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The velocity in a certain flow field is given by the equation.

\[V=3yz^2i+xz^2j+yk\]

  • Determine the expression for the three rectangular components of acceleration.

This problem familiarizes us with the rectangular components of a vector. The concept required to solve this problem is derived from basic dynamic physics which includes, velocity vector, acceleration, and rectangular coordinates.

Rectangular components are defined as the components or regions of a vector in any corresponding perpendicular axis. Thus rectangular components of acceleration would be the velocity vectors with respect to the time taken by the object.

Expert Answer

As per the statement, we are given a velocity vector which illustrates the rate of change of the displacement of an object. The absolute value of a velocity vector provides the speed of the object while the unit vector gives its direction.

From the given expression of velocity, it can be deduced that:

$u = 3yz^2$, $v = xz$, $w = y$

Now the three rectangular components of acceleration are: $a_x$, $a_y$, and $a_z$.

The formula to find the $a_x$ component of acceleration is given as:

\[ a_x = \dfrac{\partial u}{\partial t} + u \dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y} + w \dfrac{\partial u}{\partial z} \]

Inserting the values and solving for $a_x$:

\[ a_x = \dfrac{\partial}{\partial t} (3yz^2) + (3yz^2) \dfrac{\partial}{\partial x} (3yz^2) + (xz) \dfrac{\partial}{\partial y} (3yz^2) + y \dfrac{\partial }{\partial z} (3yz^2) \]

\[ = 0 + (xz)(3z^2) + (y)(6yz) \]

$a_x$ comes out to be:

\[ a_x = 3xz^3 + 6y^2z \]

The formula to find the $a_y$ component of acceleration is given as:

\[ a_y = \dfrac{\partial v}{\partial t} + u \dfrac{\partial v}{\partial x} + v \dfrac{\partial v}{\partial y} + w \dfrac{\partial v}{\partial z} \]

Inserting the values and solving for $a_y$:

\[ a_y = \dfrac{\partial}{\partial t} (xz) + (3yz^2) \dfrac{\partial}{\partial x} (xz) + (xz) \dfrac{\partial}{\partial y} (xz) + y \dfrac{\partial }{\partial z} (xz) \]

\[ = 0 + (3yz^2)(z) + (xz)(0) + (y)(x) \]

$a_y$ comes out to be:

\[ a_y = 3yz^3 + xy \]

Lastly $a_z$, formula for finding the $a_z$ component of acceleration is:

\[ a_z = \dfrac{\partial w}{\partial t} + u \dfrac{\partial w}{\partial x} + v \dfrac{\partial w}{\partial y} + w \dfrac{\partial w}{\partial z} \]

Inserting the values and solving for $a_z$:

\[ a_z = \dfrac{\partial}{\partial t} (y) + (3yz^2) \dfrac{\partial}{\partial x} (y) + (xz) \dfrac{\partial}{\partial y} (y) + y \dfrac{\partial }{\partial z} (y) \]

\[ = 0 + (3yz^2)(0) + (xz)(1) + (y)(0) \]

$a_z$ comes out to be:

\[ a_z = xz \]

Numerical Result

Expressions for the three rectangular components of acceleration are:

$a_x = 3xz^2 + 6y^2z$

$a_y = 3yz^3 + xy$

$a_z = xz$

Example

The velocity in a two-dimensional flow field is given by $V= 2xti – 2ytj$. Find the $a_x$ rectangular component of acceleration.

It can be found out that:

$u=2xt$ and $v=-2yt$

Applying formula:

\[a_x = \dfrac{\partial u}{\partial t} + u \dfrac{\partial u}{\partial x} + v \dfrac{\partial u}{\partial y}\]

Inserting values:

\[a_x =\dfrac{\partial}{\partial t} (2xt) + (2xt) \dfrac{\partial}{\partial x} (2xt) + (-2yt) \dfrac{\partial u}{\partial y} (2xt)\]

\[a_x = 2x + 4xt^2\]

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