由题意v0=20m/s,v=10m/s
由A到B匀减速,B到C匀速,C到D匀加速,
则tAB=
=v?v0
a1
=20s,10m/s?20m/s ?0.5m/s2
tBC=
=sBC v
=110s,1100m 10m/s
tCD=
=
v0?v a2
=20s,20m/s?10m/s 0.5m/s2
实际用总时间:t总=tAB+tBC+tCD=20s+110s+20s=150s,
因sAB=
=
v2?
v
2a1
=300m,(10m/s)2?(20m/s)2
2×(?0.5m/s2)
则sCD=
=
?v2
v
2a2
=300m,(20m/s)2?(10m/s)2
2×0.5m/s2
若一直匀速通过大桥:则t总′=
=
sAB+sBC+sCD
v0
=85s,300m+1100m+300m 20m/s
故延误的时间为:△t=t总-t总′=150s-85s=65s.
答:列车因为过桥而延误的时间是65s.