[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"$fiQbfxzi4hzK00oCkvOm60rVKFBFfyfGJBlg7BAM0PKY":3},{"answer":4,"createTime":5,"id":6,"options":7,"origin":12,"question":19,"related":20,"source":30,"type":31},[],"2025-03-11 19:06:25",178880414,[8,9,10,11],"20m\u002Fs","40m\u002Fs","100 \\, \\text{m\u002Fs}100m\u002Fs","140m\u002Fs",{"count":13,"courseId":14,"courseImg":15,"courseName":16,"workId":17,"workName":18},25,"d5c7803fa26afa603a67cdf22c563c28","https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F9c1e48361b00f3ee2086f4e259ed792b.jpg","大学物理(一)","e998b0b7f1a94273bfecfad13124fac7","","两列高铁相向而行,A车速度 60 \\, \\text{m\u002Fs}60m\u002Fs,B车速度 40 \\, \\text{m\u002Fs}40m\u002Fs,则A车上乘客观测B车的速度为",[21,32,41,50,59,68,71,80,89,99],{"answer":22,"createTime":5,"id":23,"options":24,"question":29,"source":30,"type":31},[],178880409,[25,26,27,28],"3 m\u002Fs","4 m\u002Fs","5 m\u002Fs","7 m\u002Fs","某无人机运动方程为 r(t) = (2t^2 +5)\\mathbf{i} + (3t -4)\\mathbf{j}r(t)=(2t2+5)i+(3t−4)j(单位:米),求 t=1t=1 秒时的速率","v1",0,{"answer":33,"createTime":5,"id":34,"options":35,"question":40,"source":30,"type":31},[],178880410,[36,37,38,39],"\u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F97f301f7f7a7a614401b669587f9fe5b.png\">","\u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002Fda7d0dadf3dec7098b779ec251136b2a.png\">","\u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F14c1dad17e3ddec3083251a60dee695c.png\">","\u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F0708f2371bea493124a3ffa2c6bc0e55.png\">","一个质点沿直线运动,其速度随时间的变化规律为 \u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F8fdf00f4891122ccb09aa11ba00db2b5.png\">,求该质点在 \u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002F4259ee039b5284314da3b9eefa01ae7d.png\"> 时刻的瞬时加速度",{"answer":42,"createTime":5,"id":43,"options":44,"question":49,"source":30,"type":31},[],178880411,[45,46,47,48],"研究探测器太阳能帆板的展开角度","计算探测器轨道周期与半长轴的关系","分析探测器着陆时与火星表面的相互作用","调整探测器姿态控制系统的工作状态","我国"天问一号"火星探测器在环绕火星的椭圆轨道运行时,若要分析其运动轨迹,以下哪种情况可将探测器视为质点",{"answer":51,"createTime":5,"id":52,"options":53,"question":58,"source":30,"type":31},[],178880412,[54,55,56,57],"成立,因为速度方向由轨迹切线决定","不成立,加速度方向才指向轨迹凸侧","成立,但仅适用于直线运动","不成立,速度方向可能偏离切线","某同学认为:"瞬时速度的方向一定与运动轨迹切线方向一致."这一说法是否成立",{"answer":60,"createTime":5,"id":61,"options":62,"question":67,"source":30,"type":31},[],178880413,[63,64,65,66],"位移,时间","路程,时间","位移,时刻","路程,时刻","李白的诗句"朝辞白帝彩云间,千里江陵一日还"中,"千里"和"一日"分别描述的是",{"answer":69,"createTime":5,"id":6,"options":70,"question":19,"source":30,"type":31},[],[8,9,10,11],{"answer":72,"createTime":5,"id":73,"options":74,"question":79,"source":30,"type":31},[],178880415,[75,76,77,78],"24 \\, \\text{rad\u002Fs}^2 24rad\u002Fs2","48 \\, \\text{rad\u002Fs}^2 48rad\u002Fs2","1 2 rad\u002Fs2","6 rad\u002Fs2","一质点作圆周运动,其角坐标方程为 \\theta(t) = 4t^3 + t \\, \\text{(rad)}θ(t)=4t3+t(rad),求 t = 2 \\, \\text{s}t=2s 时的角加速度",{"answer":81,"createTime":5,"id":82,"options":83,"question":88,"source":30,"type":31},[],178880416,[84,85,86,87],"增大","减小","不变","无法确定","台风中心移动路径的曲率半径突然减小,此时台风边缘某点的法向加速度会",{"answer":90,"createTime":5,"id":91,"options":92,"question":97,"source":30,"type":98},[],178880417,[93,94,95,96],"位置矢量","瞬时加速度","运动轨迹方程","速率分布函数","研究新冠病毒气溶胶在空气中的扩散时,以下哪些物理量可能被用到",1,{"answer":100,"createTime":5,"id":101,"options":102,"question":107,"source":30,"type":98},[],178880418,[103,104,105,106],"加速度大小为\u003Cimg src=\"https:\u002F\u002Ftihai-oss-cloud.itihey.com\u002Fimg\u002Fc46fcf68d563ad9c4ddeefbf141f8b9e.png\">","运动轨迹为抛物线","t=2s 时速率为5m\u002Fs","位移方向与速度方向始终一致","为降低碳排放,某城市推广电动公交车.若某公交车以 \\mathbf{v}(t) = (2t)\\mathbf{i} + 3\\mathbf{j} \\, \\text{(m\u002Fs)}v(t)=(2t)i+3j(m\u002Fs) 运动,则"]