135_pdfsam_math 54 differential equation solutions odd

135_pdfsam_math 54 differential equation solutions odd -...

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Exercises 3.4 Using equation (3.11) and noting that dv/dt =( dv/dx )( dx/dt )=( dv/dx ) v , we can determine the maximum height attained by the shell. With the above substitution, equation (3.11) becomes mv dv dx = ( mg +0 . 1 v 2 ) ,v (0) = 500 . Using separation of variables and integration, we get vdv 10 mg + v 2 = dx 10 m 1 2 ln ( 10 mg + v 2 ) = x 10 m + C 10 mg + v 2 = Ke x/ (5 m ) . Setting v = 500 when x = 0, we Fnd K = e 0 ( 10(3)(9 . 81) + (500) 2 ) = 250294 . 3 . Thus the equation of velocity as a function of distance is v 2 +10 mg = (250294 . 3) e x/ (5 m ) . The maximum height will occur when the shell’s velocity is zero, therefore x max is x max = 5(3) ln ± 0 + 10(3)(9 . 81) 250294 . 3 ² 101 . 19 (meters) . 15. The total torque exerted on the flywheel is the sum of the torque exerted by the motor and
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.

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