Answer to Question #269512 in Physics for Charles

Question

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)

Answer 1

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

acceleration

$a_x=\frac{\partial u}{\partial t }+u\frac{\partial u}{\partial x }+v\frac{\partial u}{\partial y }=0+2x\cdot2+0=4x$

$a_y=\frac{\partial v}{\partial t }+u\frac{\partial v}{\partial x }+v\frac{\partial v}{\partial y }=0+0+2y\cdot2=4y$

$a(4;3)=\sqrt{(4x)^2+(4y)^2}=4\sqrt{x^2+y^2}=4\sqrt{4^2+3^2}=20\ (m/s^2)$