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 step by step, we will follow the given equation of motion and derive the required retardation when \( x = \frac{b}{a} \).

### Step 1: Understand the equation of motion
The equation given is:
\[ t = ax^2 + bx \]
This relates time \( t \) to the displacement \( x \).

### Step 2: Find the velocity
To find the velocity \( v \), we need to differentiate \( x \) with respect to \( t \):
\[ v = \frac{dx}{dt} \]
Using the chain rule, we can express this as:
\[ v = \frac{dx}{dt} = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} \]
From the equation \( t = ax^2 + bx \), we can find \( \frac{dt}{dx} \):
\[ \frac{dt}{dx} = \frac{d}{dx}(ax^2 + bx) = 2ax + b \]
Thus, the velocity becomes:
\[ v = \frac{1}{2ax + b} \]

### Step 3: Find the acceleration
Acceleration \( a \) is the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} \]
We already have \( \frac{dx}{dt} = 2ax + b \). Now we need to differentiate \( v \) with respect to \( x \):
\[ v = \frac{1}{2ax + b} \]
Using the quotient rule:
\[ \frac{dv}{dx} = -\frac{(0)(2ax + b) - (1)(2a)}{(2ax + b)^2} = -\frac{2a}{(2ax + b)^2} \]
Now substituting this into the acceleration formula:
\[ a = \left(-\frac{2a}{(2ax + b)^2}\right)(2ax + b) = -\frac{2a}{2ax + b} \]

### Step 4: Substitute \( x = \frac{b}{a} \)
Now we substitute \( x = \frac{b}{a} \) into the acceleration expression:
\[ a = -\frac{2a}{2a\left(\frac{b}{a}\right) + b} = -\frac{2a}{2b + b} = -\frac{2a}{3b} \]

### Step 5: Retardation
Retardation is defined as the negative acceleration:
\[ \text{Retardation} = -a = \frac{2a}{3b} \]

### Conclusion
Thus, the retardation of the particle when \( x = \frac{b}{a} \) is:
\[ \text{Retardation} = \frac{2a}{3b} \]