ÌâÄ¿ÄÚÈÝ

ΪÁ˲ⶨÆøµæµ¼¹ìÉÏ»¬¿éµÄ¼ÓËٶȣ¬»¬¿éÉÏ°²×°ÁË¿í¶ÈΪ5mmµÄÕÚ¹â°å£¬ÈçͼËùʾ£®»¬¿éÔÚÇ£ÒýÁ¦×÷ÓÃÏÂÏȺóͨ¹ýÁ½¸ö¹âµçÃÅ£¬ÅäÌ×µÄÊý×ÖºÁÃë¼Æ¼Ç¼ÁËÕÚ¹â°åͨ¹ýµÚÒ»¸ö¹âµçÃŵÄʱ¼äΪ¡÷t1=20ms£¨1ms=10-3s£©£¬Í¨¹ýµÚ¶þ¸ö¹âµçÃŵÄʱ¼äΪ¡÷t2=5ms£¬ÕÚ¹â°å´Ó¿ªÊ¼ÕÚסµÚÒ»¸ö¹âµçÃŵ½ÕÚסµÚ¶þ¸ö¹âµçÃŵÄʱ¼ä¼ä¸ôΪ¡÷t=2.5s£¬Çó£º
£¨1£©»¬¿é¾­¹ýÁ½¸ö¹âµçÃŵÄËٶȴóС·Ö±ðÊǶà´ó£¿
£¨2£©»¬¿éµÄ¼ÓËٶȴóС£¿
¾«Ó¢¼Ò½ÌÍø
£¨1£©Í¨¹ýµÚÒ»¸ö¹âµçÃŵÄËÙ¶È
v1=
d
¡÷t1
=
5
20
=0.25m/s

ͨ¹ýµÚ¶þ¸ö¹âµçÃŵÄËÙ¶È
v2=
d
¡÷t2
=
5
5
=1m/s

£¨2£©Ôò¼ÓËÙ¶ÈΪ£º
a=
v2-v1
¡÷t
=
1-0.25
2.5
=0.3m/s2

´ð£º£¨1£©»¬¿é¾­¹ýÁ½¸ö¹âµçÃŵÄËٶȴóС·Ö±ðÊÇ0.25m/s£¬1m/s£®
£¨2£©»¬¿éµÄ¼ÓËٶȴóСÊÇ0.3m/s2£®
Á·Ï°²áϵÁдð°¸
Ïà¹ØÌâÄ¿

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

¾«Ó¢¼Ò½ÌÍø