高一物理

2025-05-21 03:33:27
推荐回答(1个)
回答1:

54km/h=15m/s
火车匀减速驶入小站 t1=0-v0/a1=0-15m/s /-0.5m/s²=30s
x1=0-v0²/2a1=0-(15m/s)²/2*(-0.5m/s²)=225m

火车在站内停留t2=120s, x2=0
火车匀加速开出小站t3=v0-0/a2=15m/s-0 / 0.3m/s²=50s
X3=v0²-0/2a2=(15m/s)²-0 / 2*(0.3m/s²)=375m
t总=t1+t2+t3=30s+120s+50s=200s

若火车匀速驶过小站t总’=x1+x2+x3 /v0=225m+0+375m /15m/s =40s

∴列车停靠小站而延误的时间△t=t总 –t总'=200s-40s=160s