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Let's say that a fluid is undergoing a one dimensional motion with the following velocity field: $\mathbf{U} = V(x,t)\mathbf{i}$.

The problem asks to find $V(x,t)$ given that the position of a particle initially at coordinates $(x_0,t_0)$ is $x=(x_0/t_0)t^2$.

I first wrote the langrangian derivative of $\mathbf{U}$ and equated it with the acceleration of a particle $2(x_0/t_0)$. This led me to a transport equation and ultimately to a wrong result.

So, I've been thinking about it for hours and I'm really getting nowhere. Any help is appreciated.

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U=dx/dt =2t*(x0/t0)

Now eliminate (x0/t0) from the above equation; x=(x0/t0)*t^2 (X0/t0)=x/t^2

So, U=2x/t which is in eulerian form.

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