πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一质点自原点开始沿抛物线=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动,它在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))轴上的分速度为一恒量,其值为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),求质点位于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))处的速度和加速度。题解:因vx = =dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为一常数,故ax = 0。当t = 0时,x = 0,由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (1)又由质点的抛物线方程,有=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (2)由y方向的运动方程可得该方向的速度和加速度分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (3)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)) (4)当质点位于x = m时,由上述各式可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:质点在Oxy平面内运动,其运动方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。求:(1)质点的轨迹方程;(2)在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时间内的平均速度;(2)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时的速度及切向和法向加速度。题解:(1)由参数方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去t得质点的轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))s到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))s时间内的平均速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)质点在任意时刻的速度和加速度分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))则t1 = s时的速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))切向和法向加速度分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:质点的运动方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),试求:(1)初速度的大小和方向;(2)加速度的大小和方向。题解:(1)速度分量式为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))当t = 0时,vx= 10=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),vy = 15=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),则初速度大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))设v与x轴的夹角为,则=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)加速度的分量式为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))则加速度的大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))设a与x轴的夹角为,则=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一质点具有恒定加速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,其速度为零,位置矢量=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。求:(1)在任意时刻的速度和位置矢量;(2)质点在=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))平面上的轨迹方程,并画出轨迹的示意图。题解:由加速度定义式,根据初始条件t0 = 0时v0 = 0,积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))又由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))及初始条件t = 0时,r = (10 m)i,积分可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))由上述结果可得质点运动方程的分量式,即=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去参数t,可得运动的轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))这是一个直线方程,直线斜率=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。轨迹如图所示。题:飞机以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的速度沿水平直线飞行,在离地面高为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,驾驶员要把物品投到前方某一地面目标处。问:(1)此时目标在飞机下方前多远(2)投放物品时,驾驶员看目标的视线和水平线成何角度(3)物品投出=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))后,它的法向加速度和切向加速度各为多少?题解:(1)取如图所示的坐标,物品下落时在水平和竖直方向的运动方程分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))飞机水平飞行速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),飞机离地面的高度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))m,由上述两式可得目标在飞机正下方前的距离=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)视线和水平线的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)在任意时刻物品的速度与水平轴的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))取自然坐标,物品在抛出2 s时,重力加速度的切向分量与法向分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:一足球运动员在正对球门前=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))处以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的初速率罚任意球,已知球门高为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。若要在垂直于球门的竖直平面内将足球直接踢进球门,问他应在与地面成什么角度的范围内踢出足球(足球可视为质点)题解:取图示坐标系Oxy,由运动方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去t得轨迹方程=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))以x = m,v = =dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))及  y  0代入后,可解得º   ºº   º如何理解上述角度得范围?在初速度一定的条件下,球击中球门底线或球门上缘都将对应有两个不同的投射倾角(如图所示)。如果以或<踢出足球,都将因射程不足而不能直接射入球门;由于球门高度的限制,角也并非能取与之间的任何值。当倾角取值为 < < 时,踢出的足球将越过门缘而离去,这时也球不能射入球门。因此可取的角度范围只能是解中的结果。题:设从某一点=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))以同样的速率,沿着同一竖直面内各个不同方向同时抛出几个物体。试证:在任意时刻,这几个物体总是散落在某个圆周上。题证:取物体抛出点为坐标原点,建立如图所示的坐标系。物体运动的参数方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))消去式中参数,得任意时刻的轨迹方程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))这是一个以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为圆心、vt为半径的圆方程(如图所示),它代表着所有物体在任意时刻t的位置。题:一质点在半径为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的圆周上以恒定的速率运动,质点由位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动到位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))所对的圆心角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。(1)试证位置=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))之间的平均加速度为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1));(2)当=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))分别等于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,平均加速度各为多少?并对结果加以讨论。=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题解:(1)由图可看到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),故=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))而=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))所以=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)将=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))分别代入上式,得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))上述结果表明:当=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,匀速率圆周运动的平均加速度趋于一极限值,该值即为法向加速度=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。题:一质点沿半径为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))的圆周按规律=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))运动,=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))、=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))都是常量。(1)求=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时刻的总加速度;(2)=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))为何值时总加速度在数值上等于=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)当加速度达到=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))时,质点已沿圆周运行了多少圈?题解:(1)质点作圆周运动的速率为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))其加速度的切向分量和法向分量分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))故加速度的大小为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))其方向与切线之间的夹角为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(2)要使=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))由=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))可得=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))(3)从t = 0开始到t = v/b时,质点经过的路程为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))因此质点运行的圈数为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))题:碟盘是一张表面覆盖一层信息记录物质的塑性圆片。若碟盘可读部分的内外半径分别为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))和=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))。在回放时,碟盘被以恒定的线速度由内向外沿螺旋扫描线(阿基米德螺线)进行扫描。(1)若开始时读写碟盘的角速度为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),则读完时的角速度为多少(2)若螺旋线的间距为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),求扫描线的总长度和回放时间。题分析:阿基米德螺线是一等速的螺旋线,在极坐标下,它的参数方程可表示为=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1)),式中r为极径,r为初始极径,为极角,a为常量。它的图线是等间距的,当间距为d时,常量a = d/2。因此,扫描线的总长度可通过积分=dfrac (dy)(dt)=Rtdfrac (2pi )(T)cos dfrac (2pi )(T)ti+ndfrac (2pi )(T)sin dfrac (2pi )(T)t} =(0.3pi rmcdot (s)^-1). πm·s^(-1))得到。

s时的速度和加速度分别为

题:一质点自原点开始沿抛物线运动,它在轴上的分速度为一恒量,其值为,求质点位于处的速度和加速度。

题解:因vx = 为一常数,故ax = 0。当t = 0时,x = 0,由积分可得

(1)

又由质点的抛物线方程,有

(2)

由y方向的运动方程可得该方向的速度和加速度分量分别为

(3)

(4)

当质点位于x = m时,由上述各式可得

题:质点在Oxy平面内运动,其运动方程为。求:(1)质点的轨迹方程;(2)在到时间内的平均速度;(2)时的速度及切向和法向加速度。

题解:(1)由参数方程

消去t得质点的轨迹方程

(2)在s到s时间内的平均速度

(3)质点在任意时刻的速度和加速度分别为

则t1 = s时的速度

切向和法向加速度分别为

题:质点的运动方程为和,试求:(1)初速度的大小和方向;(2)加速度的大小和方向。

题解:(1)速度分量式为

当t = 0时,vx= 10,vy = 15,则初速度大小为

设v与x轴的夹角为,则

(2)加速度的分量式为

则加速度的大小为

设a与x轴的夹角为,则

题:一质点具有恒定加速度,在时,其速度为零,位置矢量。求:(1)在任意时刻的速度和位置矢量;(2)质点在平面上的轨迹方程,并画出轨迹的示意图。

题解:由加速度定义式,根据初始条件t0 = 0时v0 = 0,积分可得

又由及初始条件t = 0时,r = (10 m)i,积分可得

由上述结果可得质点运动方程的分量式,即

消去参数t,可得运动的轨迹方程

这是一个直线方程,直线斜率,。轨迹如图所示。

题:飞机以的速度沿水平直线飞行,在离地面高为时,驾驶员要把物品投到前方某一地面目标处。问:(1)此时目标在飞机下方前多远(2)投放物品时,驾驶员看目标的视线和水平线成何角度(3)物品投出后,它的法向加速度和切向加速度各为多少?

题解:(1)取如图所示的坐标,物品下落时在水平和竖直方向的运动方程分别为

飞机水平飞行速度,飞机离地面的高度m,由上述两式可得目标在飞机正下方前的距离

(2)视线和水平线的夹角为

(3)在任意时刻物品的速度与水平轴的夹角为

取自然坐标,物品在抛出2 s时,重力加速度的切向分量与法向分量分别为

题:一足球运动员在正对球门前处以的初速率罚任意球,已知球门高为。若要在垂直于球门的竖直平面内将足球直接踢进球门,问他应在与地面成什么角度的范围内踢出足球(足球可视为质点)

题解:取图示坐标系Oxy,由运动方程

消去t得轨迹方程

以x = m,v = 及  y  0代入后,可解得

º   º

º   º

如何理解上述角度得范围?

在初速度一定的条件下,球击中球门底线或球门上缘都将对应有两个不同的投射倾角(如图所示)。如果以或<踢出足球,都将因射程不足而不能直接射入球门;由于球门高度的限制,角也并非能取与之间的任何值。当倾角取值为 < < 时,踢出的足球将越过门缘而离去,这时也球不能射入球门。因此可取的角度范围只能是解中的结果。

题:设从某一点以同样的速率,沿着同一竖直面内各个不同方向同时抛出几个物体。试证:在任意时刻,这几个物体总是散落在某个圆周上。

题证:取物体抛出点为坐标原点,建立如图所示的坐标系。物体运动的参数方程为

消去式中参数,得任意时刻的轨迹方程为

这是一个以为圆心、vt为半径的圆方程(如图所示),它代表着所有物体在任意时刻t的位置。

题:一质点在半径为的圆周上以恒定的速率运动,质点由位置运动到位置,和所对的圆心角为。

(1)试证位置和之间的平均加速度为;

(2)当分别等于、、和时,平均加速度各为多少?并对结果加以讨论。

题解:(1)由图可看到,故

而

所以

(2)将分别代入上式,得

上述结果表明:当时,匀速率圆周运动的平均加速度趋于一极限值,该值即为法向加速度。

题:一质点沿半径为的圆周按规律运动,、都是常量。(1)求时刻的总加速度;(2)为何值时总加速度在数值上等于(3)当加速度达到时,质点已沿圆周运行了多少圈?

题解:(1)质点作圆周运动的速率为

其加速度的切向分量和法向分量分别为

故加速度的大小为

其方向与切线之间的夹角为

(2)要使由可得

(3)从t = 0开始到t = v/b时,质点经过的路程为

因此质点运行的圈数为

题:碟盘是一张表面覆盖一层信息记录物质的塑性圆片。若碟盘可读部分的内外半径分别为和。在回放时,碟盘被以恒定的线速度由内向外沿螺旋扫描线(阿基米德螺线)进行扫描。(1)若开始时读写碟盘的角速度为,则读完时的角速度为多少(2)若螺旋线的间距为,求扫描线的总长度和回放时间。

题分析:阿基米德螺线是一等速的螺旋线,在极坐标下,它的参数方程可表示为,式中r为极径,r为初始极径,为极角,a为常量。它的图线是等间距的,当间距为d时,常量a = d/2。因此,扫描线的总长度可通过积分得到。