Question

12£®ÎïÌåÒÔËÙ¶ÈvÔÈËÙͨ¹ýÖ±ÏßÉϵÄA¡¢BÁ½µã£¬ËùÓÃʱ¼äΪt£»ÏÖÔÚÎïÌå´ÓAµãÓɾ²Ö¹³ö·¢£¬ÏÈ×öÔȼÓËÙÖ±ÏßÔ˶¯£¨¼ÓËٶȴóСΪa₁£©£¬µ½×î´óËÙ¶Èv_mºóÁ¢¼´×öÔȼõËÙÖ±ÏßÔ˶¯£¨¼ÓËٶȴóСΪa₂£©£¬ÖÁBµãËÙ¶ÈÇ¡ºÃ¼õΪ0£¬ËùÓÃʱ¼äÈÔΪt£¬ÔòÏÂÁÐ˵·¨ÕýÈ·µÄÊÇ£¨¡¡¡¡£©

• A£® v_mÖ»ÄÜΪ2v£¬Óëa₁¡¢a₂µÄ´óСÎÞ¹Ø
• B£® v_m¿ÉΪÐí¶àÖµ£¬Óëa₁¡¢a₂µÄ´óСÓйØ
• C£® a₁¡¢a₂±ØÐëÊÇÒ»¶¨Öµ
• D£® a₁¡¢a₂±ØÐëÂú×ãa₁•a₂•t=2£¨a₁+a₂£©v

Answer 1

·ÖÎö µ±ÎïÌåÔÈËÙͨ¹ýA¡¢BÁ½µãʱ£¬x=vt£®µ±ÎïÌåÏÈÔȼÓËÙºóÔȼõËÙͨ¹ýA¡¢BÁ½µãʱ£¬¸ù¾Ýƽ¾ùËٶȹ«Ê½£¬×ÜλÒÆx=$\frac{{v}_{m}}{2}{t}_{1}+\frac{{v}_{m}}{2}{t}_{2}=\frac{{v}_{m}}{2}t$£¬´Ó¶ø¿ÉµÃÖªv_mÓëvµÄ¹Øϵ£®
ÔȼÓËÙÔ˶¯µÄʱ¼äºÍÔȼõËÙÔ˶¯µÄʱ¼äÖ®ºÍ$t=\frac{{v}_{m}}{{a}_{1}}+\frac{{v}_{m}}{{a}_{2}}$£¬ÔÙ¸ù¾Ýv_mÓëvµÄ¹ØϵµÃ³öa₁¡¢a₂ËùÂú×ãµÄÌõ¼þ£®

½â´ð ½â£ºA¡¢µ±ÎïÌåÔÈËÙͨ¹ýA¡¢BÁ½µãʱ£¬x=vt£®µ±ÎïÌåÏÈÔȼÓËÙºóÔȼõËÙͨ¹ýA¡¢BÁ½µãʱ£¬¸ù¾Ýƽ¾ùËٶȹ«Ê½£¬×ÜλÒÆx=$\frac{{v}_{m}}{2}{t}_{1}+\frac{{v}_{m}}{2}{t}_{2}=\frac{{v}_{m}}{2}t$£¬Ôò$\frac{{v}_{m}}{2}t=vt$£¬µÃv_m=2v£®Óëa₁¡¢a₂µÄ´óСÎ޹أ®¹ÊAÕýÈ·£¬B´íÎó£®
 C¡¢ÔȼÓËÙÔ˶¯µÄʱ¼äºÍÔȼõËÙÔ˶¯µÄʱ¼äÖ®ºÍ$t=\frac{{v}_{m}}{{a}_{1}}+\frac{{v}_{m}}{{a}_{2}}$£¬¶øv_m=2v£¬´úÈëµÃ$t=\frac{2v}{{a}_{1}}+\frac{2v}{{a}_{2}}$£¬ÕûÀíµÃ$\frac{{a}_{1}•{a}_{2}}{{a}_{1}+{a}_{2}}=\frac{2v}{t}$£®¼´a₁¡¢a₂±ØÐëÂú×ãa₁•a₂•t=2£¨a₁+a₂£©v£»¹ÊC´íÎó£¬DÕýÈ·£®
¹ÊÑ¡£ºAD£®

µãÆÀ ½â¾ö±¾Ìâ¹Ø¼üÕÆÎÕÔȱäËÙÖ±ÏßÔ˶¯µÄƽ¾ùËٶȵĹ«Ê½$\overline{v}=\frac{{v}_{0}+v}{2}$£¬´Ó¶øµÃ³öÏÈÔȼÓËÙºóÔȼõËÙÔ˶¯µÄλÒÆx=$\frac{{v}_{m}}{2}{t}_{1}+\frac{{v}_{m}}{2}{t}_{2}=\frac{{v}_{m}}{2}t$£¬