The velocity along a stream line passing through the origin is given by v=2V=2√x^2+y^2 what is the velocity and acceleration at (4;3)
velocity v(4;3)=2⋅42+32=10 (m/s)v(4;3)=2\cdot\sqrt{4^2+3^2}=10\ (m/s)v(4;3)=2⋅42+32=10 (m/s)
acceleration
ax=∂u∂t+u∂u∂x+v∂u∂y=0+2x⋅2+0=4xa_x=\frac{\partial u}{\partial t }+u\frac{\partial u}{\partial x }+v\frac{\partial u}{\partial y }=0+2x\cdot2+0=4xax=∂t∂u+u∂x∂u+v∂y∂u=0+2x⋅2+0=4x
ay=∂v∂t+u∂v∂x+v∂v∂y=0+0+2y⋅2=4ya_y=\frac{\partial v}{\partial t }+u\frac{\partial v}{\partial x }+v\frac{\partial v}{\partial y }=0+0+2y\cdot2=4yay=∂t∂v+u∂x∂v+v∂y∂v=0+0+2y⋅2=4y
a(4;3)=(4x)2+(4y)2=4x2+y2=442+32=20 (m/s2)a(4;3)=\sqrt{(4x)^2+(4y)^2}=4\sqrt{x^2+y^2}=4\sqrt{4^2+3^2}=20\ (m/s^2)a(4;3)=(4x)2+(4y)2=4x2+y2=442+32=20 (m/s2)
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