mPtvvfi=−=×−=×22 15 10360251021 102266bgchbgbg..Wsmskg.(b) With tarbitrary, we use Ptm vvi=−12to solve for the speed v= v(t) as a function of time and obtain vtvPtmttib gbgbg=+=+××226621021510100 15...in SI units (vin m/s and tin s). (c) The force F(t) as a function of time is FtPtb gb g==×+15 10100 156..in SI units (Fin N and tin s). (d) The distance dthe train moved is given by 3601/23/23603000343( )1001006.7 10 m.
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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.