Question

The velocity vector of a 2D flow field is given by \(\vec{V} = 2y^2 \hat{i} + x^2t \hat{j}\).
The acceleration is
Note: There's a typo in the OCR of the question (î twice) and options. The solution assumes the standard form V = uî + vĵ.

• A. \(4x^2ty \hat{i} + (x^2 + 4xy^2t) \hat{j}\)
• B. \(4x^2ty \hat{i} - (x^2 + 4xy^2t) \hat{j}\)
• C. \(4x^2ty \hat{i} + x^2 \hat{j}\)
• D. \(x^2 \hat{j}\)

Hint:

Remember that acceleration in a fluid flow has both a time-dependent (local) part and a space-dependent (convective) part. A steady flow (\(\partial/\partial t = 0\)) can still have acceleration if the velocity changes with position.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The acceleration of a fluid particle in a flow field is described by the material derivative (or total derivative) of the velocity vector. It consists of two parts: the local acceleration (\(\frac{\partial \vec{V}}{\partial t}\)), which represents the change in velocity at a fixed point over time, and the convective acceleration (\((\vec{V} . \nabla)\vec{V}\)), which represents the change in velocity as the particle moves from one point to another in the flow field.
Step 2: Key Formula or Approach:
The acceleration vector \(\vec{a}\) is given by: \[ \vec{a} = \frac{D\vec{V}}{Dt} = \frac{\partial \vec{V}}{\partial t} + (\vec{V} . \nabla)\vec{V} \] For a 2D velocity vector \(\vec{V} = u(x,y,t)\hat{i} + v(x,y,t)\hat{j}\), the components of acceleration are: \[ a_x = \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} \] \[ a_y = \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \] Step 3: Detailed Explanation or Calculation:
From the given velocity vector \(\vec{V} = 2y^2 \hat{i} + x^2t \hat{j}\), we identify the velocity components: \(u = 2y^2\)
\(v = x^2t\)
First, we find the required partial derivatives: \[ \frac{\partial u}{\partial t} = 0, \frac{\partial u}{\partial x} = 0, \frac{\partial u}{\partial y} = 4y \] \[ \frac{\partial v}{\partial t} = x^2, \frac{\partial v}{\partial x} = 2xt, \frac{\partial v}{\partial y} = 0 \] Now, we substitute these into the acceleration component equations: x-component of acceleration (\(a_x\)): \[ a_x = \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = 0 + (2y^2)(0) + (x^2t)(4y) = 4x^2yt \] y-component of acceleration (\(a_y\)): \[ a_y = \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} = x^2 + (2y^2)(2xt) + (x^2t)(0) = x^2 + 4xy^2t \] Combining the components, the acceleration vector is: \[ \vec{a} = a_x \hat{i} + a_y \hat{j} = (4x^2yt) \hat{i} + (x^2 + 4xy^2t) \hat{j} \] Step 4: Final Answer:
The acceleration is \(\vec{a} = 4x^2ty \hat{i} + (x^2 + 4xy^2t) \hat{j}\).
Step 5: Why This is Correct:
The calculated acceleration vector matches option (A). The question in the provided image has a typo (\(\hat{i}\) twice), but based on the options, it is clear that the second component of velocity was intended to be in the \(\hat{j}\) direction.