如图2-1-10所示,某同学在做“研究匀变速直线运动”实验中,由

如图2-1-10所示,某同学在做“研究匀变速直线运动”实验中,由打点计时器得到表示小车运动过程的一条清晰纸带,纸带上两相邻计数点的时间间隔为T=0.10s,其中s1=7.05cm,s2=7.68cm,s4=8.95cm,s5=9.61cm,s6=10.26cm.则A点处瞬时速度的大小________m/s,小车运动的加速度计算表达式为__________,加速度大小是________m/s2(计算结果保留两位有效数字).

图2-1-10

答案

解析:根据当物体做匀变速直线运动时,中间时刻的瞬时速度等于平均速度,A点处瞬时速度v=

=0.86m/s;加速度a==0.46m/s2.

答案:0.86     a=

     0.46

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