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 step by step, we will use the equations of motion under uniform acceleration due to gravity.

### Step 1: Determine the distance x fallen in the first second.
The object is dropped from rest, so its initial velocity (u) is 0. The distance fallen in the first second (x) can be calculated using the second equation of motion:

\[ S = ut + \frac{1}{2} a t^2 \]

For the first second (t = 1 s), the acceleration (a) is equal to the acceleration due to gravity, which is approximately \(10 \, \text{m/s}^2\).

Substituting the values:

\[ x = 0 \cdot 1 + \frac{1}{2} \cdot 10 \cdot (1)^2 \]

\[ x = \frac{1}{2} \cdot 10 \cdot 1 = 5 \, \text{m} \]

### Step 2: Determine the total distance fallen in the first three seconds.
Now, we need to find the total distance fallen in the first three seconds (which includes both x and y). We again use the second equation of motion for t = 3 seconds:

\[ S = ut + \frac{1}{2} a t^2 \]

Substituting the values for t = 3 s:

\[ S = 0 \cdot 3 + \frac{1}{2} \cdot 10 \cdot (3)^2 \]

\[ S = \frac{1}{2} \cdot 10 \cdot 9 = 45 \, \text{m} \]

### Step 3: Calculate the distance y fallen in the next two seconds.
The distance y is the distance fallen from the end of the first second to the end of the third second. Therefore, we can express y as:

\[ y = \text{Total distance in 3 seconds} - \text{Distance in 1 second} \]

\[ y = 45 \, \text{m} - 5 \, \text{m} = 40 \, \text{m} \]

### Step 4: Find the ratio x/y.
Now that we have both x and y, we can find the ratio \( \frac{x}{y} \):

\[ \frac{x}{y} = \frac{5 \, \text{m}}{40 \, \text{m}} = \frac{1}{8} \]

### Final Answer:
The ratio \( \frac{x}{y} \) is \( \frac{1}{8} \).

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