Question

如圖所示，一輛長(zhǎng)為13m的客車沿平直公路以10m/s的速度勻速向西行駛，一輛長(zhǎng)為18m的貨車由靜止開(kāi)始以2.0m/s²的加速度由西向東勻加速行駛，已知貨車剛啟動(dòng)時(shí)兩車車頭相距200m，求兩車錯(cuò)車(即車頭相遇到車尾剛好分開(kāi))所用的時(shí)間．

Answer 1

答案：1s

解析：設(shè)貨車啟動(dòng)后經(jīng)過(guò)時(shí)間t₁兩車開(kāi)始錯(cuò)車，此時(shí)貨車和客車的位移分別為s₁、s₂，則有：

s₁＋s₂＝200m                                                                ①

由運(yùn)動(dòng)學(xué)公式得：s₁＝at                                             ②

s₂＝vt₁                                                                                                                                                    ③

聯(lián)立①、②、③式并代入數(shù)據(jù)解得：t₁＝10s

設(shè)貨車從開(kāi)始啟動(dòng)到兩車錯(cuò)車結(jié)束所用時(shí)間為t₂，此時(shí)貨車和客車的位移分別為s₃、s₄，則有：s₃＋s₄＝231m                                                           ④

由運(yùn)動(dòng)學(xué)公式得：s₃＝at                                             ⑤

s₄＝vt₂                                                                                                                                                    ⑥

聯(lián)立④、⑤、⑥式并代入數(shù)據(jù)解得：t₂＝11s

故兩車錯(cuò)車所用的時(shí)間為：Δt＝t₂－t₁＝1s.