Pre-Calc Homework Solutions 413

Pre-Calc Homework Solutions 413 - Section 10.3 35. (a)...

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35. (a) Referring to the figure, look at the circular arc from the point where t 5 0 to the point “ m ”. On one hand, this arc has length given by r 0 u , but it also has length given by vt . Setting those two quantities equal gives the result. (b) v ( t ) 5 1 2 v sin } r v 0 t } 2 i 1 1 v cos } r v 0 t } 2 j , and a ( t ) 5 1 2} v r 0 2 } cos } r v 0 t } 2 i 1 1 2} v r 0 2 } sin } r v 0 t } 2 j 52} v r 0 2 } 31 cos } r v 0 t } 2 i 1 1 sin } r v 0 t } 2 j 4 (c) From part (b) above, a ( t ) 52 1 } r v 0 } 2 2 r ( t ). So, by Newton’s second law, F 52 m 1 } r v 0 } 2 2 r . Substituting for F in the law of gravitation gives the result. (d) Set } v r T 0 } 5 2 p and solve for vT . (e) Substitute } 2 p T r 0 } for v in v 2 5 } G r M 0 } and solve for T 2 : 1 } 2 p T r 0 } 2 2 5 } G r M 0 } } 4 p T 2 2 r 0 2 } 5 } G r M 0 } } T 1 2 } 5 } 4 p G 2 M r 0 3 } T 2 5 } 4 G p M 2 } r 0 3 36. Solve both equations for t: t 5 e x 2 1 and t 5 ˇ y w 1 w 1 w . Now eliminate the
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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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