5_Dynamics 11ed Manual - We check both possibilities. a max...

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Engineering Mechanics - Dynamics Chapter 12 v 0 60 ft s = b 1 s = t 1 3s = Solution: vt () v 0 1 e b t () = at () t vt () d d = d t () 0 t t vt () d = d t 1 () 123.0 ft = at 1 () 2.99 ft s 2 = Problem 12-7 The position of a particle along a straight line is given by s p = at 3 + bt 2 + ct . Determine its maximum acceleration and maximum velocity during the time interval t 0 t t f . Given: a 1 ft s 3 = b 9 ft s 2 = c 15 ft s = t 0 0s = t f 10 s = Solution: s p at 3 bt 2 + ct + = v p t s p d d = 3 at 2 2 bt + c + = a p t v p d d = 2 t s p d d 2 = 6 at 2 b + = Since the acceleration is linear in time then the maximum will occur at the start or at the end.
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Unformatted text preview: We check both possibilities. a max max 6 at b + 6 at f 2 b + , ( ) = a max 42 ft s 2 = The maximum velocity can occur at the beginning, at the end, or where the acceleration is zero. We will check all three locations. t cr b 3 a = t cr 3 s = 3 Given:...
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This note was uploaded on 08/16/2011 for the course EGN 3321 taught by Professor Christianfeldt during the Spring '08 term at University of Central Florida.

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