The ration of the distance traversed, in successive intervals of time by a body, falling from rest, are

Options

A. 1 : 3 : 5 : 7 : 9 : …
B. 2 : 4 : 6 : 8 : 10 : …
C. 1 : 4 : 7 : 10 : 13 : …
D. None of the above

Explanation

To solve the problem of finding the ratio of the distances traversed by a body falling from rest in successive intervals of time, we can follow these steps:

### Step 1: Understand the motion of the body
When a body falls from rest under the influence of gravity, it accelerates uniformly. The acceleration due to gravity (g) is constant.

### Step 2: Use the formula for distance traveled
The distance (D) traveled by an object under uniform acceleration can be calculated using the formula:
\[ D = ut + \frac{1}{2} a t^2 \]
where:
- \( u \) = initial velocity (which is 0 since the body is falling from rest)
- \( a \) = acceleration (which is \( g \) for free fall)
- \( t \) = time

Since the body starts from rest, the formula simplifies to:
\[ D = \frac{1}{2} g t^2 \]

### Step 3: Calculate distances for successive time intervals
Let’s denote the distances traveled in the first, second, and third intervals of time as \( D_1, D_2, D_3 \), etc.

1. **For the first interval (t = t)**:
\[ D_1 = \frac{1}{2} g (t)^2 \]

2. **For the second interval (t = 2t)**:
\[ D_2 = \frac{1}{2} g (2t)^2 = \frac{1}{2} g (4t^2) = 2g t^2 \]
The distance traveled in the second interval is the total distance at \( 2t \) minus the distance at \( t \):
\[ D_2 = D(2t) - D(t) = \left(\frac{1}{2} g (2t)^2\right) - \left(\frac{1}{2} g (t)^2\right) = 2gt^2 - \frac{1}{2} gt^2 = \frac{3}{2} gt^2 \]

3. **For the third interval (t = 3t)**:
\[ D_3 = \frac{1}{2} g (3t)^2 = \frac{1}{2} g (9t^2) = \frac{9}{2} gt^2 \]
The distance traveled in the third interval is the total distance at \( 3t \) minus the distance at \( 2t \):
\[ D_3 = D(3t) - D(2t) = \left(\frac{1}{2} g (3t)^2\right) - \left(\frac{1}{2} g (2t)^2\right) = \frac{9}{2} gt^2 - 2gt^2 = \frac{5}{2} gt^2 \]

### Step 4: Find the ratio of distances
Now we can express the distances in terms of a common variable:
- \( D_1 = \frac{1}{2} gt^2 \)
- \( D_2 = \frac{3}{2} gt^2 \)
- \( D_3 = \frac{5}{2} gt^2 \)

Now, the ratios of the distances can be calculated:
\[ \text{Ratio} = D_1 : D_2 : D_3 = \frac{1}{2} gt^2 : \frac{3}{2} gt^2 : \frac{5}{2} gt^2 \]
This simplifies to:
\[ 1 : 3 : 5 \]

### Step 5: Generalize the pattern
Continuing this pattern, we can see that the distances traveled in successive intervals will follow the sequence of odd numbers:
- \( D_n \) for the nth interval will be proportional to \( 2n - 1 \).

### Conclusion
Thus, the ratio of the distances traversed in successive intervals of time by a body falling from rest is:
\[ 1 : 3 : 5 : 7 : \ldots \]