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128_pdfsam_math 54 differential equation solutions odd

128_pdfsam_math 54 differential equation solutions odd -...

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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 ) satisfies 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 ) = v ( t ) dt = g 10 ( 1 e 10 t ) dt = g 10 t + 1 10 e 10 t + C, and we find C using the initial condition x (0) = 0: 0 = g 10 0 + 1 10 e 10(0) + 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
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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 ² ( 1 − e − 50 t/ 500 ) 124...
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