加速起动位移s1=

1

2

a1t12=

1

2

×3×62m=54m

匀速行驶速度v₂=v₁+a₁t₁=（0+3×6）m/s=18m/s

匀速行驶位移s₂=v₂t₂=18×5×60m=5400m

甲乙两站的距离s=s₁+s₂+s₃=（54+5400+27）m=5481m

由

v

2t

-

v

20

=2as刹车时加速度a2=

v 23 - v 22

2s3

=

0-182

2×27

m/s2=-6m/s2

刹车时间t3=

v3-v2

a2

=

0-18

-6

s=3s

从甲站到乙站所用时间t=t₁+t₂+t₃=（6+5×60+3）s=309s

答：甲乙两站的距离为5481m；汽车从甲站到乙站所用时间309s．