The relation between time t and distance x is t=αx²+ßx, where α and β are constants. The retardation is (v is the velocity)

The correct option is 2αv³.

To find the retardation of the particle, we need to determine its acceleration from the given relation between time and distance:
$t=\alpha x^2+\beta x$

Step 1: Differentiate with respect to time (t) to find velocity (v)
Differentiating both sides of the equation with respect to $t$:
$\frac{dt}{dt}=\frac{d}{dt}(\alpha x^2+\beta x)$
$1=(2\alpha x+\beta )\frac{dx}{dt}$

Since velocity $v=\frac{dx}{dt}$, we can substitute $v$ into the equation:
$1=(2\alpha x+\beta )v$

This gives the velocity $v$ as:
$v=\frac{1}{2\alpha x+\beta }={(2\mathrm{��}x+\beta )}^{-1}$

Step 2: Differentiate velocity with respect to time (t) to find acceleration (a)
Using the chain rule to differentiate $v$ with respect to $t$:
$a=\frac{dv}{dt}=\frac{d}{dt}{(2\alpha x+\beta )}^{-1}$
$a=-{(2\alpha x+\beta )}^{-2}·\frac{d}{dt}(2\alpha x+\beta )$
$a=-{(2\alpha x+\beta )}^{-2}·\left(2\alpha \frac{dx}{dt}\right)$

Substitute $\frac{dx}{dt}=v$ and ${(2\alpha x+\beta )}^{-1}=v$ into the expression:
$a=-v^2·\left(2\alpha v\right)$
$a=-2\alpha v^3$

Step 3: State the retardation
Retardation is the magnitude of deceleration, which is represented by negative acceleration:
$\text{Retardation}=-a=-(-2\alpha v^3)=2\alpha v^3$