The instantaneous velocity of a particle moving in a straight line is given as a function of time by: ` v=v_0 (1-((t)/(t_0))^(n)) ` where `t_0` is a constant and n is a positive integer . The average velocity of the particle between t=0 and ` t=t_0` is ` (3v_0)/(4 ) `.Then , the value of n is

Explanation

To solve the problem, we need to find the value of \( n \) given the instantaneous velocity function and the average velocity condition. Let's break down the solution step by step.

### Step 1: Write down the given instantaneous velocity function
The instantaneous velocity of the particle is given by:
\[ v = v_0 \left( 1 - \left( \frac{t}{t_0} \right)^n \right) \]

### Step 2: Express displacement in terms of velocity
We know that instantaneous velocity can be expressed as:
\[ v = \frac{dx}{dt} \]
Thus, we can write:
\[ dx = v_0 \left( 1 - \left( \frac{t}{t_0} \right)^n \right) dt \]

### Step 3: Integrate to find displacement
Integrate both sides from \( t = 0 \) to \( t = t_0 \) to find the displacement \( x \):
\[ x = \int_0^{t_0} v_0 \left( 1 - \left( \frac{t}{t_0} \right)^n \right) dt \]
This can be split into two integrals:
\[ x = v_0 \int_0^{t_0} dt - v_0 \int_0^{t_0} \left( \frac{t}{t_0} \right)^n dt \]

### Step 4: Evaluate the integrals
1. The first integral:
\[ \int_0^{t_0} dt = t_0 \]
So, the first part becomes:
\[ v_0 t_0 \]

2. The second integral:
\[ \int_0^{t_0} \left( \frac{t}{t_0} \right)^n dt = \frac{1}{t_0^n} \int_0^{t_0} t^n dt = \frac{1}{t_0^n} \cdot \frac{t_0^{n+1}}{n+1} = \frac{t_0}{n+1} \]
Thus, the second part becomes:
\[ v_0 \cdot \frac{t_0}{n+1} \]

### Step 5: Combine the results
Putting it all together, we have:
\[ x = v_0 t_0 - v_0 \cdot \frac{t_0}{n+1} = v_0 t_0 \left( 1 - \frac{1}{n+1} \right) = v_0 t_0 \cdot \frac{n}{n+1} \]

### Step 6: Calculate the average velocity
The average velocity \( V_{avg} \) over the interval \( t = 0 \) to \( t = t_0 \) is given by:
\[ V_{avg} = \frac{x}{t_0} = \frac{v_0 t_0 \cdot \frac{n}{n+1}}{t_0} = v_0 \cdot \frac{n}{n+1} \]

### Step 7: Set the average velocity equal to the given value
We know from the problem that:
\[ V_{avg} = \frac{3v_0}{4} \]
Setting these equal gives:
\[ v_0 \cdot \frac{n}{n+1} = \frac{3v_0}{4} \]

### Step 8: Simplify the equation
Dividing both sides by \( v_0 \) (assuming \( v_0 \neq 0 \)):
\[ \frac{n}{n+1} = \frac{3}{4} \]

### Step 9: Cross-multiply to solve for \( n \)
Cross-multiplying gives:
\[ 4n = 3(n + 1) \]
Expanding the right side:
\[ 4n = 3n + 3 \]
Subtracting \( 3n \) from both sides:
\[ n = 3 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{3} \]