Question

一列長(zhǎng)為L(zhǎng)=300m的隊(duì)伍，行進(jìn)的速度為v=2m/s，通訊員以速度u=4m/s跑步從隊(duì)尾趕到隊(duì)首，又立即從隊(duì)首返回隊(duì)尾，則在此過(guò)程中隊(duì)伍前進(jìn)的路程是多少米？

Answer 1

分析：要求通訊員從離開(kāi)到重回排尾一共所需時(shí)間需要分別求出通訊員從隊(duì)伍排頭跑步到隊(duì)伍排尾所需的時(shí)間和通訊員從隊(duì)伍排尾跑步到隊(duì)伍排頭所需的時(shí)間，故需要求出通訊員從隊(duì)伍排尾跑步到隊(duì)伍排頭的過(guò)程中通訊員相對(duì)于隊(duì)伍的速度大小和通訊員從隊(duì)伍排頭跑步到隊(duì)伍排尾的過(guò)程中通訊員相對(duì)于隊(duì)伍的速度大小，而當(dāng)通訊員與隊(duì)伍的運(yùn)動(dòng)方向相同時(shí)相對(duì)速度是兩者速度的差，當(dāng)通訊員與隊(duì)伍的運(yùn)動(dòng)方向相反時(shí)相對(duì)速度是兩者速度的和．

解答：解：在通訊員從隊(duì)伍排尾跑步到隊(duì)伍排頭的過(guò)程中通訊員相對(duì)于隊(duì)伍的速度大小為V=u-v=2/s，故通訊員從隊(duì)伍排尾跑步到隊(duì)伍排頭所需的時(shí)間為t₁=

L

V

300

2

=150s，
在通訊員從隊(duì)伍排頭跑步到隊(duì)伍排尾的過(guò)程中通訊員相對(duì)于隊(duì)伍的速度大小為V′=u+v=6m/s，故通訊員從隊(duì)伍排頭跑步到隊(duì)伍排尾所需的時(shí)間為t₂=

L

V′

300

6

=50s，
故通訊員從離開(kāi)到重回排尾一共所需時(shí)間t=t₁+t₂=200s．
則在此過(guò)程中隊(duì)伍前進(jìn)的路程是S=vt=2×200=400m．
答：在此過(guò)程中隊(duì)伍前進(jìn)的路程是400m．

點(diǎn)評(píng)：解決此類題目主要是要理清相對(duì)速度關(guān)系和相對(duì)位移關(guān)系．無(wú)論是從排頭跑步到隊(duì)伍排尾還是從從隊(duì)伍排尾跑步到隊(duì)伍排頭，相對(duì)位移都是隊(duì)伍的長(zhǎng)度，當(dāng)通訊員與隊(duì)伍的運(yùn)動(dòng)方向相同時(shí)相對(duì)速度是兩者速度的差，當(dāng)通訊員與隊(duì)伍的運(yùn)動(dòng)方向相反時(shí)相對(duì)速度是兩者速度的和．