问题描述:
一个做匀加速直线运动的物体,初速度为2m/s ,它第三秒内通过的位移为4.5m,则它的加速度为多少?最佳答案:
设加速度为a,三秒内通过的位移是S1=2*3+1/2*a*3^2两秒内通过的位移是S2=2*2+1/2*a*2^2S1-S2=4.5解得a=1米每秒
来源:网络整理 免责声明:本文仅限学习分享,如产生版权问题,请联系我们及时删除。
相关文章:
下面是家里老人外出散步,女儿嘱咐老人的话,其中语言表达04-30
口语交际与语言表达。阅读下面一则笑话,然后从语文的角04-30
呼读音有哪些04-30
Water can absorb and give off a lot of heat withou04-30
Ten years ago the population of our village was __04-30
This magazine is very _____ with young people, who04-30
John was dismissed last week because of his _____ 04-30
Studies show that people are more _____to suffer f04-30