Streamline and time, particle x = 1 to 2

Given: $\vec V = \hat i\,A/x + \hat j\,Ay/x^2$ with $A = 2$ m²/s.

Streamline through (1, 3)

The streamline satisfies $dy/dx = v/u$:

$$\frac{dy}{dx} = \frac{Ay/x^2}{A/x} = \frac{y}{x}\;\Rightarrow\;\int\frac{dy}{y} = \int\frac{dx}{x}\;\Rightarrow\; y = Cx$$

Using $(x, y) = (1, 3)$: $C = 3$, so the streamline is $y = 3x$.

Time for a particle to move from x = 1 m to x = 2 m

Along its path $dx/dt = u = A/x$, so $x\,dx = A\,dt$:

$$t = \int_1^2\frac{x}{A}dx = \frac{x_2^2 - x_1^2}{2A} = \frac{4 - 1}{2\times 2} = 0.75\ \text{s}$$

(The flow is steady, so the particle path and the streamline coincide; the particle moves along $y = 3x$, from $(1, 3)$ to $(2, 6)$.)

Answer: streamline $y = 3x$; time $= 0.75$ s.