推荐回答（6个）
两个都错，答案也有错的时候！这个题本身有问题。
设B运动停止的时候，t1=V/a=20S。运动位移LB=V²/2a=40m
在这段时间里A运动的位移LA=V0t+at²/2=80m，此时的速度V1=V0+at=6m/s
此时A距B还有：S=LB+L-LA=39.6m。说明B停止运动后，A在慢慢的跑上去。在这里说下这题出的有问题，题中说恰好追上，什么叫恰好追上？恰好追上的意思是A加速，B减速，当AB速度相同时或LA-LB=L时，这两种情况下的时候，叫恰好。显然题目不可能是这两结果。所以题目有问题。
言归正题：在剩余的路程中，S=V1+at1²/2，代入S和V1得t1=6S
所以t=20+6=26s。
S=v0t+0.5at²
得到SA=2t+0.5×0.2t²=2t+0.1t² m
SB=4t-0.5×0.2t²=4t-0.1t²
追上说明SA-SB=79.6
即0.2t²-2t=79.6
化简得到
t²-10t-398=0
解得t≈25.6s
解：设经过t时间之后A车追上B车。假设A车在追上B车前，B车的速度还未降为0.
A车经过的路程：SA=VA*t+1/2*aA*t^2
B车经过的路程：SB=VB*t+1/2*aB*t^2
SA=SB+L
VA*t+1/2*aA*t^2=VB*t+1/2*aB*t^2+L
由已知得：aA=0.2m/s^2,aB=-0.2m/s^2,VA=2m/s,VB=4m/s,L=79.6m
代入方程：
2t+1/2*0.2*t^2=4t-1/2*0.2*t^2+79.6
0.2t^2-2t=79.6
t=根号(423)+5约等于25.57s
此时的VB=4-0.2*25.57<0，所以假设不成立，即在A追上B之前，B已经停止运动
设经过t1时间B停止运动：VB=4-0.2*t1=0，得：t1=20s
此时B经过的距离为SB=1/2*0.2*t1^2=0.1*400=40(m)
设经过T时间，A、B相遇
VA*T+1/2*aA*T^2=SB+L
2T+0.1T^2=40+79.6
T^2+20T-1196=0
(T+10)^2=1296
T=36-10=26s
即经过26s后相遇
可以画图，可知追上时A的位移xA－xB＝L。那么可以带入，VAt＋½at²＝VBt－½at²＋L，解得，t。不过是不是数字有点问题…我怎么算不开…t约为25.6?
B车的速度为0时用时20秒，此时B车行使路程为40m；而20秒内，A车的行使距离小于40+79.6，所以B车停驶后A车仍然继续行使才能赶上B车。所以赶上B车的用时实际就是A车行使了（40+79.6）的时间
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