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Question

(C) Given the relation: $t = ax^2 + bx$. Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2ax + b$. Since velocity $v = \frac{dx}{dt}$

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$-2av^3$

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(C) Given the relation: $t = ax^2 + bx$. Differentiating both sides with respect to $x$: $\frac{dt}{dx} = 2ax + b$. Since velocity $v = \frac{dx}{dt}$, we have $\frac{dt}{dx} = \frac{1}{v}$. Therefore, $\frac{1}{v} = 2ax + b$, which implies $v = (2ax + b)^{-1}$. Acceleration $a_{acc} = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}$. Differentiating $v$ with respect to $x$: $\frac{dv}{dx} = -1(2ax + b)^{-2} \cdot (2a) = -2a(2ax + b)^{-2}$. Substituting $(2ax + b) = \frac{1}{v}$: $\frac{dv}{dx} = -2a \cdot (\frac{1}{v})^{-2} = -2av^2$. Thus, acceleration $a_{acc} = v \cdot (-2av^2) = -2av^3$.

The relation between time $t$ and distance $x$ of a particle is $t = ax^2 + bx$, where $a$ and $b$ are constants. If $v$ is the velocity of the particle, then its acceleration is

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