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
答
由速度位移公式v2−v02=2ax得,a=
=
v2−v02
2x
m/s2=−0.25m/s2.25−100 2×150
列车停止还需的时间t0=
=0−v a
s=20s.−5 −0.25
则30s内前进的距离等于20s内前进的距离.
x=vt0+
at02=5×20−1 2
×0.25×400m=50m.故A正确,B、C、D错误.1 2
故选A.