A freely falling body traveled xm in `n^(th)` second distance travelled in `n-1^(th)` second is

Options

A. x
B. x+g
C. x-g
D. 2x+3g

Answer

To solve the problem of finding the distance traveled by a freely falling body in the \( (n-1)^{th} \) second, we will use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We need to find the distance traveled by a freely falling body in the \( (n-1)^{th} \) second, given that it travels \( x \) meters in the \( n^{th} \) second.

2. **Using the Equation of Motion**:
The distance traveled by a freely falling body in \( n \) seconds can be expressed using the equation:
\[ s_n = u \cdot n + \frac{1}{2} a n^2 \]
Here, \( u = 0 \) (initial velocity), and \( a = g \) (acceleration due to gravity). Thus, the equation simplifies to:
\[ s_n = \frac{1}{2} g n^2 \]

3. **Distance Traveled in \( n \) Seconds**:
We are given that the distance traveled in \( n \) seconds is \( x \):
\[ s_n = x = \frac{1}{2} g n^2 \]

4. **Distance Traveled in \( (n-1)^{th} \) Second**:
To find the distance traveled in the \( (n-1)^{th} \) second, we need to calculate the distance traveled in \( (n-1) \) seconds:
\[ s_{n-1} = \frac{1}{2} g (n-1)^2 \]

5. **Calculating the Distance in \( (n-1)^{th} \) Second**:
The distance traveled in the \( (n-1)^{th} \) second is given by:
\[ \text{Distance in } (n-1)^{th} \text{ second} = s_n - s_{n-1} \]
Substituting the values:
\[ s_n - s_{n-1} = \left(\frac{1}{2} g n^2\right) - \left(\frac{1}{2} g (n-1)^2\right) \]

6. **Expanding \( s_{n-1} \)**:
\[ s_{n-1} = \frac{1}{2} g (n^2 - 2n + 1) = \frac{1}{2} g n^2 - g n + \frac{1}{2} g \]

7. **Finding the Difference**:
\[ \text{Distance in } (n-1)^{th} \text{ second} = \left(\frac{1}{2} g n^2\right) - \left(\frac{1}{2} g n^2 - g n + \frac{1}{2} g\right) \]
Simplifying this gives:
\[ = g n - \frac{1}{2} g = g(n - \frac{1}{2}) \]

8. **Relating to \( x \)**:
Since we know \( x = \frac{1}{2} g n^2 \), we can express \( g \) in terms of \( x \):
\[ g = \frac{2x}{n^2} \]

9. **Final Calculation**:
Substituting \( g \) back into the distance formula:
\[ \text{Distance in } (n-1)^{th} \text{ second} = x - g = x - \frac{2x}{n^2} \]
Thus, the distance traveled in the \( (n-1)^{th} \) second is:
\[ = x - g = x - \frac{2x}{n^2} \]

10. **Conclusion**:
The distance traveled in the \( (n-1)^{th} \) second is:
\[ \text{Distance} = x - g \]

### Final Answer:
The distance traveled in the \( (n-1)^{th} \) second is \( x - g \).