问题详情：
在刚刚结束的校运动会上，一年九班获男子4×100m接力第一名。设该队*乙两同学在直跑道上进行交接棒，他们在奔跑时有相同的最大速度，最大速度为vm=8m/s，乙从静止开始全力奔跑需跑出s0=16m才能达到最大速度，这一过程可看做是匀加速直线运动，现在*持棒以最大速度向乙奔来，乙在接力区伺机全力奔出．
（1）求乙加速过程的加速度a的大小
（2）若要求乙接棒时奔跑的速度达到v=6m/s，求乙在接力区须奔出的距离s
（3）如果乙是傻傻站着接棒，接到棒后才从静止开始全力奔跑，求这样会比乙恰好达到最大速度时接棒浪费的时间Δt
【回答】
(1)乙从静止加速到最大速度vm2=2as0························································· （2分）
得a= 2m/s2·························································································· （1分）
(2)乙从静止加速到6m/s
v2=2as·································································································· （2分）
得s=9 m······························································································· （1分）
(3若乙以最大速度跑出16m的时间为
t1＝2 s·································································································· （2分）
若乙以匀加速跑出16m
x＝
at22得t2=4s················································································· （2分）
Δt= t2- t1＝2 s······················································································· （2分）
知识点：匀变速直线运动的研究单元测试
题型：计算题
