26 Motion in One DimensionSection 2.3AccelerationP2.11Choose the positive direction to be the outward direction, perpendicular to the wall.vvatfi=+: avt==−−×=×−∆∆22 025 0350 10134 1034....mss±ms2af.P2.12(a)Acceleration is constant over the first ten seconds, so at the end,at=+=+()=02 0010 020 0.s2ch.Then a=0 so vis constant from t=10 0. s to t=15 0. s. And over the last five seconds thevelocity changes toat=+=+()=20 03 005 005 00..s2.(b)In the first ten seconds,xxvtatfii=+ +=++()=12001220010022sm2.Over the next five seconds the position changes tovtat=+()+=125000 2002m mssm.And at t=20 0. s,vtat=+()+−()=12123 005 0026222mm ssm ssm2..afch.*P2.13(a)The average speed during a time interval ∆tis vt=distance traveled
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .