The distance travelled by a body falling from reset in the frist, second and third seconds are in the ratio

Options

A. `1 : 2 : 3`
B. `1 : 3: 5`
C. `1 : 4: 9 `
D. none of the above

Explanation

To find the distances traveled by a body falling from rest in the first, second, and third seconds, we can use the equations of motion under uniform acceleration due to gravity.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A body falls from rest, meaning its initial velocity \( u = 0 \).
- We need to calculate the distances covered in the first second (\( s_1 \)), the second second (\( s_2 \)), and the third second (\( s_3 \)).

2. **Distance in the First Second (\( s_1 \))**:
- The formula for distance traveled under constant acceleration is given by:
\[ s = ut + \frac{1}{2} a t^2 \]
- For the first second (\( t = 1 \) s):
\[ s_1 = 0 \cdot 1 + \frac{1}{2} g (1^2) = \frac{1}{2} g \]

3. **Distance in the First Two Seconds (\( s_2 \))**:
- For the total distance traveled in the first two seconds (\( t = 2 \)):
\[ s_{total} = 0 \cdot 2 + \frac{1}{2} g (2^2) = \frac{1}{2} g \cdot 4 = 2g \]
- To find \( s_2 \) (distance traveled during the second second), we subtract the distance traveled in the first second from the total distance in two seconds:
\[ s_2 = s_{total} - s_1 = 2g - \frac{1}{2} g = \frac{4g}{2} - \frac{1}{2} g = \frac{3g}{2} \]

4. **Distance in the First Three Seconds (\( s_3 \))**:
- For the total distance traveled in the first three seconds (\( t = 3 \)):
\[ s_{total} = 0 \cdot 3 + \frac{1}{2} g (3^2) = \frac{1}{2} g \cdot 9 = \frac{9g}{2} \]
- To find \( s_3 \) (distance traveled during the third second), we subtract the distance traveled in the first two seconds from the total distance in three seconds:
\[ s_3 = s_{total} - s_{total \, (2s)} = \frac{9g}{2} - 2g = \frac{9g}{2} - \frac{4g}{2} = \frac{5g}{2} \]

5. **Finding the Ratios**:
- Now we have:
\[ s_1 = \frac{1}{2} g, \quad s_2 = \frac{3}{2} g, \quad s_3 = \frac{5}{2} g \]
- The ratio \( s_1 : s_2 : s_3 \) can be expressed as:
\[ \frac{1/2 g}{1/2 g} : \frac{3/2 g}{1/2 g} : \frac{5/2 g}{1/2 g} = 1 : 3 : 5 \]

### Final Answer:
The distances traveled by the body in the first, second, and third seconds are in the ratio \( 1 : 3 : 5 \).

---