图是用打点计时器打出一系列点的纸带．纸带固定在一个做直线运动

问题问答题

图是用打点计时器打出一系列点的纸带．纸带固定在一个做直线运动的小车后面．A、B、C、D、E为选好的点，这些点间有四个点未画出，图中相邻两点间的距离已标出．则由图可判断小车作______运动；在纸带各点中速度最大值为______m/s，最小值为______m/s．

如果当时电网中交变电流的频率是f=49Hz，则打出的对应点与电网中交变电流的频率正常时（频率是f=50Hz）比较间距将变______；（填“大”或“小”）如果做实验的同学并不知道，那么由此测量计算出的加速度值比实际值偏______．（填“大”或“小”）

答案

由于相邻时间间隔内的位移差△x=4mm为常数，因此小车做匀加速直线运动；

相邻计数点间的时间间隔为0.1s，

根据匀变速直线运动的推论公式△x=aT²

得：△x=0.004m=a（0.1s）²，

解得：a=0.4m/s²．

根据时间中点的速度等于该过程中的平均速度，可知B点的速度为：

vB=

xAC

(24.0+28.0)mm

2×0.1s

=0.26m/s

开始A点的速度最小，E点的速度最大，根据v=v₀+at，得：

A点的速度为：v_A=v_B-aT=0.26-0.4×0.1=0.22m/s

E点的速度为：v_E=v_B+a×3T=0.26+0.4×0.3=0.38m/s

如果在某次实验中，交流电的频率为49Hz，那么实际周期大于0.02s，则打出的对应点与电网中交变电流的频率正常时（频率是f=50Hz）比较间距将变大，

根据运动学公式△x=at²得，所以测量的加速度值与真实的加速度值相比是偏大的．

故答案为：匀加速直线，0.22，0.38，大，大．

完形填空
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