A particle is falling freely under gravity. In first t second it covers distance `x_1` and in the next t second, it covers distance `x_2,` then t is given by

Options

A. `sqrt((x_2-x_1)/g)`
B. `sqrt((x_2+x_1)/g)`
C. `sqrt((2(x_2-x_1))/g)`
D. `sqrt((2(x_2+x_1))/g)`

Explanation

To solve the problem of a particle falling freely under gravity, we will use the equations of motion. Let's break down the solution step by step.

### Step 1: Understand the Problem
We have a particle falling freely under gravity. In the first time interval of \( t \) seconds, it covers a distance \( x_1 \), and in the next \( t \) seconds, it covers a distance \( x_2 \).

### Step 2: Use the Equation of Motion
The equation of motion for distance \( s \) covered under uniform acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
For a freely falling object, the initial velocity \( u = 0 \) and the acceleration \( a = g \) (acceleration due to gravity). Thus, the equation simplifies to:
\[ s = \frac{1}{2} g t^2 \]

### Step 3: Calculate Distance \( x_1 \)
For the first \( t \) seconds, the distance \( x_1 \) covered is:
\[ x_1 = \frac{1}{2} g t^2 \]

### Step 4: Calculate Distance \( x_2 \)
In the next \( t \) seconds, the particle has already been falling for \( t \) seconds. The distance covered in the next \( t \) seconds can be calculated by considering the total distance fallen in \( 2t \) seconds:
\[ \text{Total distance in } 2t \text{ seconds} = \frac{1}{2} g (2t)^2 = \frac{1}{2} g \cdot 4t^2 = 2g t^2 \]
The distance \( x_2 \) covered in the second \( t \) seconds is then:
\[ x_2 = \text{Total distance in } 2t \text{ seconds} - \text{Distance in first } t \text{ seconds} \]
\[ x_2 = 2g t^2 - x_1 = 2g t^2 - \frac{1}{2} g t^2 = \frac{3}{2} g t^2 \]

### Step 5: Find the Relation Between \( x_1 \) and \( x_2 \)
Now we have:
\[ x_1 = \frac{1}{2} g t^2 \]
\[ x_2 = \frac{3}{2} g t^2 \]

### Step 6: Calculate \( x_2 - x_1 \)
Now, we can find the difference between \( x_2 \) and \( x_1 \):
\[ x_2 - x_1 = \frac{3}{2} g t^2 - \frac{1}{2} g t^2 = g t^2 \]

### Step 7: Solve for \( t \)
From the equation \( x_2 - x_1 = g t^2 \), we can express \( t \) as:
\[ t^2 = \frac{x_2 - x_1}{g} \]
Taking the square root gives:
\[ t = \sqrt{\frac{x_2 - x_1}{g}} \]

### Final Answer
Thus, the time \( t \) is given by:
\[ t = \sqrt{\frac{x_2 - x_1}{g}} \]