The velocity of a particle is given as vec{v} | vedclass

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(B) The acceleration vector is given by the material derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = (\vec{v} \cdot 
abla) \vec{v}$.Given $\

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$9$

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(B) The acceleration vector is given by the material derivative of velocity: $\vec{a} = \frac{d\vec{v}}{dt} = (\vec{v} \cdot 
abla) \vec{v}$.Given $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$,we calculate the components:$a_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} + v_z \frac{\partial v_x}{\partial z} = (-x)(-1) + (2y)(0) + (-z)(0) = x$.$a_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z} = (-x)(0) + (2y)(2) + (-z)(0) = 4y$.$a_z = v_x \frac{\partial v_z}{\partial x} + v_y \frac{\partial v_z}{\partial y} + v_z \frac{\partial v_z}{\partial z} = (-x)(0) + (2y)(0) + (-z)(-1) = z$.Thus,$\vec{a} = x\hat{i} + 4y\hat{j} + z\hat{k}$.At point $(1, 2, 4)$,$\vec{a} = 1\hat{i} + 4(2)\hat{j} + 4\hat{k} = 1\hat{i} + 8\hat{j} + 4\hat{k}$.The magnitude is $|\vec{a}| = \sqrt{1^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 	ext{ m/s}^2$.

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

The velocity of a particle is given as $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k} \text{ m/s}$. The magnitude of acceleration at point $(1, 2, 4)$ is . . . . . . $\text{m/s}^2$.

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