Question

7．小汽車(chē)正前方s處有一輛正在以速度v₀行駛的載重卡車(chē)，此時(shí)小汽車(chē)因有急事需追趕載重卡車(chē)，于是通知載重卡車(chē)司機(jī)，卡車(chē)司機(jī)立即以加速度大小為a₁做勻減速運(yùn)動(dòng)，同時(shí)小汽車(chē)從靜止開(kāi)始以加速度大小為a₂做勻加速運(yùn)動(dòng)，試求汽車(chē)需多長(zhǎng)時(shí)間可追上載重卡車(chē)．

Answer 1

分析 在汽車(chē)的追擊和相遇的問(wèn)題中，當(dāng)有汽車(chē)在做減速運(yùn)動(dòng)時(shí)，一定要注意汽車(chē)的運(yùn)動(dòng)時(shí)間的判斷，貨車(chē)的運(yùn)動(dòng)的時(shí)間和小汽車(chē)的運(yùn)動(dòng)的時(shí)間很可能是不一致的，所以要分情況來(lái)討論．

解答 解：分兩種情況討論：
由于載重卡車(chē)經(jīng)$\frac{{v}_{0}^{\;}}{{a}_{1}^{\;}}$時(shí)間停下，在這段時(shí)間內(nèi)小汽車(chē)的位移為$\frac{1}{2}{a}_{2}^{\;}（\frac{{v}_{0}^{\;}}{{a}_{1}^{\;}}）_{\;}^{2}$
（1）若$\frac{1}{2}{a}_{2}^{\;}（\frac{{v}_{0}^{\;}}{{a}_{1}^{\;}}）_{\;}^{2}＞s+\frac{{v}_{0}^{2}}{2{a}_{1}^{\;}}$，則小汽車(chē)在載重卡車(chē)停下前追上，有
$\frac{1}{2}{a}_{2}^{\;}{t}_{\;}^{2}=s+{v}_{0}^{\;}t-\frac{1}{2}{a}_{1}^{\;}{t}_{\;}^{2}$
解得：$t=\frac{{v}_{0}^{\;}+\sqrt{{v}_{0}^{2}+2（{a}_{1}^{\;}+{a}_{2}^{\;}）s}}{{a}_{1}^{\;}+{a}_{2}^{\;}}$
（2）若$\frac{1}{2}{a}_{2}^{\;}（\frac{{v}_{0}^{\;}}{{a}_{1}^{\;}}）_{\;}^{2}＜s+\frac{{v}_{0}^{2}}{2{a}_{1}^{\;}}$，則小汽車(chē)在載重卡車(chē)停下后追上，有
$\frac{1}{2}{a}_{2}^{\;}{t}_{\;}^{2}=s+\frac{{v}_{0}^{2}}{2{a}_{1}^{\;}}$
解得：$t=\sqrt{\frac{2s}{{a}_{2}^{\;}}+\frac{{v}_{0}^{2}}{{a}_{1}^{\;}{a}_{2}^{\;}}}=\sqrt{\frac{{v}_{0}^{2}+2{a}_{1}^{\;}s}{{a}_{1}^{\;}{a}_{2}^{\;}}}$
答：Ⅰ、小汽車(chē)在載重卡車(chē)停下前追上，則$t=\frac{{v}_{0}^{\;}+\sqrt{{v}_{0}^{2}+2（{a}_{1}^{\;}+{a}_{2}^{\;}）s}}{{a}_{1}^{\;}+{a}_{2}^{\;}}$
Ⅱ、小汽車(chē)在載重卡車(chē)停下后追上，則$t=\sqrt{\frac{{v}_{0}^{2}+2{a}_{1}^{\;}s}{{a}_{1}^{\;}{a}_{2}^{\;}}}$

點(diǎn)評(píng) 對(duì)汽車(chē)的這種追擊問(wèn)題，在分析問(wèn)題的時(shí)候一定要注意汽車(chē)的剎車(chē)的時(shí)間，從而先判斷出車(chē)的運(yùn)動(dòng)的情況．