一辆值勤的警车停在平直公路上的A点,当警员发现从他旁边以v=9m/s的速度匀速驶过的货车有违章行为时,决...
2024-07-02 20:31:57 学考宝 作者:佚名
题目
一辆值勤的警车停在平直公路上的A点,当警员发现从他旁边以v=9m/s的速度匀速驶过的货车有违章行为时,决定前去追赶。警车启动时货车已运动到B点,A、B两点相距x0=48 m,警车从A点由静止开始向右做匀加速运动,到达B点后开始做匀速运动.已知警车从开始运动到追上货车所用的时间t=32 s,求:
(1) 警车加速运动过程所用的时间t1和加速度a的大小
(2) 警车追上货车之前的最远距离x
答案和解析
(1)设警车加速过程所用的时间为t1,加速度大小为a
则x0=at·························································································· (1分)
at1(t-t1)=v t···························································································· (2分)
得t1=8 s·································································································· (1分)
a=1.5m/s2····························································································· (1分)
(2)设经t0警车与货车共速,此时警车追上货车之前最远
t0==6s································································································· (1分)
t0时间内货车运动的距离x1=v t0=54m······················································ (1分)
t0时间内警车运动的距离x2=t0=27 m···················································· (1分)
此时相距x= x0+ x1- x2=75 m······································································ (2分)