问题
计算题
如图所示,在一次警车A追击劫匪车B时,两车同时由静止向同一方向加速行驶,经过30 s追上.两车各自的加速度为aA=15 m/s2,aB=10 m/s2,各车最高时速分别为vA=45 m/s,vB=40 m/s,问追上时各行驶多少路程?原来相距多远?
答案
162.5 m
以A为坐标原点,Ax为正向,令L为警车追上匪劫车所走过的全程,l为匪劫车走过的全程.则两车原来的间距为ΔL=L-l
设两车加速用的时间分别为tA1、tB1,以最大速度匀速运动的时间分别为tA2、tB2,则
vA=aAtA1,
tA1="3" s,tA2="30" s-3 s="27" s
同理tB1="4" s,tB2="30" s-4 s="26" s
警车在0~tA1时段内做匀加速运动,L1=ya
,在3~30 s时段内做匀速运动;L2=vA(t-tA1),追上匪劫车的全部行程为L=L1+L2=yaA
+vA(t-tA1)= y×15×32 m+45×(30-3) m="1282.5" m
同理匪车被追上时的全部行程为
l=l1+l2=yaB
+vB(t-tB1)="1120" m
两车原来相距
ΔL="L-l=1282.5" m-1120 m="162.5" m.
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