Question #269512

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)

1
Expert's answer
2021-11-21T17:31:40-0500


velocity v(4;3)=242+32=10 (m/s)v(4;3)=2\cdot\sqrt{4^2+3^2}=10\ (m/s)


acceleration


ax=ut+uux+vuy=0+2x2+0=4xa_x=\frac{\partial u}{\partial t }+u\frac{\partial u}{\partial x }+v\frac{\partial u}{\partial y }=0+2x\cdot2+0=4x


ay=vt+uvx+vvy=0+0+2y2=4ya_y=\frac{\partial v}{\partial t }+u\frac{\partial v}{\partial x }+v\frac{\partial v}{\partial y }=0+0+2y\cdot2=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)











Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

Charles
22.11.21, 07:55

Thank you.

LATEST TUTORIALS
APPROVED BY CLIENTS