A freely falling body travelled x m in `n^("th")` second. Then 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, given that it travels x meters in the nth second, we can follow these steps:

### Step 1: Understand the motion of a freely falling body
A freely falling body is under the influence of gravity, which accelerates it at a constant rate (g). The distance traveled by the body in a given second can be expressed using the equations of motion.

### Step 2: Use the formula for distance traveled in the nth second
The distance traveled by a freely falling body in the nth second is given by the formula:
\[ S_n = \frac{1}{2} g (2n - 1) \]
where \( g \) is the acceleration due to gravity, and \( n \) is the second in which the distance is measured.

### Step 3: Set the equation for the nth second
According to the problem, the distance traveled in the nth second is given as:
\[ S_n = x \]
Thus, we can write:
\[ x = \frac{1}{2} g (2n - 1) \]

### Step 4: Find the distance traveled in the (n-1)th second
The distance traveled in the (n-1)th second can be calculated using the formula:
\[ S_{n-1} = \frac{1}{2} g (2(n-1) - 1) \]
This simplifies to:
\[ S_{n-1} = \frac{1}{2} g (2n - 2 - 1) = \frac{1}{2} g (2n - 3) \]

### Step 5: Substitute for g using the expression from step 3
From the equation \( x = \frac{1}{2} g (2n - 1) \), we can express \( g \) in terms of \( x \):
\[ g = \frac{2x}{2n - 1} \]

### Step 6: Substitute g into the equation for S_{n-1}
Now, substituting \( g \) into the equation for \( S_{n-1} \):
\[ S_{n-1} = \frac{1}{2} \left(\frac{2x}{2n - 1}\right) (2n - 3) \]

### Step 7: Simplify the expression
\[ S_{n-1} = \frac{x(2n - 3)}{2(2n - 1)} \]
Now, we can simplify this further:
\[ S_{n-1} = x - g \]
This means:
\[ S_{n-1} = x - \frac{2x}{2n - 1} \]

### Final Answer
Thus, the distance traveled in the (n-1)th second is:
\[ S_{n-1} = x - g \]