Equations of Motion: Practice Problem & Solution

Solution Explained:

To solve this problem, we apply the core principles of Equations of Motion. Understanding the underlying formula is key to arriving at the correct answer below:

The distance in the \(n^{\text{th}}\) second is \(S_n = u + \frac{a}{2}(2n-1)\). Since \(u=0\), \(S_n \propto 2n-1\). Thus, \(\frac{S_n}{S_{n+1}} = \frac{2n-1}{2(n+1)-1} = \frac{2n-1}{2n+1}\).