{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

132_pdfsam_math 54 differential equation solutions odd

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

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

Chapter 3 With the chute open, the parachutist falls 1946 . 38 m. It takes t seconds, where t satisfies x 2 ( t ) = 1946 . 38. Solving yields 5 g 6 t + 5 6 20 5 g 6 ( 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 . If we let 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 . 1) v . Solving this equation by separation of variables, we have
This is the end of the preview. Sign up to access the rest of the document.

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 ² ≈ 13 . 2 sec . 128...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online