优质解答
| 1 m/s 3 m/s 2  |
| 解法一(用位移公式和速度公式法求解):如图所示,设质点从位置A时开始计时,s 1 ="8" m,t="2" s,s 2 ="20" m 根据位移公式:s 1 =v A ·t+  at 2 ①s 2 =v B ·t+ at 2 ②v B =v A +at ③ 由①②③得:v A ="1" m/s,a="3" m/s 2 . 解法二(用平均速度求解):  AB = m/s="4" m/s BC = m/s="10" m/s AC = m/s="7" m/s又因为 = , AB = BC = , AC = 解得:v A ="1" m/s,v B ="7" m/s a=  m/s 2 ="3" m/s 2 .解法三:物体做匀变速直线运动,满足Δs=aT 2 所以a=  m/s 2 ="3" m/s 2 由s 1 =v A t+ at 2 得v A ="1" m/s. |
| 1 m/s 3 m/s 2 |
| 解法一(用位移公式和速度公式法求解):如图所示,设质点从位置A时开始计时,s 1 ="8" m,t="2" s,s 2 ="20" m 根据位移公式:s 1 =v A ·t+ at 2 ①s 2 =v B ·t+ at 2 ②v B =v A +at ③ 由①②③得:v A ="1" m/s,a="3" m/s 2 . 解法二(用平均速度求解): AB = m/s="4" m/s BC = m/s="10" m/s AC = m/s="7" m/s又因为 = , AB = BC = , AC = 解得:v A ="1" m/s,v B ="7" m/s a= m/s 2 ="3" m/s 2 .解法三:物体做匀变速直线运动,满足Δs=aT 2 所以a= m/s 2 ="3" m/s 2 由s 1 =v A t+ at 2 得v A ="1" m/s. |