128_pdfsam_math 54 differential equation solutions odd

128_pdfsam_math 54 differential equation solutions odd - ....

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 3 With given data, the force due to gravity is F 1 = mg =5 g and the air resistance force is F 2 = 50 v . Therefore, the velocity v ( t )sat isFes m dv dt = F 1 + F 2 =5 g 50 v dv dt = g 10 v, v (0) = 0 . Separating variables yields dv 10 v g = dt 1 10 ln | 10 v g | = t + C 1 v ( t )= g 10 + Ce 10 t . Substituting the initial condition, v (0) = 0, we get C = g/ 10, and so v ( t )= g 10 ( 1 e 10 t ) . Integrating this equation yields x ( t )= Z v ( t ) dt = Z g 10 ( 1 e 10 t ) dt = g 10 ± t + 1 10 e 10 t ² + C, and we Fnd C using the initial condition x (0) = 0: 0= g 10 ± 0+ 1 10 e 10(0) ² + C C = g 100 x ( t )= g 10 t + g 100 ( e 10 t 1 ) =(0 . 981) t +(0 . 0981) e 10 t 0 . 0981 (m) . When the object hits the ground, x ( t ) = 1000 m. Thus we solve
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: . 981) t + (0 . 0981) e 10 t . 0981 = 1000 , which gives ( t is nonnegative!) t 1019 . 468 1019 sec. 3. or this problem, m = 500 kg, v = 0, g = 9 . 81 m/sec 2 , and b = 50 kg/sec. We also see that the object has 1000 m to fall before it hits the ground. Plugging these variables into equation (6) on page 111 of the text gives the equation x ( t ) = (500)(9 . 81) 50 t + 500 50 (500)(9 . 81) 50 ( 1 e 50 t/ 500 ) 124...
View Full Document

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.

Ask a homework question - tutors are online