火车初速度为10m/s,关闭油门后前进150m,速度减为5m/s,再经过30s,火车前进的距离为(  ) A.50m B.37.5m C.150m D.43.5m

问题描述:

火车初速度为10m/s,关闭油门后前进150m,速度减为5m/s,再经过30s,火车前进的距离为(  )
A. 50m
B. 37.5m
C. 150m
D. 43.5m

由速度位移公式v2−v02=2ax得,a=v2−v022x=25−1002×150m/s2=−0.25m/s2.列车停止还需的时间t0=0−va=−5−0.25s=20s.则30s内前进的距离等于20s内前进的距离.x=vt0+12at02=5×20−12×0.25×400m=50m...