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Pre-Calc Homework Solutions 239

Pre-Calc Homework Solutions 239 - Section 6.1 d 2s dt 2 d...

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54. We first solve } d dt 2 2 s } 5 2 k with the initial conditions s 9 (0) 5 44 and s (0) 5 0. E } d dt 2 2 s } dt 5 2 k } d d s t } 5 2 kt 1 C 1 Initial condition: s 9 (0) 5 44 44 5 ( 2 k )(0) 1 C 1 44 5 C 1 Velocity: } d d s t } 5 2 kt 1 44 E } d d s t } dt 5 E ( 2 kt 1 44) dt s 5 2 } 2 k } t 2 1 44 t 1 C 2 Initial condition: s (0) 5 0 0 5 2 } 2 k } (0) 2 1 44(0) 1 C 2 0 5 C 2 Position: s 5 2 } 2 k } t 2 1 44 t Now, } d d s t } 5 2 kt 1 44 5 0 when t 5 } 4 k 4 } , so it takes } 4 k 4 } seconds to stop, and we require: s 1 } 4 k 4 } 2 5 45 2 } 2 k } 1 } 4 k 4 } 2 2 1 44 1 } 4 k 4 } 2 5 45 } 96 k 8 } 5 45 k 5 } 9 4 6 5 8 } < 21.5 It requires a constant deceleration of approximately 21.5 ft/sec 2 . 55. E } d dt 2 2 s } dt 5 E 2 5.2 dt } d d s t } 5 2 5.2 t 1 C 1 Initial condition: } d d s t } 5 0 when t 5 0 0 5 2 5.2(0) 1 C 1 0 5 C 1 Velocity: } d d s t } 5 2 5.2 t E } d d s t } dt 5 E 2 5.2 dt s 5 2 2.6 t 2 1 C 2 Initial condition: s 5 4 when t 5 0 4 5 2 2.6(0) 2 1 C 2 4 5 C 2 Position: s ( t ) 5 2 2.6 t 2 1 4 Solving s ( t ) 5 0, we have t 2 5 } 2 4 .6 } , so the positive solution is t < 1.240 sec. They took about 1.240 sec to fall. 56. Solving } d dt 2 2 s } 5 a , s (0) 5 s 0 , and v (0) 5 v 0 : E } d dt 2 2 s } dt 5 E a dt } d d s t } 5 at 1 C 1 Initial condition: s 9 (0) 5 v 0 v 0 5 ( a )(0) 1 C 1 v 0 5 C 1 Velocity: } d d s t } 5 at 1 v 0 E } d d s t } dt 5 E ( at 1 v 0 ) dt s 5 } a 2 } t 2 1 v 0 t 1 C 2 Initial condition: s (0) 5 s 0 s 0 5 } a 2 } (0) 2 1 ( v 0 )(0) 1 C 2 s 0 5 C 2 Position: s 5 } a 2 } t 2 1 v 0 t 1 s 0 57. We use the method of Example 7. V 5 } 1 3 } p r 2 h 5 } 1 3 } p 1 } 2 5 } h 2 2 h 5 } 4 7 p 5 } h 3 } d d V t } 5 } d d t } 1 } 4 7 p 5 } h 3 2 2 } 1 6 } ˇ h w 5 } 4 2 p 5 } h 2 } d d h
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