## Distance travelled in n'th second

A body slides down a frictionless inclined plane starting from rest. If \(S_n\) and \(S_{n+1}\) be the distance travelled by the body during \(n^{th}\) and \((n + 1)^{th}\) seconds, then the ratio \(\dfrac{S_{n+1}}{S_n}\) is:

\(\dfrac{{2n-1}}{2n+1}\)

\(\dfrac{{2n}}{2n+1}\)

\(\dfrac{{2n+1}}{2n-1}\)

\(\dfrac{{2n}}{2n-1}\)

When a body is rolling down a frictionless plane, its acceleration is \(g\sin \theta\) and the distance travelled in (total) ’n’ seconds using Newton’s equation of motion will be: \[\begin{aligned} s &= ut + \dfrac{1}{2} at^2\\ &= 0 + \dfrac{1}{2} (g \sin \theta) n^2\\ &= \dfrac{1}{2} (g \sin \theta) n^2 \end{aligned}\]

Similarly, the distance travelled in (total) ’(n-1)’ seconds: \[s = \dfrac{1}{2} (g \sin \theta) (n-1)^2\] which means, the distance travelled DURING \(n^{th}\) second will be: \[\begin{aligned} S_n &= \dfrac{1}{2} (g \sin \theta) \left[n^2 - n^2 +2n - 1 \right]\\ &= \dfrac{1}{2} (g \sin \theta) (2n - 1) \end{aligned}\]

Replacing n by (n+1), \[\begin{aligned} S_{n+1} &= \dfrac{1}{2} (g \sin \theta) (2(n+1) - 1)\\ &= \dfrac{1}{2} (g \sin \theta) (2n+1) \end{aligned}\]

Taking ratio: \[\dfrac{S_{n+1}}{S_{n}} = \dfrac{2n+1}{2n - 1}\] Ans: **C**