Answer:
Explanation:
The value of a will be zero as it is provided that the particle is on the x-axis.
Calculate the velocity of particles along x-axis.
[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]
Substitute 0 for y.
[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]
Here,
[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]
Calculate the magnitude of vector V .
[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]
Substitute
[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]
[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]
The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]
Calculate the direction of flow.
[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]
Here, θ is the flow from positive x-axis in a counterclockwise direction.
Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.
[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]
The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.
