试题与答案

经检测汽车A的制动性能:以标准速度20m/s在平直公路上行使时,制动后40s停下

题型:计算题

题目:

经检测汽车A的制动性能:以标准速度20m/s在平直公路上行使时,制动后40s停下来。现A在平直公路上以20m/s的速度行使发现前方180m处有一货车B以6m/s的速度同向匀速行使,司机立即制动,能否发生撞车事故?

答案:

由于S>S+180,所以能发生撞车事故

解:汽车加速度

汽车与货车速度相等时,距离最近,对汽车有:

vo-at=vt   得t=28s

vo2-vt2=2aS   得S=364m

而S=vt=168m

且S>S+180

所以能发生撞车事故

【标签】直线运动追及和相遇问题

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