81. The problem consists of two constant-acceleration parts: part 1 with v0= 0, v= 6.0 m/s, x= 1.8 m, and x0= 0 (if we take its original position to be the coordinate origin); and, part 2 with v0= 6.0 m/s, v= 0, and a2= –2.5 m/s2(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– x0= 12(v0+ v)t1¡1.8 m – 0 = 12(0 + 6.0 m/s) t1so that t1= 0.6 s. And Eq. 2-11 is used to obtain the time for the second part 022vv at=+¡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.