130_pdfsam_math 54 differential equation solutions odd

130_pdfsam_math 54 differential equation solutions odd -...

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Chapter 3 5 dv dt = F 1 + F 2 =5 g 10 v dv dt = g 2 v, v (0) = 50 . Solving this iniial value problem yields v ( t )= g 2 + Ce 2 t ; 50 = v (0) = g 2 + Ce 2(0) C = 100 g 2 v ( t )= g 2 + 100 g 2 e 2 t . We now integrate v ( t ) to obtain the equation of the motion of the object: x ( t )= Z v ( t ) dt = Z ± g 2 + 100 g 2 e 2 t ² dt = g 2 t 100 g 4 e 2 t + C, where C is such that x (0) = 0. Computing 0= x (0) = g 2 (0) 100 g 4 e 2(0) + C C = 100 g 4 , we answer the Frst question in this problem, that is, x ( t )= g 2 t 100 g 4 e 2 t + 100 g 4 4 . 905 t +22 . 5475 22 . 5475 e 2 t . Answering the second question, we solve the equation x ( t ) = 500 to Fnd time t when the object passes 500 m, and so strikes the ground. 4 . 905 t +22 . 5475 22 . 5475 e 2 t = 500 t 97 . 34 (sec) . 7.
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Unformatted text preview: Since the air resistance force has di±erent coeﬃcients of proportionality for closed and for opened chute, we need two di±erential equations describing the motion. Let x 1 ( t ), x 1 (0) = 0, denote the distance the parachutist has fallen in t seconds, and let v 1 ( t ) = dx/dt denote her velocity. With m = 75, b = b 1 = 30 N-sec/m, and v = 0 the initial value problem (4) on page 111 of the text becomes 75 dv 1 dt = 75 g − 30 v 1 ⇒ dv 1 dt + 2 5 v 1 = g, v 1 (0) = 0 . 126...
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