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 find the displacement of the particle during the 5th second of motion given the varying accelerations. Let's break down the steps:

### Step 1: Understand the accelerations
The acceleration during the nth second is given by:
\[ a_n = \frac{a_0}{2^{n-1}} \]
For the 5th second, the acceleration is:
\[ a_5 = \frac{a_0}{2^{5-1}} = \frac{a_0}{16} \]

### Step 2: Calculate the velocities at the end of each second
1. **At the end of the 1st second (t = 1s)**:
\[ v_1 = u + a_1 \cdot t = 0 + a_0 \cdot 1 = a_0 \]

2. **At the end of the 2nd second (t = 2s)**:
\[ v_2 = v_1 + a_2 \cdot t = a_0 + \frac{a_0}{2} \cdot 1 = a_0 + \frac{a_0}{2} = \frac{3a_0}{2} \]

3. **At the end of the 3rd second (t = 3s)**:
\[ v_3 = v_2 + a_3 \cdot t = \frac{3a_0}{2} + \frac{a_0}{4} \cdot 1 = \frac{3a_0}{2} + \frac{a_0}{4} = \frac{6a_0}{4} + \frac{a_0}{4} = \frac{7a_0}{4} \]

4. **At the end of the 4th second (t = 4s)**:
\[ v_4 = v_3 + a_4 \cdot t = \frac{7a_0}{4} + \frac{a_0}{8} \cdot 1 = \frac{7a_0}{4} + \frac{a_0}{8} = \frac{14a_0}{8} + \frac{a_0}{8} = \frac{15a_0}{8} \]

### Step 3: Calculate the displacement during the 5th second
The displacement during the 5th second can be calculated using the formula:
\[ D_5 = v_4 + \frac{1}{2} a_5 \cdot t^2 \]
Where \( t = 1 \) second for the 5th second.

Substituting the values:
\[ D_5 = v_4 + \frac{1}{2} \cdot \frac{a_0}{16} \cdot 1^2 \]
\[ D_5 = \frac{15a_0}{8} + \frac{1}{2} \cdot \frac{a_0}{16} = \frac{15a_0}{8} + \frac{a_0}{32} \]

### Step 4: Find a common denominator and simplify
The common denominator for \( \frac{15a_0}{8} \) and \( \frac{a_0}{32} \) is 32:
\[ \frac{15a_0}{8} = \frac{15a_0 \cdot 4}{32} = \frac{60a_0}{32} \]
So,
\[ D_5 = \frac{60a_0}{32} + \frac{a_0}{32} = \frac{61a_0}{32} \]

### Step 5: Compare with the given expression
We are given that:
\[ D_5 = \frac{K a_0}{32} \]
Comparing both expressions:
\[ \frac{61 a_0}{32} = \frac{K a_0}{32} \]
This implies:
\[ K = 61 \]

### Final Answer
The value of \( K \) is \( \boxed{61} \).