The relation between time `t` and distance `x` of a particle is given by `t=Ax^(2)+Bx`, where `A` and `B` are constants. Then the
`(A)` velocity is given by `v=2Ax+B`
`(B)` velocity is given by `v=(2Ax+B)^(-1)`
`(C )` retardation is given by `2Av^(3)`
`(D)` retardation is given by `2Bv^(2)`
Select correct statement `:-`

Options

A. Only `C`
B. Only `D`
C. Only `A` & `D`
D. Only `B` & `C`

Explanation

To solve the problem, we need to analyze the given relation between time \( t \) and distance \( x \) of a particle, which is expressed as:

\[ t = Ax^2 + Bx \]

where \( A \) and \( B \) are constants. We will derive the expressions for velocity and retardation from this equation.

### Step 1: Differentiate the equation with respect to time \( t \)

We start by differentiating both sides of the equation with respect to time \( t \):

\[ \frac{dt}{dt} = \frac{d}{dt}(Ax^2 + Bx) \]

This simplifies to:

\[ 1 = \frac{d}{dt}(Ax^2) + \frac{d}{dt}(Bx) \]

### Step 2: Apply the chain rule

Using the chain rule, we differentiate \( Ax^2 \) and \( Bx \):

\[ 1 = 2Ax \frac{dx}{dt} + B \frac{dx}{dt} \]

Here, \( \frac{dx}{dt} \) is the velocity \( v \). Thus, we can rewrite the equation as:

\[ 1 = (2Ax + B)v \]

### Step 3: Solve for velocity \( v \)

Now, we can solve for \( v \):

\[ v = \frac{1}{2Ax + B} \]

This means we can also express it as:

\[ v = (2Ax + B)^{-1} \]

This matches with option (B).

### Step 4: Find Retardation

To find the retardation (negative acceleration), we need to differentiate the velocity \( v \) with respect to time \( t \). We start from:

\[ v = (2Ax + B)^{-1} \]

### Step 5: Differentiate \( v \)

Using the chain rule again, we differentiate \( v \):

\[ \frac{dv}{dt} = -\frac{1}{(2Ax + B)^2} \cdot \frac{d}{dt}(2Ax + B) \]

Now, we need to differentiate \( 2Ax + B \):

\[ \frac{d}{dt}(2Ax + B) = 2A \frac{dx}{dt} = 2Av \]

Substituting this back into the equation for \( \frac{dv}{dt} \):

\[ \frac{dv}{dt} = -\frac{1}{(2Ax + B)^2} \cdot (2Av) \]

### Step 6: Express retardation

Since retardation is defined as the negative of acceleration, we have:

\[ \text{Retardation} = -\frac{dv}{dt} = \frac{2Av}{(2Ax + B)^2} \]

### Step 7: Substitute for \( v \)

From our earlier expression for \( v \):

\[ v = (2Ax + B)^{-1} \]

Substituting this into the retardation equation gives:

\[ \text{Retardation} = 2Av^3 \]

This matches with option (C).

### Conclusion

The correct options are:
- (B) velocity is given by \( v = (2Ax + B)^{-1} \)
- (C) retardation is given by \( 2Av^3 \)