An object is dropped from rest. If it falls distance x during first second and additional distance y in next two seconds. Then ratio of `x/y` is

Options

A. `1 : 4`
B. `1 : 1`
C. `8 : 1`
D. `1 : 8`

Answer

To solve the problem, we need to find the distances x and y as described in the question and then calculate the ratio \( \frac{x}{y} \).

### Step-by-step Solution:

1. **Understanding the Motion**:
The object is dropped from rest, so its initial velocity \( u = 0 \). The distance fallen under gravity in time \( t \) can be calculated using the formula:
\[ s = ut + \frac{1}{2}gt^2 \]
where \( g \) is the acceleration due to gravity.

2. **Calculating Distance x**:
For the first second (i.e., \( t = 1 \) second):
\[ x = u \cdot 1 + \frac{1}{2}g \cdot (1^2) = 0 + \frac{1}{2}g = \frac{g}{2} \]

3. **Calculating Total Distance at t = 3 seconds**:
Now, we need to find the total distance fallen from \( t = 0 \) to \( t = 3 \) seconds:
\[ \text{Total distance} = u \cdot 3 + \frac{1}{2}g \cdot (3^2) = 0 + \frac{1}{2}g \cdot 9 = \frac{9g}{2} \]

4. **Finding Distance y**:
The distance \( y \) is the additional distance fallen in the next two seconds (from \( t = 1 \) to \( t = 3 \)):
\[ y = \text{Total distance at } t = 3 - \text{Distance at } t = 1 \]
The distance at \( t = 1 \) is \( x = \frac{g}{2} \), so:
\[ y = \frac{9g}{2} - \frac{g}{2} = \frac{8g}{2} = 4g \]

5. **Calculating the Ratio \( \frac{x}{y} \)**:
Now we can find the ratio \( \frac{x}{y} \):
\[ \frac{x}{y} = \frac{\frac{g}{2}}{4g} = \frac{g}{2} \cdot \frac{1}{4g} = \frac{1}{8} \]

6. **Final Result**:
Therefore, the ratio \( \frac{x}{y} \) is:
\[ \frac{x}{y} = \frac{1}{8} \]