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p0416 - y = y o when t = 0 we have C = y o − x o − U τ...

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386 CHAPTER 4. KINEMATICS 4.16 Chapter 4, Problem 16 By definition, the velocity components in the Lagrangian description are u = X x t ~ x o ,y o = U and v = X y t ~ x o ,y o = x t + τ Hence, we can solve for x and y from the following differential equations: dx = Udt and dy = x t + τ dt Integrating the equation for x ,wef ind x = x o + Ut whereweassume x = x o at t =0 . Substituting for x , the equation for y becomes dy = x o + Ut t + τ dt = x o U τ + U ( t + τ ) t + τ dt = w x o U τ t + τ + U W dt Therefore, integrating once gives y =( x o U τ ) f n ( t + τ )+ Ut + C where C
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Unformatted text preview: y = y o when t = 0 , we have C = y o − ( x o − U τ ) f n τ Hence, the solution for y is y = y o + Ut + ( x o − U τ ) f n w t + τ τ W Thus, the particle location is given by x = x o + Ut and y = y o + Ut + ( x o − U τ ) f n w t + τ τ W Therefore, the Lagrangian description of this flow is r = [ x o + Ut ] i + } y o + Ut + ( x o − U τ ) f n w t + τ τ W] j...
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