ÌâÄ¿ÄÚÈÝ

¸ßËÙ¹«Â·¸øÈËÃÇ´øÀ´·½±ã£¬µ«ÊÇÒòΪÔÚ¸ßËÙ¹«Â·ÉÏÐÐÊ»µÄ³µÁ¾Ëٶȴó£¬ÎíÌìÍùÍù³öÏÖÊ®¼¸Á¾³µ×·Î²Á¬ÐøÏàײµÄʹʣ®Èç¹ûijÌìÓб¡Îí£¬½Î³µÔÚij¸ßËÙ¹«Â·µÄÐÐÊ»ËÙÂÊΪ72km/h£¬Éè˾»úµÄ·´Ó¦Ê±¼äΪ0.5s£®
£¨1£©½Î³µÔÚ·´Ó¦Ê±¼ätÄÚÇ°½øµÄ¾àÀ룮
£¨2£©Èô¼ÝʻԱͻȻ·¢ÏÖÕýÇ°·½42mÓÐÒ»Á¾×ÔÐгµÒÔ4m/sµÄËÙ¶È×öͬ·½ÏòµÄÔÈËÙÖ±ÏßÔ˶¯£¬½Î³µÒÔ4m/s2¼ÓËÙ¶ÈÖƶ¯£¬Äܱ£Ö¤°²È«Âð£¿
£¨1£©½Î³µÔÚ·´Ó¦Ê±¼äÄÚ×öÔÈËÙÖ±ÏßÔ˶¯£¬ÔÚ·´Ó¦Ê±¼äÄÚµÄλÒÆx=v0¡÷t=10m
£¨2£©Éèɲ³µºó¾­¹ýtsºóÁ½³µµÄËÙ¶ÈÏàµÈ       t=
v×Ô-
v ½Î
-a
=4s           
½Î³µtsÄÚµÄλÒÆ           x½Î=
v×Ô 2-v½Î 2
-2a½Î 
=48m       
½Î³µµÄ×ÜλÒÆ              X=x+x½Î=58m              
×ÔÐгµµÄ×ÜλÒÆ          x×Ô=v×Ô£¨t+0.5£©=18m        
ËùÒÔ¡÷x=58-18=40m£¼42m 
¹Ê¿ÉÖª ½Î³µÒÔ4m/s2¼ÓËÙ¶ÈÖƶ¯£¬°²È«£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

Î¥·¨ºÍ²»Á¼ÐÅÏ¢¾Ù±¨µç»°£º027-86699610 ¾Ù±¨ÓÊÏ䣺58377363@163.com

¾«Ó¢¼Ò½ÌÍø