Ratio of Distances in nth Second – Rankers Physics

Equations of Motion: Practice Problem & Solution

A body starts from rest, what is the ratio of the distance travelled by the body during the \(4^{\text{th}}\) and \(3^{\text{rd}}\) second? (1993)
7/5
5/7
7/3
3/7

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:

Concept: Distance covered in the \(n^{text{th}}\) second for uniformly accelerated motion.
Formula: \(S_n = u + \frac{a}{2}(2n - 1)\). Since it starts from rest, \(u=0\).
Solution: \(S_4 = \frac{a}{2}(2(4) - 1) = \frac{7a}{2}\), \(S_3 = \frac{a}{2}(2(3) - 1) = \frac{5a}{2}\). Ratio \(S_4:S_3 = 7a/2 : 5a/2 = 7:5\).

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