35. (a) From the figure, we see that x0= –2.0 m. From Table 2-1, we can apply x– x0= v0t+ 12at2with t= 1.0 s, and then again with t= 2.0 s. This yields two equations for the two unknowns, v0and a:()()()202010.02.0 m1.0 s1.0 s216.0 m2.0 m2.0 s2.0 s .2va−−=+=+Solving these simultaneous equations yields the results
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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.