9. Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as xvt at=+0122(where x00=). We choose to analyze the third and fifth points, obtaining 202010.2m(1.0 s)(1.0 s)210.8m(2.0 s)(2.0 s)2vaSimultaneous solution of the equations leads to 00v=and a=040.ms2. We now have two ways to finish the problem. One is to compute force from
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.