ÌâÄ¿ÄÚÈÝ

2£®ÈçͼËùʾ£¬Ò»Áмòгºá²¨ÑØxÖáÕý·½Ïò´«²¥£¬ÔÚt1=0ʱ¿Ì²¨´«²¥µ½x=2.0m´¦µÄÖʵãC£®ÔÚt2=0.1sʱ¿Ì£¬x=1.0m´¦µÄÖʵãBµÚÒ»´ÎÔ˶¯µ½¸º·½Ïò×î´óλÒÆ´¦£¬Ôò£¨¡¡¡¡£©
A£®ÖʵãC¿ªÊ¼Õñ¶¯Ê±µÄÔ˶¯·½ÏòÑØyÖḺ·½Ïò
B£®¸Ã¼òгºá²¨µÄ²¨ËÙµÈÓÚ5m/s
C£®ÔÚt1¡«t2ʱ¼äÄÚ£¬ÖʵãBͨ¹ýµÄ·³ÌΪ4.0cm
D£®ÔÚt2ʱ¿Ì£¬ÕâÁв¨¸ÕºÃ´«²¥µ½Î»ÓÚx=3.0m´¦µÄÖʵãD

·ÖÎö ¼òгºá²¨ÑØxÖáÕý·½Ïò´«²¥£¬ÓÉ¡°ÉÏÏÂÆ·¨¡±ÅжÏÖʵãC¿ªÊ¼Õñ¶¯Ê±µÄÔ˶¯·½Ïò£®¸ù¾Ýt1=0ʱ¿ÌÖʵãBµÄÕñ¶¯·½ÏòºÍËüµÚÒ»´ÎÔ˶¯µ½¸º·½Ïò×î´óλÒÆ´¦µÄʱ¼äÇóµÃÖÜÆÚ£¬¶Á³ö²¨³¤£¬¼´¿ÉÇóµÃ²¨ËÙ£®¸ù¾Ýʱ¼äÓëÖÜÆڵĹØϵÇóÔÚt1¡«t2ʱ¼äÄÚÖʵãBͨ¹ýµÄ·³Ì£®ÓÉx=vtÇ󲨴«²¥µÄ¾àÀ룮

½â´ð ½â£ºA¡¢¼òгºá²¨ÑØxÖáÕý·½Ïò´«²¥£¬ÓÉ¡°ÉÏÏÂÆ·¨¡±Öª£¬ÖʵãC¿ªÊ¼Õñ¶¯Ê±µÄÔ˶¯·½ÏòÑØyÖáÕý·½Ïò£®¹ÊA´íÎó£®
B¡¢ÔÚt1=0ʱ¿Ì£¬ÖʵãBÕýÏòÏÂÔ˶¯£¬¸ù¾ÝÔÚt2=0.1sʱ¿Ì£¬ÖʵãBµÚÒ»´ÎÔ˶¯µ½¸º·½Ïò×î´óλÒÆ´¦£¬ÓÐ $\frac{T}{4}$=0.1s£¬µÃ T=0.4s
ÓÉͼ֪²¨³¤  ¦Ë=2m£¬ËùÒÔ²¨ËÙΪ v=$\frac{¦Ë}{T}$=$\frac{2}{0.4}$=5m/s£¬¹ÊBÕýÈ·£®
C¡¢ÒòΪt2-t1=0.1s=$\frac{T}{4}$£¬ËùÒÔÔÚt1¡«t2ʱ¼äÄÚ£¬ÖʵãBͨ¹ýµÄ·³ÌΪһ¸öÕñ·ù£¬Îª2.0cm£¬¹ÊC´íÎó£®
D¡¢ÔÚt2ʱ¼äÄÚ²¨´«²¥µÄ¾àÀë x=vt2=5¡Á0.1m=0.5m£¬ËùÒÔÔÚt2ʱ¿Ì£¬ÕâÁв¨¸ÕºÃ´«²¥µ½C¡¢DµÄÖе㣬¹ÊD´íÎó£®
¹ÊÑ¡£ºB

µãÆÀ ±¾ÌâÓɲ¨¶¯Í¼Ïó¶Á³ö²¨³¤¡¢Óɲ¨µÄ´«²¥·½ÏòÅжÏÖʵãµÄÕñ¶¯·½Ïò¡¢ÓÉʱ¼äÓëÖÜÆÚ¹ØϵÇóÖʵãͨ¹ýµÄ·³Ì¶¼Êdz£¹æÎÊÌ⣮¡°ÉÏÏÂÆ·¨¡±ÊDz¨µÄͼÏóÖг£Óõķ½·¨£¬ÒªÊìÁ·ÕÆÎÕ£®

Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø