If a freely falling body travels in the last second, a distance equal to the distance travelled by it in the first three second, the time of the travel is

Options

A. 6 sec
B. 5 sec
C. 4 sec
D. 3 sec

Explanation

To solve the problem, we need to find the total time of travel \( T \) for a freely falling body, given that the distance traveled in the last second is equal to the distance traveled in the first three seconds.

### Step-by-step Solution:

1. **Understanding the Problem**:
We need to find the total time \( T \) such that the distance traveled in the last second is equal to the distance traveled in the first three seconds.

2. **Distance in the First 3 Seconds**:
The distance traveled by a freely falling body in \( t \) seconds is given by the formula:
\[ S = \frac{1}{2} g t^2 \]
For the first 3 seconds, the distance \( S_3 \) is:
\[ S_3 = \frac{1}{2} g (3)^2 = \frac{1}{2} g \cdot 9 = \frac{9g}{2} \]

3. **Distance in the Last Second**:
The distance traveled in the last second of motion can be calculated as the difference between the distance traveled in \( T \) seconds and the distance traveled in \( T-1 \) seconds:
\[ S_{\text{last}} = S_T - S_{T-1} = \frac{1}{2} g T^2 - \frac{1}{2} g (T-1)^2 \]
Expanding \( S_{T-1} \):
\[ S_{T-1} = \frac{1}{2} g (T^2 - 2T + 1) = \frac{1}{2} g T^2 - gT + \frac{1}{2} g \]
Thus, the distance in the last second becomes:
\[ S_{\text{last}} = \frac{1}{2} g T^2 - \left(\frac{1}{2} g T^2 - gT + \frac{1}{2} g\right) = gT - \frac{1}{2} g \]

4. **Setting Up the Equation**:
According to the problem, the distance traveled in the last second is equal to the distance traveled in the first three seconds:
\[ gT - \frac{1}{2} g = \frac{9g}{2} \]
Dividing through by \( g \) (assuming \( g \neq 0 \)):
\[ T - \frac{1}{2} = \frac{9}{2} \]

5. **Solving for \( T \)**:
Rearranging the equation gives:
\[ T = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \]
Therefore, the total time of travel \( T \) is:
\[ T = 5 \text{ seconds} \]

### Final Answer:
The total time of travel is \( 5 \) seconds.