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

**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:

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

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