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(A) Given the relation: $t = \alpha x^2 + \beta x$Differentiating with respect to $x$:$\frac{dt}{dx} = 2\alpha x + \beta$Since velocity $v = \frac{dx}

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$2\alpha v^3$

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(A) Given the relation: $t = \alpha x^2 + \beta x$Differentiating with respect to $x$:$\frac{dt}{dx} = 2\alpha x + \beta$Since velocity $v = \frac{dx}{dt}$,we have:$v = \frac{1}{2\alpha x + \beta} \implies 2\alpha x + \beta = \frac{1}{v}$Acceleration $a$ is given by $a = v \frac{dv}{dx}$.First,find $\frac{dv}{dx}$ by differentiating $v = (2\alpha x + \beta)^{-1}$ with respect to $x$:$\frac{dv}{dx} = -1(2\alpha x + \beta)^{-2} \cdot (2\alpha) = -\frac{2\alpha}{(2\alpha x + \beta)^2}$Substitute $(2\alpha x + \beta) = \frac{1}{v}$ into the expression:$\frac{dv}{dx} = -2\alpha \cdot v^2$Now,calculate acceleration:$a = v \cdot (-2\alpha v^2) = -2\alpha v^3$Retardation is the negative of acceleration:$	ext{Retardation} = -a = 2\alpha v^3$

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. The retardation of the particle is:

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