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

Given:
Velocity vector: v = −x î + 2y ĵ − z k̂

Step 1: Use formula for acceleration
a = (v · ∇)v

Step 2: Components of velocity
v_x = −x, v_y = 2y, v_z = −z

Step 3: Compute acceleration components

a_x = v_x ∂(−x)/∂x + v_y ∂(−x)/∂y + v_z ∂(−x)/∂z
= (−x)(−1) + (2y)(0) + (−z)(0) = x

a_y = v_x ∂(2y)/∂x + v_y ∂(2y)/∂y + v_z ∂(2y)/∂z
= (−x)(0) + (2y)(2) + (−z)(0) = 4y

a_z = v_x ∂(−z)/∂x + v_y ∂(−z)/∂y + v_z ∂(−z)/∂z
= (−x)(0) + (2y)(0) + (−z)(−1) = z

Step 4: Acceleration vector
a = x î + 4y ĵ + z k̂

Step 5: At (1,1,4)
a = î + 4ĵ + 4k̂

Step 6: Magnitude
\[ |a| = \sqrt{1^2 + 4^2 + 4^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \]

Final Answer: √33