Pre-Calc Homework Solutions 239

Pre-Calc Homework - Section 6.1 d 2s dt 2 d 2s dt 2 239 54 We first solve s(0 2 k with the initial conditions 0 56 Solving d s dt dt 2 ds at dt 2 a

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Unformatted text preview: Section 6.1 d 2s dt 2 d 2s dt 2 239 54. We first solve s (0) 2 k with the initial conditions 0. 56. Solving d s dt dt 2 ds at dt 2 a, s(0) a dt s0, and v(0) v0: 44 and s(0) k C1 d s dt dt 2 ds kt dt C1 v0 Initial condition: s (0) 44 v0 v0 (a)(0) C1 ds dt Initial condition: s (0) 44 44 ( k)(0) C1 ds dt C1 C1 Velocity: kt 44 ds dt dt a 2 s t 2 at v0 Velocity: ds dt dt (at v0t v0) dt C2 s0 C2 ( kt k 2 t 2 k 2 (0) 2 44) dt C2 0 C2 s 44t Initial condition: s(0) s0 s0 a 2 (0) 2 Initial condition: s(0) 0 0 44(0) (v0 )(0) C2 a 2 t 2 C2 k 2 t 2 Position: s 44t 0 when t 44 , so it takes k v0t s0 Position: s Now, ds dt 57. We use the method of Example 7. V 1 2 r h 3 dV dt 1 h 6 25 24 25 dt 24 25 t 24 1 2 2 h h 3 5 d 4 3 h dt 75 4 2 dh h 25 dt dh h 3/2 dt 3/2 dh h dt dt 2 5/2 h C 5 4 3 h 75 kt 44 44 seconds to stop, and we require: k s k 44 2 2 k 44 k 44 k 968 k 45 45 45 968 45 44 k 21.5 It requires a constant deceleration of approximately 21.5 ft/sec 2. 55. d 2s dt dt 2 ds 5.2t dt Initial condition: h 25 (0) 24 10 when t 0. 5.2 dt C1 ds dt C 0 when t 0 25 t 24 125t 48 Initial condition: 0 0 5.2(0) C1 ds dt 2 2 (10)5/2 C 5 2 (10)5/2 5 2 5/2 2 h (10)5/2 5 5 C1 h 5/2 10 5/2 105/2 105/2 2/5 2/5 h5/2 5.2t 5.2 dt h Velocity: ds dt dt 125t 48 125t 48 The height is given by h volume is given by 4 when t 0 V 4 3 h 75 4 75 125t 48 s 2.6t C2 C2 2.6t 2 4 4 , so the positive 2.6 125t 48 105/2 and the Initial condition: s 4 4 2.6(0)2 C2 105/2 6/5 . Position: s(t) Solving s(t) solution is t 0, we have t 2 1.240 sec. They took about 1.240 sec to fall. ...
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Spring '08 term at University of Florida.

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