一道关于物理初二速度问题的题!急需!

2024年11月18日 17:44
有3个网友回答
网友(1):

过程是这样的:小华听到火车鸣笛的时刻,火车头已经进入山洞口一段距离,那么设没进入山洞口的火车长度为 L ,设山洞长为 x ,设火车速度为 V ,那么:
根据这个条件:"小华恰好座在车尾,从听到笛声到车尾出洞,小华共数出85次车身振动,所用的时间是一分45秒"
也就是说:火车用了一分45秒走过了山洞的长度+小华听到鸣笛声时火车没进入山洞那段。也就是L + x = (85-1)*12.5
其中85-1是因为第一次振的时候才开始计数,到第二次振一共走过12.5m
以此类推一共走了84*12.5=1050m
走过这1050m需要1分45秒,所以火车速度 V = 1050m/105s = 10m/s

下面求山洞长:
1.小华听到笛声所用时间 s = 175 m/(340 + 10)m/s = 0.5s
解释:当小华听到笛声的时候,经过了火车长度175m,速度为声速还应该加上火车本身的速度。
2.从火车头算,0.5秒,火车走过了10m/s * 0.5s = 5m
3.L = 175-5 = 170m
4.用这个公式 L + x = (85-1)*12.5 =1050
推导出 x = 880 m
所以山洞为880m

这么整的数一看就是给中学生设计的,另外,你画个图就明白得很快了。

网友(2):

洞的长度+火车的长度=火车的速度×时间
即洞的长度+火车的长度=12.5m×(84-1)=1037.5m
洞的长度=1037.5-175=862.5m
火车的速度=1037.5/(105+37/68)=9.83m/s
这是我在物理学那知道的答案!

网友(3):

好像是这样:山洞的长度是862.45,当时火车的速度是9.73米/秒
我爸是物理老师,他是这么做的

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