Question

11．一質(zhì)點(diǎn)從A點(diǎn)開始做初速度為零的勻加速直線運(yùn)動，加速度的大小為a，B，C，D三點(diǎn)是質(zhì)點(diǎn)運(yùn)動路徑上三點(diǎn)，且BC=x₁，CD=x₂．質(zhì)點(diǎn)通過BC間所用時(shí)間與經(jīng)過CD間所用時(shí)間相等，則質(zhì)點(diǎn)經(jīng)過C點(diǎn)的速度為（　�。�

• A． $\frac{{x}_{1}+{x}_{2}}{2}$$\sqrt{\frac{a}{{x}_{2}-{x}_{1}}}$
• B． $\frac{{x}_{1}+{x}_{2}}{4}$$\sqrt{\frac{a}{{x}_{2}-{x}_{1}}}$
• C． $\frac{{x}_{2}-{x}_{1}}{2}$$\sqrt{\frac{a}{{x}_{2}+{x}_{1}}}$
• D． $\frac{{x}_{2}-{x}_{1}}{4}$$\sqrt{\frac{a}{{x}_{2}+{x}_{1}}}$

Answer 1

分析 根據(jù)連續(xù)相等時(shí)間內(nèi)的位移之差是一恒量求出通過BC間所用時(shí)間與經(jīng)過CD間所用時(shí)間，根據(jù)某段時(shí)間內(nèi)的平均速度等于中間時(shí)刻的瞬時(shí)速度，即可求出質(zhì)點(diǎn)經(jīng)過C點(diǎn)時(shí)的速度．

解答 解：設(shè)質(zhì)點(diǎn)從B到C所用時(shí)間為T，則${x}_{2}^{\;}-{x}_{1}^{\;}=a{T}_{\;}^{2}$，因此$T=\sqrt{\frac{{x}_{2}^{\;}-{x}_{1}^{\;}}{a}}$，則從B到D的時(shí)間為$2\sqrt{\frac{{x}_{2}^{\;}-{x}_{1}^{\;}}{a}}$，質(zhì)點(diǎn)經(jīng)過C點(diǎn)的速度${v}_{C}^{\;}=\frac{{x}_{1}^{\;}+{x}_{2}^{\;}}{2T}$=$\frac{{x}_{1}^{\;}+{x}_{2}^{\;}}{2}\sqrt{\frac{a}{{x}_{1}^{\;}-{x}_{2}^{\;}}}$，故A正確，BCD錯(cuò)誤；
故選：A

點(diǎn)評 解決本題的關(guān)鍵掌握勻變速直線運(yùn)動的公式和推論，并能靈活運(yùn)用．