Water flows through a pipe with a velocity given by  m/s, where j is the unit vector in the y direction, t (> 0) is in seconds, and x and y are in meters. The magnitude of total acceleration at the point (x, y) = (1, 1) at t= 2s is m/s²

The correct answer is 3.

From the provided image, the velocity vector of the fluid flow is given by:
$\overrightarrow{V}=\left(\frac{4}{t}+x+y\right)\hat{j}$
Here, the velocity components in the x, y, and z directions are:
$u=0$
$v=\frac{4}{t}+x+y$
$w=0$

The total acceleration in a two-dimensional flow is given by its components. The acceleration component in the x-direction is:
$a_x=\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+v\frac{\partial u}{\partial y}=0$

The acceleration component in the y-direction is:
$a_y=\frac{\partial v}{\partial t}+u\frac{\partial v}{\partial x}+v\frac{\partial v}{\partial y}$

Since $u=0$, this simplifies to:
$a_y=\frac{\partial v}{\partial t}+v\frac{\partial v}{\partial y}$

Now we calculate the partial derivatives:
$\frac{\partial v}{\partial t}=\frac{\partial }{\partial t}\left(\frac{4}{t}+x+y\right)=-\frac{4}{t^2}$
$\frac{\partial v}{\partial y}=\frac{\partial }{\partial y}\left(\frac{4}{t}+x+y\right)=1$

Substituting these back into the expression for $a_y$:
$a_y=-\frac{4}{t^2}+\left(\frac{4}{t}+x+y\right)·1$
$a_y=-\frac{4}{t^2}+\frac{4}{t}+x+y$

Evaluate the acceleration component $a_y$ at the point $(x,y)=(1,1)$ and time $t=2\text{s}$:
$a_y=-\frac{4}{2^2}+\frac{4}{2}+1+1$
$a_y=-1+2+1+1=3$

Since $a_x=0$, the magnitude of total acceleration $a$ is:
$a=\sqrt{a_x^2+a_y^2}=\sqrt{0^2+3^2}=3{\text{m/s}}^2$