The relation between time t and displacement x is `t = alpha x^2 + beta x,` where `alpha and beta` are constants. The retardation is

Options

A. `2 alpha v^3`
B. `2 beta v^3`
C. `2 alpha beta v^3`
D. `2 beta^2 v^3`

Explanation

To find the retardation given the relation between time \( t \) and displacement \( x \) as \( t = \alpha x^2 + \beta x \), we will follow these steps:

### Step 1: Differentiate the equation with respect to \( x \)
We start with the equation:
\[ t = \alpha x^2 + \beta x \]
Differentiating both sides with respect to \( x \):
\[ \frac{dt}{dx} = 2\alpha x + \beta \]

### Step 2: Find the expression for velocity
Velocity \( v \) is defined as:
\[ v = \frac{dx}{dt} \]
From the previous step, we can express \( \frac{dx}{dt} \) as:
\[ \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2\alpha x + \beta} \]

### Step 3: Differentiate velocity to find acceleration
Now, we differentiate \( v \) with respect to \( t \) to find acceleration \( a \):
\[ a = \frac{dv}{dt} = \frac{d}{dt}\left(\frac{1}{2\alpha x + \beta}\right) \]
Using the chain rule, we have:
\[ a = -\frac{1}{(2\alpha x + \beta)^2} \cdot \frac{d(2\alpha x + \beta)}{dt} \]
Now, we need to find \( \frac{d(2\alpha x + \beta)}{dt} \):
\[ \frac{d(2\alpha x + \beta)}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v \]
Substituting this back into the expression for acceleration:
\[ a = -\frac{2\alpha v}{(2\alpha x + \beta)^2} \]

### Step 4: Substitute \( v \) back into the equation
We already found that:
\[ v = \frac{1}{2\alpha x + \beta} \]
Substituting \( v \) into the acceleration equation:
\[ a = -\frac{2\alpha \cdot \frac{1}{2\alpha x + \beta}}{(2\alpha x + \beta)^2} \]
This simplifies to:
\[ a = -\frac{2\alpha}{(2\alpha x + \beta)^3} \]

### Final Result
Thus, the retardation (which is the negative acceleration) is given by:
\[ \text{Retardation} = \frac{2\alpha}{(2\alpha x + \beta)^3} \]