# 131&amp;133 - 30 sin 40 a 30 cos 40 a 40 2 n 2 t 2 s m...

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TAM 212 Spring, 2004 1.131 Given: 3 ct s = , c = 1ft/s 3 . At t = 2s, 2 2 n 2 t / 15 a a s ft = + = a Required: The radius of curvature at that time. Solution: We are given that at t = 2s, 2 2 n 2 t / 15 a a s ft = + = a , so ( 29 2 2 2 n 2 t / 225 a a s ft = + where s a t = and r 2 s a t = From 3 ct s = , 3 2 2 / 3 3 s ft t ct s = = , and ( 29 3 / 6 6 s ft t ct s = = . At t = 2s, s ft s / 12 = and 2 / 12 s ft s = Putting these together: ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 2 2 2 2 2 / 9 / / 12 / 225 / / 12 / 2 1 s ft s ft s ft s ft s ft = = + r r ( 29 ft s t 16 2 = = r 1.133 Given: (translating from the figure) At a certain instant: s m t / 2 e v = , and ( 29 ( 29
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Unformatted text preview: 30 sin / 40 a ; 30 cos / 40 a ; / 40 2 n 2 t 2 s m s m s m a =-= = Problem(a): Find (at this instant) dt d v Solution: s & = v , so s dt d & & = v ; 2 / ) 30 cos( * 40 s m a s t-= = & & Problem(b): Find (at this instant) ρ Solution: ρ = 2 s a n & , so ( 29 2 2 2 / ) 30 (sin 40 / 2 s m s m a s n = = ρ & 2 / 6 . 34 s m s-= & & m 2 . = ρ...
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