ÌâÄ¿ÄÚÈÝ

¾Æºó¼Ý³µÑÏÖØÍþв¹«ÖÚ½»Í¨°²È«¡£Èô½«¼ÝʻԱ´ÓÊÓ¾õ¸Ð֪ǰ·½Î£ÏÕµ½Æû³µ¿ªÊ¼Öƶ¯µÄʱ¼ä³ÆÎª·´Ó¦Ê±¼ä£¬½«·´Ó¦Ê±¼äºÍÖÆ¶¯Ê±¼äÄÚÆû³µÐÐÊ»µÄ×ܾàÀë³ÆÎª¸ÐÖªÖÆ¶¯¾àÀë¡£¿ÆÑ§Ñо¿·¢ÏÖ£¬·´Ó¦Ê±¼äºÍ¸ÐÖªÖÆ¶¯¾àÀëÔÚ¼ÝʻԱÒû¾ÆÇ°ºó»á·¢ÉúÃ÷ÏԱ仯¡£Ò»¼ÝʻԱÕý³£¼Ý³µºÍ¾Æºó¼Ý³µÊ±£¬¸Ð֪ǰ·½Î£ÏÕºóÆû³µÔ˶¯µÄv-tͼÏß·Ö±ðÈçͼ¼×¡¢ÒÒËùʾ¡£Çó£º

£¨1£©Õý³£¼ÝʻʱµÄ¸ÐÖªÖÆ¶¯¾àÀës£»

£¨2£©¾Æºó¼ÝʻʱµÄ¸ÐÖªÖÆ¶¯¾àÀë±ÈÕý³£¼ÝʻʱÔö¼ÓµÄ¾àÀ릤s¡£

1£©Éè¼ÝʻԱÒû¾ÆÇ°¡¢ºóµÄ·´Ó¦Ê±¼ä·Ö±ðΪt1¡¢t2£¬ÓÉͼÏ߿ɵÃ

                  t1 = 0.5 s     t2 = 1.5 s           ¢Ù (2·Ö)                           

Æû³µ¼õËÙµÄʱ¼ä   t3 £½ 4.0 s    ³õËÙ¶È v0=30m/s        ¢Ú (2·Ö)

ÓÉͼÏß¿ÉÖª       s = v0 t1 +                  ¢Û (4·Ö)

ÓÉ¢Ù¢Ú¢ÛʽµÃ  s = 75 m                              ¢Ü (2·Ö)

£¨2£© ¦¤s £½ v0(t2£­t1)                                                  ¢Ý (4·Ö)

ÓÉ¢Ù¢Ú¢Ý Ê½µÃ            ¦¤s = 30 m                             £¨2·Ö£©

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø