The velocity of a particle moving along a straight line is given by the relation `v^(2)=ax^(2) +2bx+c`. The acceleration varies as ...........

Answer

To find the relation of acceleration based on the given velocity equation \( v^2 = ax^2 + 2bx + c \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**:
We start with the equation for velocity:
\[ v^2 = ax^2 + 2bx + c \]

2. **Differentiate with respect to time**:
To find acceleration, we need to differentiate the velocity with respect to time. Since \( v \) is a function of \( x \), we apply the chain rule:
\[ \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
Here, \( \frac{dx}{dt} = v \) (the definition of velocity).

3. **Differentiate \( v^2 \)**:
Differentiate both sides of the equation \( v^2 = ax^2 + 2bx + c \) with respect to \( x \):
\[ \frac{d(v^2)}{dx} = \frac{d(ax^2 + 2bx + c)}{dx} \]
Using the chain rule on the left side:
\[ 2v \frac{dv}{dx} = 2ax + 2b \]

4. **Solve for \( \frac{dv}{dx} \)**:
Rearranging the equation gives:
\[ \frac{dv}{dx} = \frac{2ax + 2b}{2v} = \frac{ax + b}{v} \]

5. **Substitute \( \frac{dv}{dx} \) into the acceleration formula**:
Now, substituting back into the acceleration formula:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = \frac{dv}{dx} \cdot v \]
This gives:
\[ a = \left(\frac{ax + b}{v}\right) \cdot v = ax + b \]

6. **Final expression for acceleration**:
Therefore, the acceleration \( a \) varies as:
\[ a = ax + b \]

### Final Result:
The acceleration varies as \( ax + b \).