dv 2i-4j d t d t a=4ms2方向沿轴的负方向 (3)F=2+(9-2(2-47) =4t-4(19-22)=4(2-18) =8(t+3)(t-3)=0 1=0(s),t2=3(s)两矢量垂直i tj dt dr v     = = 2 − 4 j dt dv a    = = −4 -2 a = 4 ms 方向沿y轴的负方向 （3） r v  ti ( t )j ( i tj)       2 19 2 2 4 2  = + −  − 4 4 (19 2 ) 4 (2 18) 2 2 = t − t − t = t t − = 8t(t +3)(t −3) = 0 0 ( ) , 3 ( ) 1 2 t = s t = s 两矢量垂直