Magnitude of Acceleration for Position Dependent Velocity Vector

To find the magnitude of the acceleration of the particle, we use the expression for total derivative of velocity with respect to time for a position-dependent velocity field $\vec{v}(x, y, z)$:

$\vec{a} = \frac{d\vec{v}}{dt} = v_x \frac{\partial \vec{v}}{\partial x} + v_y \frac{\partial \vec{v}}{\partial y} + v_z \frac{\partial \vec{v}}{\partial z}$

The velocity components of the particle are given as: $v_x = \frac{dx}{dt} = -x$ $v_y = \frac{dy}{dt} = 2y$ $v_z = \frac{dz}{dt} = -z$

Now, we calculate the individual components of acceleration $a_x$, $a_y$, and $a_z$:

1. x-component of acceleration ($a_x$): $a_x = v_x \frac{\partial v_x}{\partial x} + v_y \frac{\partial v_x}{\partial y} + v_z \frac{\partial v_x}{\partial z}$ $a_x = (-x)(-1) + (2y)(0) + (-z)(0) = x$

2. y-component of acceleration ($a_y$): $a_y = v_x \frac{\partial v_y}{\partial x} + v_y \frac{\partial v_y}{\partial y} + v_z \frac{\partial v_y}{\partial z}$ $a_y = (-x)(0) + (2y)(2) + (-z)(0) = 4y$

3. z-component of acceleration ($a_z$): $a_z = v_x \frac{\partial v_z}{\partial x} + v_y \frac{\partial v_z}{\partial y} + v_z \frac{\partial v_z}{\partial z}$ $a_z = (-x)(0) + (2y)(0) + (-z)(-1) = z$

Thus, the acceleration vector as a function of position is: $\vec{a} = x\hat{i} + 4y\hat{j} + z\hat{k}$

Evaluating the acceleration vector at the point $(x, y, z) = (1, 2, 4)$: $\vec{a} = (1)\hat{i} + 4(2)\hat{j} + (4)\hat{k} = 1\hat{i} + 8\hat{j} + 4\hat{k}$

The magnitude of the acceleration vector $|\vec{a}|$ is: $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{1^2 + 8^2 + 4^2}$ $|\vec{a}| = \sqrt{1 + 64 + 16} = \sqrt{81} = 9 \text{ m/s}^2$

Correct Option: B