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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 = \fra

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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 $v = \frac{1}{2\alpha x + \beta}$. Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$. Differentiating $v = (2\alpha x + \beta)^{-1}$ with respect to $x$: $\frac{dv}{dx} = -1(2\alpha x + \beta)^{-2} \cdot (2\alpha) = -\frac{2\alpha}{(2\alpha x + \beta)^2}$. Substituting this into the acceleration formula: $a = v \cdot \left( -\frac{2\alpha}{(2\alpha x + \beta)^2} \right)$. Since $v = \frac{1}{2\alpha x + \beta}$,then $v^2 = \frac{1}{(2\alpha x + \beta)^2}$. Therefore,$a = v \cdot (-2\alpha \cdot v^2) = -2\alpha v^3$. The retardation is the magnitude of negative acceleration,which is $2\alpha v^3$.

The relation between time $t$ and distance $x$ is given by $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,then what is the retardation?

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