The relation between time t and displacement | vedclass

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(D) Given the relation: $t = \alpha x^2 + \beta x$. Differentiating both sides with respect to $t$: $1 = 2 \alpha x \left( \frac{dx}{dt} \right) + \be

Vedclass · Accepted Answer

$2 \alpha v^3$

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(D) Given the relation: $t = \alpha x^2 + \beta x$. Differentiating both sides with respect to $t$: $1 = 2 \alpha x \left( \frac{dx}{dt} ight) + \beta \left( \frac{dx}{dt} ight)$. Since $v = \frac{dx}{dt}$,we have: $1 = (2 \alpha x + \beta) v \implies v = \frac{1}{2 \alpha x + \beta}$. 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 $x$: $\frac{dv}{dx} = -1(2 \alpha x + \beta)^{-2} \cdot (2 \alpha) = -2 \alpha v^2$. Thus,$a = v (-2 \alpha v^2) = -2 \alpha v^3$. Retardation is defined as negative acceleration: $	ext{Retardation} = -a = -(-2 \alpha v^3) = 2 \alpha v^3$.

The relation between time $t$ and displacement $x$ is $t = \alpha x^2 + \beta x$,where $\alpha$ and $\beta$ are constants. If $v$ is the velocity,the retardation is:

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