A particle moves according to the equation `t = ax^(2)+bx`, then the retardation of the particle when `x = (b)/(a)` is

Options

A. `(a)/(b^(3))`
B. `(2)/(9) (a)/(b^(3))`
C. `(2)/(27) (a)/(b^(3))`
D. none of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given equation
The motion of the particle is described by the equation:
\[ t = ax^2 + bx \]
where \( t \) is time, \( x \) is the position, and \( a \) and \( b \) are constants.

**Hint:** Identify the variables and constants in the equation.

### Step 2: Differentiate the equation with respect to time
To find the velocity \( v \), we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 2ax + b \]
Using the chain rule, we have:
\[ \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2ax + b} \]
Thus, the velocity \( v \) is given by:
\[ v = \frac{dx}{dt} = \frac{1}{2ax + b} \]

**Hint:** Remember that velocity is the reciprocal of the derivative of time with respect to position.

### Step 3: Find acceleration
Acceleration \( a \) can be expressed as:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
We already have \( \frac{dx}{dt} \) from the previous step. Now we need to find \( \frac{dv}{dx} \).

First, we differentiate \( v \):
\[ v = \frac{1}{2ax + b} \]
Using the quotient rule:
\[ \frac{dv}{dx} = -\frac{2a}{(2ax + b)^2} \]
Now substituting \( \frac{dx}{dt} \):
\[ a = v \cdot \frac{dv}{dx} = \frac{1}{2ax + b} \cdot \left(-\frac{2a}{(2ax + b)^2}\right) \]
This simplifies to:
\[ a = -\frac{2a}{(2ax + b)^3} \]

**Hint:** Use the chain rule to relate acceleration to velocity and position.

### Step 4: Substitute \( x = \frac{b}{a} \)
Now we need to find the retardation when \( x = \frac{b}{a} \):
Substituting \( x = \frac{b}{a} \) into the acceleration equation:
\[ a = -\frac{2a}{(2a(\frac{b}{a}) + b)^3} \]
This simplifies to:
\[ a = -\frac{2a}{(2b + b)^3} = -\frac{2a}{(3b)^3} = -\frac{2a}{27b^3} \]

**Hint:** Substitute the value of \( x \) carefully and simplify.

### Step 5: Identify the retardation
The negative sign indicates that this is retardation. Therefore, the magnitude of the retardation is:
\[ \text{Retardation} = \frac{2a}{27b^3} \]

**Hint:** Remember that retardation is the magnitude of acceleration with a negative sign indicating deceleration.

### Final Answer
The retardation of the particle when \( x = \frac{b}{a} \) is:
\[ \frac{2a}{27b^3} \]