The speed v of a particle moving along a straight line is given by `a+bv^(2)=x^(2)` where x is its distance from the origin. The acceleration of the particle is

Options

A. `(x)/(b)`
B. `(x)/(ab)`
C. abx
D. ax

Explanation

To find the acceleration of the particle moving along a straight line given the equation \( a + bv^2 = x^2 \), we will follow these steps:

### Step 1: Write down the given equation
The speed \( v \) of the particle is given by:
\[ a + bv^2 = x^2 \]

### Step 2: Differentiate both sides with respect to time \( t \)
To find the acceleration, we need to differentiate the entire equation with respect to time \( t \):
\[ \frac{d}{dt}(a + bv^2) = \frac{d}{dt}(x^2) \]

### Step 3: Differentiate each term
- The derivative of \( a \) (a constant) is \( 0 \).
- For \( bv^2 \), we apply the chain rule:
\[ \frac{d}{dt}(bv^2) = b \cdot 2v \cdot \frac{dv}{dt} = 2bv \frac{dv}{dt} \]
- For \( x^2 \), we again apply the chain rule:
\[ \frac{d}{dt}(x^2) = 2x \frac{dx}{dt} = 2xv \]
(since \( \frac{dx}{dt} = v \), the velocity).

Putting this all together, we have:
\[ 0 + 2bv \frac{dv}{dt} = 2xv \]

### Step 4: Simplify the equation
We can simplify the equation:
\[ 2bv \frac{dv}{dt} = 2xv \]
Dividing both sides by \( 2v \) (assuming \( v \neq 0 \)):
\[ b \frac{dv}{dt} = x \]

### Step 5: Solve for acceleration
The acceleration \( a \) is defined as \( \frac{dv}{dt} \):
\[ \frac{dv}{dt} = \frac{x}{b} \]

Thus, the acceleration of the particle is:
\[ a = \frac{x}{b} \]

### Final Answer
The acceleration of the particle is:
\[ \frac{x}{b} \]

---