Question

A steady, incompressible, two-dimensional (in the x-y plane) velocity field is given by V→=(0.523−1.88x+3.94y)i→+(−2.44+1.26x+1.88y)j→ . Calculate the acceleration at the point (x,y) = (-1, 3.5).

Answer 1

`\vec a = -19,096 \cdot i + 25.818 \cdot j`

Step-by-step explanation:

According to the definition of the derivative of vectorial sum and the physical concept of acceleration, the required expression is:

`\vec a = (d v_(x))/(dt) \cdot i + (d v_(y))/(dt) \cdot j + (d v_(z))/(dt) \cdot k`

`\vec a = (-1.88\cdot (dx)/(dt) + 3.94 \cdot (dy)/(dt))\cdot i + (1.26 \cdot (dx)/(dt) + 1.88 \cdot (dy)/(dt))\cdot j`

Where:

`(dx)/(dt) = 0.523 - 1.88 \cdot x + 3.94 \cdot y\\(dy)/(dt) = - 2.44 + 1.26 \cdot x + 1.88 \cdot y\\`

The acceleration at given point is calculated as follows:

`(dx)/(dt) = 16.193\\(dy)/(dt) = 2.88`

`\vec a = -19,096 \cdot i + 25.818 \cdot j`