(1)A、操作时,应先接通电源和释放小车.故A错误.
B、开始释放小车前,小车应靠近打点计时器.故B正确.
C、测量时应舍去纸带上的密集点,然后取计数点.故C正确.
D、若点迹比较密,可以每5个打印点设为一个计数点,若点迹较疏,不需要每5个打印点设为一个计数点.故D错误.
本题选错误的,故选:AD.
(2)某段时间内的平均速度等于中间时刻的瞬时速度,则vB=
=
x1+x2
2T
m/s=0.140m/s.2.80×10-2
0.2
vC=
=
x2+x3
2T
m/s=0.179m/s.(1.60+1.98)×10-2
0.2
vD=
=
x3+x4
2T
m/s=0.218m/s.(1.98+2.38)×10-2
0.2
vE=
=
x4+x5
2T
m/s=0.259m/s.(2.38+2.79)×10-2
0.2
vF=
=
x5+x6
2T
m/s=0.299m/s.(2.79+3.18)×10-2
0.2
(3)速度时间图线如图所示.
(4)根据v-t图线可以求出加速度,图线的斜率等于小车的加速度,a=
=△v △t
=0.4m/s2.0.3-0.1 0.5
故答案为:(1)AD
(2)0.140,0.179,0.218,0.259,0.299
(3)如图所示
(4)能.求v-t图中的斜率.0.4m/s2