132_pdfsam_math 54 differential equation solutions odd

132_pdfsam_math 54 differential equation solutions odd - v...

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Chapter 3 With the chute open, the parachutist falls 1946 . 38 m. It takes t seconds, where t satisFes x 2 ( t ) = 1946 . 38. Solving yields 5 g 6 t + 5 6 ± 20 5 g 6 ² ( 1 e 6 t / 5 ) = 1946 . 38 t 236 . 884 (sec) . Therefore, the parachutist will hit the ground after t + t 241 . 1 seconds. 9. This problem is similar to Example 1 on page 110 of the text with the addition of a buoyancy force of magnitude (1 / 40) mg .I fw el e t x ( t ) be the distance below the water at time t and v ( t ) the velocity, then the total force acting on the object is F = mg bv 1 40 mg. We are given m = 100 kg, g =9 . 81 m/sec 2 ,and b = 10 kg/sec. Applying Newton’s Second Law gives 100 dv dt = (100)(9 . 81) 10 v 10 4 (9 . 81) dv dt =9 . 56 (0
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Unformatted text preview: v ( t ) = 95 . 65 + Ce t/ 10 . Since v (0) = 0, we Fnd C = 95 . 65 and, hence, v ( t ) = 95 . 65 95 . 65 e t/ 10 . Integrating yields x ( t ) = 95 . 65 t 956 . 5 e t/ 10 + C 1 . Using the fact that x (0) = 0, we Fnd C 1 = 956 . 5. Therefore, the equation of motion of the object is x ( t ) = 95 . 65 t 956 . 5 e t/ 10 956 . 5 . To determine when the object is traveling at the velocity of 70 m/sec, we solve v ( t ) = 70. That is, 70 = 95 . 65 95 . 65 e t/ 10 = 95 . 65 ( 1 e t/ 10 ) t = 10 ln 1 70 95 . 65 13 . 2 sec . 128...
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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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