A particle starts moving from rest such that it has uniform acceleration `a_0` during the 1st second of motion, uniform acceleration `(a_0)/(2) ` during the 2nd second of motion, uniform acceleration `(a_0)/(4)` during the 3rd second of motion, and so on, such that it has uniform acceleration `(a_0)/(2^(n-1)) ` during the nth second of motion If the displacement of the particle during the 5th second of motion is`D_5 =(Ka_0)/(32) ` , the value of K is

Explanation

To solve the problem, we need to calculate the displacement of the particle during the 5th second of motion, given that it starts from rest and has varying uniform acceleration during each second.

### Step-by-Step Solution:

1. **Understanding the Acceleration:**
The particle has an acceleration of:
- \( a_0 \) during the 1st second,
- \( \frac{a_0}{2} \) during the 2nd second,
- \( \frac{a_0}{4} \) during the 3rd second,
- \( \frac{a_0}{8} \) during the 4th second,
- \( \frac{a_0}{16} \) during the 5th second.

2. **Initial Conditions:**
The particle starts from rest, so the initial velocity \( u = 0 \).

3. **Calculating Final Velocities:**
- **For the 1st second (0 to 1s):**
\[ v_1 = u + a_0 \cdot 1 = 0 + a_0 = a_0 \]
- **For the 2nd second (1 to 2s):**
\[ v_2 = v_1 + \frac{a_0}{2} \cdot 1 = a_0 + \frac{a_0}{2} = \frac{3a_0}{2} \]
- **For the 3rd second (2 to 3s):**
\[ v_3 = v_2 + \frac{a_0}{4} \cdot 1 = \frac{3a_0}{2} + \frac{a_0}{4} = \frac{7a_0}{4} \]
- **For the 4th second (3 to 4s):**
\[ v_4 = v_3 + \frac{a_0}{8} \cdot 1 = \frac{7a_0}{4} + \frac{a_0}{8} = \frac{15a_0}{8} \]
- **For the 5th second (4 to 5s):**
\[ v_5 = v_4 + \frac{a_0}{16} \cdot 1 = \frac{15a_0}{8} + \frac{a_0}{16} = \frac{31a_0}{16} \]

4. **Calculating Displacement during the 5th Second:**
The displacement during the nth second can be calculated using:
\[ D_n = u_n + \frac{1}{2} a_n \]
where \( u_n \) is the initial velocity at the start of the nth second and \( a_n \) is the acceleration during the nth second.

For the 5th second:
- Initial velocity \( u_5 = v_4 = \frac{15a_0}{8} \)
- Acceleration \( a_5 = \frac{a_0}{16} \)

Therefore, the displacement during the 5th second is:
\[ D_5 = u_5 + \frac{1}{2} a_5 = \frac{15a_0}{8} + \frac{1}{2} \cdot \frac{a_0}{16} \]
\[ = \frac{15a_0}{8} + \frac{a_0}{32} \]
To add these fractions, convert \( \frac{15a_0}{8} \) to a common denominator of 32:
\[ = \frac{60a_0}{32} + \frac{a_0}{32} = \frac{61a_0}{32} \]

5. **Finding the Value of K:**
We are given that:
\[ D_5 = \frac{Ka_0}{32} \]
From our calculation:
\[ D_5 = \frac{61a_0}{32} \]
Therefore, comparing both expressions:
\[ \frac{Ka_0}{32} = \frac{61a_0}{32} \]
This implies:
\[ K = 61 \]

### Final Answer:
The value of \( K \) is \( 61 \).