Question

一列隊(duì)伍長(zhǎng)l=90m，行進(jìn)速度v₁=1.5m/s，為了傳達(dá)一個(gè)命令，通訊員從隊(duì)伍排尾跑步到隊(duì)伍排頭，其速度v₂=3m/s，然后又立即以與隊(duì)伍行進(jìn)速度大小相同的速度返回排尾，則通訊員從離開(kāi)到重回排尾一共所需時(shí)間t=    s．

Answer 1

【答案】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=v₂-v₁=1.5m/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===60s，
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′=v₃+v₁=3m/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===30s，
故通訊員從離開(kāi)到重回排尾一共所需時(shí)間t=t₁+t₂=90s．
故本題的答案為90s．
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．