Question Details

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

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Correct Answer :

√ 33

Solution :

The correct answer is √ 33.

Step-by-step Explanation:

We are given the velocity vector of a particle as:
$\vec{v} = v_{x} \hat{i} + v_{y} \hat{j} + v_{z} \hat{k} = - x \hat{i} + 2 y \hat{j} - z \hat{k}$

This gives the individual velocity components:
$v_{x} = - x$
$v_{y} = 2 y$
$v_{z} = - z$

The components of the acceleration vector $\vec{a}$ in a steady flow field are given by the convective acceleration:
$a_{x} = v_{x} ∂ v_{x} ∂ x + v_{y} ∂ v_{x} ∂ y + v_{z} ∂ v_{x} \partial z$
$a_{y} = v_{x} ∂ v_{y} \partial x + v_{y} ∂ v_{y} \partial y + v_{z} ∂ v_{y} \partial z$
$a_{z} = v_{x} ∂ v_{z} \partial x + v_{y} ∂ v_{z} \partial y + v_{z} \frac{\partial v_{z}}{\partial z}$

Let us compute the partial derivatives:
For $v_{x} = - x$ :
$∂ v_{x} ∂ x = - 1, \frac{\partial v_{x}}{\partial y} = 0, \frac{\partial v_{x}}{\partial z} = 0$

For $v_{y} = 2 y$ :
$\frac{\partial v_{y}}{\partial x} = 0, \frac{\partial v_{y}}{\partial y} = 2, \frac{\partial v_{y}}{\partial z} = 0$

For $v_{z} = - z$ :
$\frac{\partial v_{z}}{\partial x} = 0, \frac{\partial v_{z}}{\partial y} = 0, \frac{\partial v_{z}}{\partial z} = - 1$

Now, substitute these derivatives into the acceleration component equations:
$a_{x} = (- x) (- 1) + 0 + 0 = x$
$a_{y} = 0 + (2 y) (2) <+> 0 = 4 y$
$a_{z} = 0 <+> 0 + (- z) (- 1) = z$

So, the acceleration vector is:
$\vec{a} = x \hat{i} + 4 y \hat{j} + z \hat{k}$

We need to find the magnitude of acceleration at the coordinate (1, 1, 4).
Substituting $x = 1$ , $y = 1$ , and $z = 4$ into the acceleration components:
$a_{x} = 1$
$a_{y} = 4 (1) = 4$
$a_{z} = 4$

The magnitude of the acceleration vector is:
$| \vec{a} | = \sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2}}$
$| \vec{a} | = \sqrt{1^{2} + 4^{2} + 4^{2}}$
$| \vec{a} | = \sqrt{1 + 16 + 16} = \sqrt{33}$