 1) = 3 × 5 = 15 m. The option 18 m wrongly uses 2n instead of (2n − 1). The option 12 m uses (2n − 3) by mistake. The option 27 m computes total distance in three seconds (½×6×9), not the distance in the third second alone. Plausibility check: total distance in 3 s is 27 m and in 2 s is 12 m, and their difference 27 − 12 = 15 m exactly equals the distance in the third second, confirming the formula.

This hard difficulty physics question is from the chapter kinematics, covering the topic of distance in nth second. It appeared in the 2025 exam.

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