If v = x^2 - 5x + 4,find the acceleration of | vedclass

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(D) The acceleration $a$ of a particle is given by the relation $a = v \frac{dv}{dx}$.Given the velocity $v = x^2 - 5x + 4$.First,calculate the deriva

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(D) The acceleration $a$ of a particle is given by the relation $a = v \frac{dv}{dx}$.Given the velocity $v = x^2 - 5x + 4$.First,calculate the derivative of $v$ with respect to $x$:$\frac{dv}{dx} = \frac{d}{dx}(x^2 - 5x + 4) = 2x - 5$.Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:$a = (x^2 - 5x + 4)(2x - 5)$.We need to find the acceleration when the velocity $v = 0$:$x^2 - 5x + 4 = 0$$(x - 1)(x - 4) = 0$So,$x = 1$ or $x = 4$.If $x = 1$,$a = (0)(2(1) - 5) = 0$.If $x = 4$,$a = (0)(2(4) - 5) = 0$.Therefore,when the velocity is zero,the acceleration is $0$.

Column-$I$	Column-$II$
$(1)$ Acceleration is positive	$(a)$ Speed of the particle decreases
$(2)$ Acceleration is negative	$(b)$ Speed of the particle increases
	$(c)$ Speed of the particle keeps changing

If $v = x^2 - 5x + 4$,find the acceleration of the particle when the velocity of the particle is zero.

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