Question

火车以v=54km/h,的速度沿平直轨道运动,因故需在某小路暂停t=1min,进站刹车时的加速度大小a1=0.3m/s^

Answer 1

100s.

速度v=54km/h=15m/s,刹车用时为 t1=15/0.3=50s,前进s1=0.5*0.3*50^2=375m；

加速用时t2= 15/0.5=30s,前进s2=0.5*0.5*30^2=225m.

即该过程共前进s=375+225=600m,用时t=50+60+30=140s,

若不耽误600m只需用时600/15=40s.

所以暂停耽误时间为140-40=100s