1、运动到静止的时间t
速度为+,a=-6
Vt-V0=at
0-60=-6*t
t=10s
着陆后12s内滑行的距离就是10s的距离
s=V0t+0.5att=60*10-0.5*6*10*10=300m
2、前六秒的距离
s=V0t+0.5att=60*6-0.5*6*6*6=252m
最后4秒内滑行的距离=300-252=48m
飞机着陆后做匀速直线运动,速度逐渐减小,加速度的大小是6m/s 2,初速度是60m/s,求
飞机着陆后做匀速直线运动,速度逐渐减小,加速度的大小是6m/s 2,初速度是60m/s,求
1）着陆后12s内滑行的距离 2）最后4秒内滑行的距离
1）着陆后12s内滑行的距离 2）最后4秒内滑行的距离
物理人气：919 ℃时间：2020-04-09 10:44:58
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