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(C) The velocity of the particle is given by $v = 3x^2 - 4x$. Acceleration $a$ is defined as the rate of change of velocity with respect to time,which

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$(3x^2 - 4x)(6x - 4)$

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(C) The velocity of the particle is given by $v = 3x^2 - 4x$. Acceleration $a$ is defined as the rate of change of velocity with respect to time,which can be expressed in terms of position $x$ as $a = v \cdot \frac{dv}{dx}$. First,differentiate $v$ with respect to $x$: $\frac{dv}{dx} = \frac{d}{dx}(3x^2 - 4x) = 6x - 4$. Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula: $a = (3x^2 - 4x)(6x - 4)$. Thus,the correct expression for acceleration is $(3x^2 - 4x)(6x - 4)$.

The velocity of a particle is given by the equation $v(x) = 3x^2 - 4x$,where $x$ is the distance covered by the particle. The expression for its acceleration is

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