在平直公路上以12m/s的速度行驶的汽车,因故以4m/s2的加速度紧急刹车,求汽车刹车后4s内的位移.
人气:276 ℃ 时间:2020-06-12 07:15:09
解答
汽车刹车做匀减速直线运动,已知v
0=12m/s,a=-4m/s
2,t=4s
设经时间t汽车速度减为零,由v=v
0+at,得
t=
=3s,即t=3s时,汽车速度刚好减为零,
则汽车刹车后4s内的位移为s=v
0t+
at2=12×3-
×4×3
2=18m
答:汽车刹车后4s内的位移是18m.
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