Question

A particle is moving such that its velocity vector at co-ordinate $(x,y,z)$ is $\vec{v} = -x\hat{i} + 2y\hat{j} - z\hat{k}$. Find magnitude of acceleration at $(1,1,4)$.

Answer 1

Step 1: Write down the velocity components.
$v_x = -x$, $v_y = 2y$, $v_z = -z$.

Step 2: Find the acceleration components using the chain rule ($a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$).
$a_x = v_x \frac{\partial v_x}{\partial x} = (-x) \cdot (-1) = x$
$a_y = v_y \frac{\partial v_y}{\partial y} = (2y) \cdot (2) = 4y$
$a_z = v_z \frac{\partial v_z}{\partial z} = (-z) \cdot (-1) = z$

Step 3: Calculate acceleration components at the point $(1,1,4)$.
$a_x = 1$, $a_y = 4(1) = 4$, $a_z = 4$.

Step 4: Calculate the magnitude of the acceleration vector $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$.
$|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} = \sqrt{1^2 + 4^2 + 4^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \text{ m/s}^2$.

Final Answer: $\sqrt{33} \text{ m/s}^2$.