[题目]直线AB以大小为v_1的速度沿垂直于AB的方向向下移动,而直线CD以大小为v_2的速度沿垂直于CD的方向向右下方移动,两直线交角为θ,求它们的交点P的速度大小(图1)。 [分析]面对这样一道竞赛题,有很多同学会毫不思考地用矢量合成法,直接求出交点P的速度为 v_p=(v_1~2+v_2~2+2v_1v_2cos(180°-θ))~(1/2) 理由很简单,因为速度为矢量,遵守平行四边 [Subject] The straight line AB moves downward in the direction perpendicular to AB at a speed of size v_1, and the straight line CD moves right and down in a direction perpendicular to the CD at a velocity of size v_2. The intersection angle of the two lines is θ. Find them. The speed of intersection P (Figure 1). [Analysis] In the face of such a contest, many students will use the vector synthesis method without thinking, and directly find the speed of the intersection P as v_p=(v_1~2+v_2~2+2v_1v_2cos(180°-θ)) The reason for ~(1/2) is very simple, because the speed is a vector, obeys the parallel four sides