1) [tex]a_x=4.287+2.772x\\a_y=-5.579+2.772y[/tex]
2) 8.418
Explanation:
1)
The two components of the velocity field in x and y for the field in this problem are:
[tex]u=1.85+2.05x+0.656y[/tex]
[tex]v=0.754-2.18x-2.05y[/tex]
The x-component and y-component of the acceleration field can be found using the following equations:
[tex]a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}[/tex]
[tex]a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}[/tex]
The derivatives in this problem are:
[tex]\frac{du}{dt}=0[/tex]
[tex]\frac{dv}{dt}=0[/tex]
[tex]\frac{du}{dx}=2.05[/tex]
[tex]\frac{du}{dy}=0.656[/tex]
[tex]\frac{dv}{dx}=-2.18[/tex]
[tex]\frac{dv}{dy}=-2.05[/tex]
Substituting, we find:
[tex]a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x[/tex]
And
[tex]a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y[/tex]
2)
In this part of the problem, we want to find the acceleration at the point
(x,y) = (-1,5)
So we have
x = -1
y = 5
First of all, we substitute these values of x and y into the expression for the components of the acceleration field:
[tex]a_x=4.287+2.772x\\a_y=-5.579+2.772y[/tex]
And so we find:
[tex]a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281[/tex]
And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:
[tex]a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418[/tex]
