Answer to Question #130133 in Mechanics | Relativity for Ariyo Emmanuel

Question

A steady, incompressible, two-dimensional flow field defined by the velocity vector is expressed as

V = (u,v) = (0.5 + 0.8x)i + (1.5 - 0.8y)j

where the x- and y-coordinates are in meters and the magnitude of velocity is in m/s.

Calculate the material acceleration at the point (x = 2 m and y = 4 m)


Select one:
a. ax = 1.68 m/s2; ay = 1.36 m/s2

b. ax = 0.72 m/s2; ay = 1.68 m/s2
c. ax = 1.68 m/s2; ay=0.72 m/s2
d. ax = 2.45 m/s2; ay = 1.72 m/s2

Answer 1

Use equation for material acceleration in Cartesian coordinates

$a_x = \frac{∂u}{∂t} + u\frac{∂u}{∂x} + v\frac{∂u}{∂y} + w\frac{∂u}{∂z}$

$a_x = 0 + (0.5 + 0.8x)0.8 + (1.5 – 0.8)0 + 0 = 0.4 + 0.64x$

$a_y = 0 + (0.5 + 0.8x)0 + (1.5 – 0.8)(-0.8) + 0 = -1.2 + 0.64y$

At the point (x = 2 m, y = 4 m)

$a_x = 0.4 + (0.64\times2) = 1.68 m/s^2$

$a_y = -1.2 + (0.64\times4) = 1.36 m/s^2$

Answer: a. a_x = 1.68 m/s²; a_y = 1.36 m/s²