I know that what I am asking is easy but I can't understand how to deal and relate.
My question is this that we write that total acceleration of fluid particle as: $$\vec{a}=\frac{\partial \vec{V}}{\partial t}+\big(\vec{V}\bullet \vec{\nabla}\big)\vec{V}$$ But in natural coordinates system we can also write that total acceleration of fluid particle as: $$\vec{a}=\frac{\partial \vec{V}}{\partial t}+\vec{V}\frac{\partial \vec{V}}{\partial s}$$ where "s" is arclength parameter. My question is How $$\frac{\partial \vec{V}}{\partial t}+|\vec{V}|\frac{\partial \vec{V}}{\partial s}=\frac{\partial \vec{V}}{\partial t}+\big(\vec{V}\bullet \vec{\nabla}\big)\vec{V}$$ Can anyone explain in detail.
Thanks in advance
First, a quibble: Your final LHS should read
$\displaystyle \frac{\partial \vec{V}}{\partial t}+|\vec{V}|\frac{\partial \vec{V}}{\partial s}$
And that might be the clue you need to your answer, in fact. $|\vec{V}|\frac{\partial}{\partial s}$ will equal $(\vec{V}\cdot\nabla)$ if and only if the "natural coordinate" is measuring length along the curve of the fluid element's path. Another way to say it is that $\vec{V}$ must be tangent to the curve whose length $s$ measures.
To see this, imagine that $\vec{V}$ does indeed flow along the curved measured by $s$. That means that if I draw tangent vectors all along that curve, they all point in the direction of local $\vec{V}$. This means that $\vec{V}$ dotted with the gradient operator is the same as $|\vec{V}|$ multiplied by the change in length along the curve.
Looked at in another way,
$\displaystyle \vec{V}\cdot\nabla = |\vec{V}|\hat{v}\cdot\nabla$
where $\hat{v}$ is the unit vector of $\vec{V}$. Now, for any unit vector $\hat{u}$, the dot product $\hat{u}\cdot\nabla$ give the change per unit length along the direction of $\hat{u}$. So the dot product $\hat{v}\cdot\nabla$ gives us the change per unit length along the velocity. We know this change per unit length must be $\partial/\partial s$. And so
$\displaystyle \hat{v}\cdot\nabla = \frac{\partial}{\partial s}$
Therefore
$\displaystyle \vec{V}\cdot\nabla = |\vec{V}|\hat{v}\cdot\nabla = |\vec{V}|\frac{\partial}{\partial s}$