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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)
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