Question

the velocity vector in a fluid flow is given by v=4x³i-10x²y²j+2tk find the velocity and acceleration of a fluid particle at (2,1,3) at time t=1​

Answer 1

Given: Velocity vector = 4x³i-10x²y²j+2tk

           Position of the particle = (2,1,3)

           Time = 1

To find: The velocity and acceleration

Solution:

The velocity component of u, v and w are u = 4x³, y= - 10x², w = 2t

For the point (2,1,3), we have x = 2, y = 1, and z = 3 at time t = 1

Hence the velocity component at (2,1,3) are -

u = 4×(3)³ = 32 units

v = -10(1)² = -10 units

w = 2×1 = 2 units

Therefore, vector v at (2,1,3) = 32i - 40j + 2k

Resultant velocity = $\sqrt{u^2 + v^2 + w^2$

                             = $\sqrt{32^2 + (-40)^2 + 2^2$

                            = 51.26 units

The resultant velocity is equal to 51.26 units.

Acceleration is given by =

aₓ = udu/dx + vdu/dy + wdu/dz + du/dt

$a_{y}$ =  udv/dx + vdv/dy + wdv/dz + dv/dt

$a_{z}$ =  udw/dx + vdw/dy + wdw/dz + dw/dt

And from velocity component we have,

du/dx = 12x², du/dy = 0,                    du/dz = 0, and du/dt = 0

dv/dx = -20xy, dv/dy = -10x²,             dv/dz = 0, dv/dt = 0

dw/dx = 0, dw/dy = 0,                        dw/dz = 0, dw/dt = 2.1

Substituting the values, the acceleration component at (2,1,3) at time t = 1 are,

aₓ =  4x³(12x²) + (-10x²)(0) + 2t(0)  

   = 48x⁵

   = 48(2)⁵ = 1536 units

$a_{y}$ = 4x³(-20xy) + (-10x²y)(-10x²) + 2t(0)

   = -80x⁴y + 100x⁴y

    = 20x⁴y

   = 20(2)⁴(1)

  = 320 units

$a_{z}$ = 4x³(0) + (-10x²y)(0) + 2t(0) + 2.1

   = 2.0 units

Resultant acceleration A = $\sqrt{(1536)^2 + (320)^2 + (2)^2$

                                         = 1568.9 units

The resultant acceleration will be 1568.9 units.