1. We are given the relation between time and distance:

$t = mx^2 + nx$

where $m$ and $n$ are constants.

We need to find the retardation in terms of velocity $v$.

2. Velocity is

$v = \frac{dx}{dt}$

Since $t$ is given as a function of $x$, first find $\dfrac{dt}{dx}$:

$\frac{dt}{dx} = 2mx + n$

Therefore,

$v = \frac{dx}{dt} = \frac{1}{\frac{dt}{dx}} = \frac{1}{2mx+n}$

3. Express $x$ in terms of $v$.

From

$v = \frac{1}{2mx+n}$

we get

$2mx+n = \frac{1}{v}$

4. Now find acceleration:

$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = v\frac{dv}{dx}$

Differentiate

$v = (2mx+n)^{-1}$

with respect to $x$:

$\frac{dv}{dx} = -(2m)(2mx+n)^{-2}$

So,

$a = v\left[-2m(2mx+n)^{-2}\right]$

But

$(2mx+n)^{-1} = v \quad \Rightarrow \quad (2mx+n)^{-2} = v^2$

Hence,

$a = v(-2m v^2) = -2mv^3$

5. Since retardation is the magnitude of negative acceleration,

$\text{retardation} = 2mv^3$

6. Checking options:

• A: $2mv^3$ ✅
• B: $2mnv^3$ ❌
• C: $2nv^3$ ❌
• D: $2n^2v^3$ ❌

So the correct option is:

$\boxed{A}$