A particle starts sliding down a frictionless inclined plane. If \[{S_n}\] is the distance travelled by it from time \[t = \left( {n - 1} \right)\sec \], to\[t = n\sec \] , find the ratio of \[\dfrac{{{S_n}}}{{{S_{n + 1}}}}\].
A. \[\dfrac{{2n - 1}}{{2n + 1}}\]
B. \[\dfrac{{2n + 1}}{{2n}}\]
C. \[\dfrac{{2n}}{{2n + 1}}\]
D. \[\dfrac{{2n + 1}}{{2n - 1}}\]

Hint: Before we start addressing the problem, we need to know about the inclined plane. An inclined plane is a simple machine that consists of a sloping surface, usually used for raising heavy bodies.

Formula Used:
To find the distance covered in \[{n^{th}}\] sec, we have,
\[{S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)\]
Where, u is the initial velocity and a is acceleration.

Complete step by step solution:
Consider a particle that starts sliding down the frictionless inclined plane. If \[{S_n}\]is the distance travelled by it from time \[t = \left( {n - 1} \right)\sec \], to \[t = n\sec \] , then we need to find the ratio of \[\dfrac{{{S_n}}}{{{S_{n + 1}}}}\]. Here, let us say a ball is sliding down, so it will have some acceleration a.

Now to find the distance covered in \[{n^{th}}\]sec, we have,
\[{S_n} = u + \dfrac{a}{2}\left( {2n - 1} \right)\]
Similarly, the distance covered in \[\left( {n + 1} \right)\sec \] is,
\[{S_{n + 1}} = u + \dfrac{a}{2}\left( {2\left( {n + 1} \right) - 1} \right)\]
We know that before it starts sliding down its initial velocity is zero, since it was at rest. Therefore, u becomes zero.

Now, taking the ratio of \[\dfrac{{{S_n}}}{{{S_{n + 1}}}}\]we get,
\[\dfrac{{{S_n}}}{{{S_{n + 1}}}} = \dfrac{{\dfrac{a}{2}\left( {2n - 1} \right)}}{{\dfrac{a}{2}\left( {2\left( {n + 1} \right) - 1} \right)}}\]
\[\Rightarrow \dfrac{{{S_n}}}{{{S_{n + 1}}}} = \dfrac{{\left( {2n - 1} \right)}}{{\left( {2\left( {n + 1} \right) - 1} \right)}}\]
\[\Rightarrow \dfrac{{{S_n}}}{{{S_{n + 1}}}} = \dfrac{{\left( {2n - 1} \right)}}{{\left( {2n + 2 - 1} \right)}}\]
\[\therefore \dfrac{{{S_n}}}{{{S_{n + 1}}}} = \dfrac{{\left( {2n - 1} \right)}}{{\left( {2n + 1} \right)}}\]
Therefore, the ratio of \[\dfrac{{{S_n}}}{{{S_{n + 1}}}}\] is \[\dfrac{{\left( {2n - 1} \right)}}{{\left( {2n + 1} \right)}}\].

Hence, option A is the correct answer.

Additional information: In an inclined plane, the force that is required to move an object up the incline is less than the weight being raised without including friction. If the slope is steeper, or incline, more force is required to lift the object.

Note:In this problem it is important to remember the formula for the distance covered by a particle which is sliding down from an inclined plane. Then it will be easier to find the solution.