(1)C点的瞬时速度等于B到D的平均速度,得:
v C =
OD-OB
2t =
(19.83-7.54)×1 0 2
2×0.1 m/s=0.615 m/s
(2)同理,可以计算出ABD等点的速度: v A =
OB
2t =0.377 m/s
v B =
OC-OA
2t =0.496 m/s
v D =
OE-OC
2t =0.733 m/s
v E =
OF-OD
2t =0.852 m/s
在v-t图中标出坐标的刻度,将ABCDE五个点在坐标系中标出,然后用一条直线将这五个点连接起来,即可.
(3)从图线上可以看到,BE两点都落在直线上,故:
a=k=
△v
△t =
0.852-0.496
0.3 =1.19 m/s 2
故答案为:(1)0.615m/s(2)答案如图
(3)1.19m/s 2(误差范围在0.02范围都算正确)