# For a steady flow, the velocity field is $\vec V = \left( { - {x^2} + 3y} \right)\hat i + \left( {2xy} \right)\hat j$ . The magnitude of the acceler

18 views
in General
closed
For a steady flow, the velocity field is $\vec V = \left( { - {x^2} + 3y} \right)\hat i + \left( {2xy} \right)\hat j$ . The magnitude of the acceleration of a particle at (1, -1) is
1. 2
2. 1
3. $2\sqrt 5$
4. 0

by (37.3k points)
selected

Correct Answer - Option 3 : $2\sqrt 5$

Explanation:

$\vec V = \left( { - {x^2} + 3y} \right)\hat i + 2xy\hat j$

u = - x2 + 3y; v = 2xy

${a_x} = u\frac{{\partial u}}{{\partial x}} + v\frac{{\partial u}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( { - 2x} \right) + 2xy\left( 3 \right)$

ax = [-(1)2 + 3 (-1)] (-2(1))+6(1)(-1)

a1 = 8 – 6

ax = 2

${a_y} = u\frac{{\partial v}}{{\partial x}} + v\frac{{\partial v}}{{\partial y}} = \left( { - {x^2} + 3y} \right)\left( {2y} \right) + \left( {2xy} \right)\left( {2x} \right)$

${a_y} = \left[ { - {{\left( 1 \right)}^2} + 3\left( { - 1} \right)} \right]\left[ {2\left( { - 1} \right)\left] + \right[4{{\left( 1 \right)}^2}\left( { - 1} \right)} \right]$

ay = 8 – 4

ay = 4

$a = \sqrt {a_x^2 + a_y^2\;} = \sqrt {{2^2} + {4^2}} = \sqrt {20}$

$= 2\sqrt 5$