The relationship between time $ t $ and distance $ x $ for a moving body is given by $ t = mx^2 + nx $, where $ m $ and $ n $ are constants. To determine the retardation (negative acceleration) of the motion, let's follow the steps to derive it:

Given:

$ t = mx^2 + nx $

First, differentiate $ t $ with respect to $ x $:

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

Since velocity $ v $ is defined as $ \frac{dx}{dt} $, we can write:

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

Thus,

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

Next, to find the acceleration $ a $ (which is the derivative of velocity with respect to time), we start with the chain rule:

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

Since $ v = \frac{dx}{dt} $, substituting $ x $ gives:

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

Rewrite it:

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

Differentiate $ v $ with respect to $ x $:

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

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

Then,

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

Substitute $ v = \frac{1}{2mx + n} $ into the expression:

$ \frac{dv}{dt} = -\frac{2m}{(2mx + n)^2} \cdot \frac{1}{2mx + n} $

Simplify the expression:

$ \frac{dv}{dt} = -2m \left( \frac{1}{2mx + n} \right)^3 $

$ a = -2m v^3 $

So, the retardation (negative acceleration) is:

$ a = -2m v^3 $