Question

A body starts from rest and moves with a uniform acceleration. The ratio of the distance covered by the body in the \( n^{th} \) second of its motion to the total distance travelled in \( n \) seconds is

• A. \( \frac{2n - 1}{n^2} \)
• B. \( \frac{n}{n^2 + 1} \)
• C. \( \frac{n}{n^2 + 1} \)
• D. \( \frac{n^2}{2n + 1} \)

Hint:

To find the distance covered in the \( n^{th} \) second, use the formula \( S_n = u + \frac{a}{2} \left( 2n - 1 \right) \), where \( u \) is the initial velocity.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need the ratio of displacement in a specific interval (\( n^{th} \) second) to total displacement from the start (\( n \) seconds).
Step 2: Key Formula or Approach:
1. Displacement in \( n^{th} \) second: \( S_n = u + \frac{a}{2}(2n - 1) \).
2. Total displacement in \( n \) seconds: \( S_{total} = un + \frac{1}{2}an^2 \).
Step 3: Detailed Explanation:
Since the body starts from rest, \( u = 0 \).
Distance in \( n^{th} \) second: \( S_n = \frac{a}{2}(2n - 1) \).
Total distance in \( n \) seconds: \( S_{total} = \frac{1}{2}an^2 \).
Ratio:
\[ \text{Ratio} = \frac{S_n}{S_{total}} = \frac{\frac{a}{2}(2n - 1)}{\frac{1}{2}an^2} \]
\[ \text{Ratio} = \frac{2n - 1}{n^2} = \frac{2n}{n^2} - \frac{1}{n^2} = \frac{2}{n} - \frac{1}{n^2} \]
Step 4: Final Answer:
The ratio is \( \frac{2}{n} - \frac{1}{n^2} \).