problem02_22 solution

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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2.22: a) The acceleration is found from Eq. (2.13), which x v 0 = 0; ( 29 ( 29 ( 29 ( 29 , s m 0 . 32 ) ft 307 ( 2 ) hr mi 173 ( ) ( 2 2 ft 3.281 m 1 2 hr mi 1 s m 4470 . 0 0 2 = = - = x x v a x x where the conversions are from Appendix E. b) The time can be found from the above acceleration, ( 29 . s 42 . 2 s m 0 . 32 ) hr mi 173 ( 2 hr mi 1 s m 4470 . 0 = = = x x a v t The intermediate calculation may be avoided by using Eq. (2.14), again with
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Unformatted text preview: v x = 0, ( 29 ( 29 ( 29 . s 42 . 2 ) hr mi 173 ( ft 307 ( 2 ) ( 2 hr mi 1 s m 4470 . ft 3.281 m 1 = =-= x v x x t...
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