A body falls freely form rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of

Options

A. 3 s
B. 5 s
C. 7 s
D. 9s

Explanation

To solve the problem step by step, we will use the equations of motion for a freely falling body.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- A body is falling freely from rest.
- The distance covered in the last second of its motion is equal to the distance covered in the first three seconds.

2. **Identify the Variables:**
- Let \( t \) be the total time of fall.
- The initial velocity \( u = 0 \) (since the body starts from rest).
- The acceleration \( a = g \) (acceleration due to gravity).

3. **Distance Covered in the First 3 Seconds:**
- The distance covered in the first \( n \) seconds is given by:
\[ S_n = ut + \frac{1}{2} a t^2 \]
- For the first 3 seconds (\( t = 3 \)):
\[ S_3 = 0 \cdot 3 + \frac{1}{2} g (3^2) = \frac{1}{2} g \cdot 9 = \frac{9g}{2} \]

4. **Distance Covered in the Last Second:**
- The distance covered in the \( n \)-th second is given by:
\[ S_n = u + \frac{1}{2} a (2n - 1) \]
- For the last second (\( n = t \)):
\[ S_t = 0 + \frac{1}{2} g (2t - 1) = \frac{1}{2} g (2t - 1) \]

5. **Setting Up the Equation:**
- According to the problem, the distance covered in the last second is equal to the distance covered in the first three seconds:
\[ S_t = S_3 \]
- Thus, we have:
\[ \frac{1}{2} g (2t - 1) = \frac{9g}{2} \]

6. **Canceling \( g \) and Solving for \( t \):**
- Cancel \( \frac{1}{2} g \) from both sides:
\[ 2t - 1 = 9 \]
- Rearranging gives:
\[ 2t = 10 \implies t = 5 \]

7. **Final Answer:**
- The body has fallen for a total time of \( t = 5 \) seconds.