《2025年名校题库七年级数学上册》

9. (武侯区期末)已知$A$,$B$,$C$,$D$四点在数轴上的位置如图所示,它们对应的数分别为$a$,$b$,$c$,$d$,且$\vert b\vert=\vert c\vert = 6$,$AB= \frac{3}{2}BC= \frac{9}{5}CD$.动点$P$,$Q同时分别从点A$,$D$出发,相向而行,点$P的运动速度为每秒4$个单位长度,点$Q的运动速度为每秒2$个单位长度,线段$BC$所在部分为“交换区”,规则为:点$P从点B$进入“交换区”,其运动速度变为点$Q$原来的运动速度,点$Q从点C$进入“交换区”,其运动速度变为点$P$原来的运动速度,点$P和点Q$出“交换区”之后都分别以各自原来的运动速度继续前行,设运动的时间为$t$秒.
(1)分别求$a$,$d$的值;
(2)当$P$,$Q$两点相遇时,求$t$的值及相遇点在数轴上所对应的数;
(3)当点$P在点Q的左侧且满足BP = CQ$时,求$t$的值.

答案： 解:
(1)$\because |b| = |c| = 6$,点B、点C分别在原点的两侧,$\therefore b = - 6$,$c = 6$,$\therefore BC = 12$.又$\because AB = \frac{3}{2}BC = \frac{9}{5}CD$,$\therefore AB = 18$,$CD = 10$,$\therefore a = - 24$,$d = 16$.
(2)由题意可得,点P从点A到点B需$\frac{9}{2}$秒,点Q从点D到点C需5秒,$\therefore$点P、点Q在线段BC上相遇.$\because AB = 18$,$CD = 10$,$AD = |-24 - 16| = 40$,$\therefore 18 + 2(t - \frac{9}{2}) + 10 + 4(t - 5) = 40$,解得$t = \frac{41}{6}$,$\therefore$此时相遇点表示的数为$-24 + 18 + 2×(\frac{41}{6} - \frac{9}{2}) = -\frac{4}{3}$,故相遇点在数轴上所对应的数为$-\frac{4}{3}$.
(3)由题意可得,点P从点A到点B需$\frac{9}{2}$秒,点Q从点D到点C需5秒,两点从起点到相遇点的运动时间为$\frac{41}{6}$秒.当点P在点A,B之间,点Q在点C,D之间,即$0 ＜ t ＜ \frac{9}{2}$时,点P对应的数为$-24 + 4t$,点Q对应的数为$16 - 2t$,点B对应的数为 - 6,点C对应的数为6,$\therefore BP = 18 - 4t$,$CQ = 10 - 2t$.$\because BP = CQ$,$\therefore 18 - 4t = 10 - 2t$,$\therefore t = 4$.当点P在点B,C之间,点Q在点C,D之间,即$\frac{9}{2} ＜ t ＜ 5$时,点P对应的数为$2t - 15$,点Q对应的数为$16 - 2t$,$\therefore BP = 2t - 9$,$CQ = 10 - 2t$.$\because BP = CQ$,$\therefore 2t - 9 = 10 - 2t$,$\therefore t = \frac{19}{4}$.当点P,Q都在点B,C之间,且在相遇前,即$5 ＜ t ＜ \frac{41}{6}$时,点P对应的数为$2t - 15$,点Q对应的数为$-4t + 26$,$\therefore BP = 2t - 9$,$CQ = 4t - 20$.$\because BP = CQ$,$\therefore 2t - 9 = 4t - 20$,$\therefore t = \frac{11}{2}$.综上所述,t的值为4或$\frac{19}{4}$或$\frac{11}{2}$.