A body falls freely from 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. 9 s

Explanation

To solve the problem, we need to find the total time a body falls freely from rest, given that the distance covered in the last second of its motion is equal to the distance covered in the first three seconds.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The body is falling freely from rest, which means its initial velocity \( u = 0 \).
- The distance covered in the last second of its motion is equal to the distance covered in the first three seconds.

2. **Distance Covered in the First Three Seconds**:
- The formula for the distance covered under uniform acceleration (due to gravity \( g \)) is:
\[ s = ut + \frac{1}{2}gt^2 \]
- For the first three seconds (\( t = 3 \) seconds):
\[ s_3 = 0 \cdot 3 + \frac{1}{2}g(3^2) = \frac{1}{2}g \cdot 9 = \frac{9g}{2} \]

3. **Distance Covered in the Last Second**:
- The distance covered in the \( n \)-th second is given by:
\[ s_n = u + \frac{1}{2}g(2n - 1) \]
- Since \( u = 0 \), this simplifies to:
\[ s_n = \frac{1}{2}g(2n - 1) \]

4. **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:
\[ s_n = s_3 \]
- Therefore:
\[ \frac{1}{2}g(2n - 1) = \frac{9g}{2} \]

5. **Canceling \( g \)**:
- Since \( g \) is common on both sides, we can cancel it out:
\[ 2n - 1 = 9 \]

6. **Solving for \( n \)**:
- Rearranging the equation gives:
\[ 2n = 10 \implies n = 5 \]

7. **Finding the Total Time**:
- Since \( n \) represents the last second of the motion, the total time of fall is:
\[ t = n = 5 \text{ seconds} \]

### Final Answer:
The body has fallen for a total time of **5 seconds**.

---