ch02-p081 - 81. The problem consists of two...

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81. The problem consists of two constant-acceleration parts: part 1 with v 0 = 0, v = 6.0 m/s, x = 1.8 m, and x 0 = 0 (if we take its original position to be the coordinate origin); and, part 2 with v 0 = 6.0 m/s, v = 0, and a 2 = –2.5 m/s 2 (negative because we are taking the positive direction to be the direction of motion). (a) We can use Eq. 2-17 to find the time for the first part x x 0 = 1 2 ( v 0 + v ) t 1 ¡ 1.8 m – 0 = 1 2 (0 + 6.0 m/s) t 1 so that t 1 = 0.6 s. And Eq. 2-11 is used to obtain the time for the second part 02 2 vv a t =+ ¡ 0 = 6.0 m/s + (–2.5 m/s
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This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.

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