A particle is moving in a straight line and relationn between time and displacement is `t=ax^(2)+betax`. If retardation is found to be proportional to the `v^(n)`, where v is instantaneous velocity, find the value of n.

Explanation

To solve the problem, we need to analyze the given relationship between time (t) and displacement (x) and find the value of n, where retardation is proportional to \( v^n \).

### Step-by-Step Solution:

1. **Given Relation**:
The relation between time and displacement is given as:
\[ t = ax^2 + \beta x \]

2. **Differentiate to Find Velocity**:
We differentiate \( t \) with respect to \( x \) to find \( \frac{dt}{dx} \):
\[ \frac{dt}{dx} = \frac{d}{dx}(ax^2 + \beta x) = 2ax + \beta \]
Now, to find velocity \( v \), we take the reciprocal:
\[ v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2ax + \beta} \]

3. **Differentiate to Find Acceleration**:
Acceleration \( a \) is defined as \( \frac{dv}{dt} \). We need to differentiate \( v \) with respect to \( t \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{2ax + \beta}\right) \]
Using the chain rule:
\[ a = -\frac{1}{(2ax + \beta)^2} \cdot \frac{d(2ax + \beta)}{dt} \]
Now, we need to find \( \frac{d(2ax + \beta)}{dt} \):
\[ \frac{d(2ax + \beta)}{dt} = 2a \frac{dx}{dt} = 2a v \]
Therefore, substituting back:
\[ a = -\frac{2a v}{(2ax + \beta)^2} \]

4. **Express Acceleration in Terms of Velocity**:
We can express \( a \) in terms of \( v \):
\[ a = -\frac{2a v}{(2ax + \beta)^2} \]
Notice that \( (2ax + \beta) \) can be replaced with \( \frac{1}{v} \):
\[ a = -\frac{2a v}{\left(\frac{1}{v}\right)^2} = -2a v^3 \]

5. **Relate Acceleration to Retardation**:
Since retardation is negative acceleration, we can write:
\[ |a| = 2a v^3 \]
This implies that retardation is proportional to \( v^3 \):
\[ |a| \propto v^3 \]

6. **Conclusion**:
From the above derivation, we conclude that the value of \( n \) is:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).