t=1+12+t3=4.5s,故C错误,D正确。9. AD【解析】墨盒在水平方向受到的摩擦力F=Amng,由牛顿第二定律得墨盒的加速度大小为Fm=4ms2,墨盒从左端滑上传送带后先做匀减速运动,设速度减为零时的位移是x,则一2ax=0-v,代入数据解得x=8m<12m,所以墨盒速度减为零后反向做匀加速运动,最终从传送带的最左端离开,A正确:假设墨盒反向做匀地速动后能与传送带达到共速,此过程的位移为+2mk8m,以墨盒后传透带可达到共速,之后向左匀速运动到传送带的最左端,想盒离传送带瞬间的速度大小为v=4ms,B错误:墨盒向右减速的时间为1=2∠2,墨盒向左加速的时间为,==1s,墨盒向左匀速运动的时间为3=1.5,所以墨盒在传送带上运动的总时间为

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