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 problem.
The body starts from rest and moves with uniform acceleration. Let the acceleration be \( a \), and the total distance covered in \( n \) seconds is given by the formula:
\[ S = \frac{1}{2} a n^2. \]
The distance covered in the \( n^{th} \) second is given by the formula:
\[ S_n = u + \frac{a}{2} \left( 2n - 1 \right), \]
where \( u = 0 \) (since the body starts from rest).

Step 2: Substituting into the formula for distance in the \( n^{th} \) second.
Substituting \( u = 0 \) into the formula for \( S_n \):
\[ S_n = \frac{a}{2} \left( 2n - 1 \right). \]

Step 3: Finding the ratio.
Now, we need to find the ratio of the distance covered in the \( n^{th} \) second to the total distance covered in \( n \) seconds:
\[ \text{Ratio} = \frac{S_n}{S} = \frac{\frac{a}{2} \left( 2n - 1 \right)}{\frac{1}{2} a n^2} = \frac{2n - 1}{n^2}. \]

Step 4: Final answer.
Thus, the ratio is:
\[ \boxed{\frac{2n - 1}{n^2}}. \]