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(B) Given the relation: $t = m x^{2} + n x$.Differentiating with respect to $x$:$\frac{dt}{dx} = 2mx + n$.Since velocity $v = \frac{dx}{dt}$,we have $

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$2 m v^{3}$

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(B) Given the relation: $t = m x^{2} + n x$.Differentiating with respect to $x$:$\frac{dt}{dx} = 2mx + n$.Since velocity $v = \frac{dx}{dt}$,we have $\frac{1}{v} = \frac{dt}{dx} = 2mx + n$.Thus,$v = (2mx + n)^{-1}$.Now,differentiate $v$ with respect to $t$ to find acceleration $a$:$a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}$.$\frac{dv}{dx} = -1(2mx + n)^{-2} \cdot (2m) = -2m(2mx + n)^{-2}$.Since $(2mx + n) = \frac{1}{v}$,then $(2mx + n)^{-2} = v^{2}$.Therefore,$a = v \cdot (-2m \cdot v^{2}) = -2mv^{3}$.Retardation is the negative of acceleration,so retardation $= 2mv^{3}$.

The relation between time $t$ and distance $x$ for a moving body is given as $t = m x^{2} + n x$,where $m$ and $n$ are constants. The retardation of the motion is:

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