jeffsheh
幼苗
共回答了15个问题采纳率:100% 举报
(1)根据△x=aT 2 ,可得 a=
△x
T 2 =
BC-AB
T 2 =
30-18
2 2 m/ s 2 =3m/ s 2
(2)根据 v 中时 =
.
v =
x
t ,可得 v B =
AC
2T =
18+30
2×2 m/s=12m/s
根据v t =v 0 +at,可得v C =v B +at=12+3×2m/s=18m/s
根据v t =v 0 +at,可得v B =v A +at⇒v A =v B -at=12-3×2m/s=6m/s
小球通过A、B、C三点时的速度分别是v A =6m/s,v B =12m/s,v C =18m/s
(3)根据v t 2 -v 0 2 =2ax 则v A 2 -0=2ax
⇒(6m/s) 2 -0=2×3m/s 2 ×x OA
x OA =6m.
故OA两点的距离为6m.
1年前
8