section 5.9 solutions

A m 2 kl32 xm 3 l 3 5 b m 1 kl3 xm 1 l 3 4 29

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Unformatted text preview: g = 1 t2 − t 1 t2 t1 at dt = v ( t1 ) + v ( t 2 ) at1 + at2 = 2 2 √ 4 7+2 =√ 3 7−3 √ 7−1 27. (a) M = 2 kL3/2 , xM = 3 L 3 5 (b) M = 1 kL3 , xM = 1 L 3 4 29. xM2 = (2M − M1 )L/8M2 31. x = 2M ± kL2 2kL f (x) dx = 0. ANSWERS TO ODD-NUMBERED EXERCISES A-49 33. see answer to Exercise 31, Section 5.3 35. If f and g take on the same average on every interval [a, x], then x 1 x−a f (t )dt = a 1 x−a x g (t ) dt . a Multiplication by (x − a) gives x x f (t ) dt = a g (t ) dt . a Differentiation with respect to x gives f (x) = g (x). This shows...
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This document was uploaded on 11/11/2013.

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