34
解得m＝．
15
14333
＋＝，15－5×＝12，即C（，12）． ··················································· 5分
315555 方法二：
BC对应的函数关系式为y＝45x－15．
AC对应的函数关系式为y＝－5x＋15． ···························································· 3分
3
BC与AC的交点C的坐标为（，12）． ························································· 5分
5
3
点C的实际意义为王老师在出发 h后，在距离车站12 km处接到甲． ··················· 6分
5
（2）预设方案2：
方法一：
设王老师把乙放下后，再经过n h与甲相遇． (45＋5)n＝45a－5a．
4
解得n＝a． ································································································ 7分
55454
由于王老师骑摩托车一共行驶 h，可得方程15－5(a＋a)＝45×[－(a＋
6565
a)]． ······························································································································ 9分
5
解得a＝． ······························································································ 10分
16 方法二：
根据题意，得点B（a，15－45a）． ································································· 7分
9
求得点C（a，15－9a）． ············································································· 8分
5 所以CD对应的函数关系式为y＝－45x＋72a＋15． ············································· 9分
55
将（，0）代入，解得a＝．······································································ 10分
61615
（说明：未经过任何说明，直接判断点E坐标为（，0），从而解得a＝只
216
得1分）
12
优化方案
（3）本题答案不唯一，以下方法供参考． y/km A 15 C 10
5 B E D 251 O 1 x/h 363
······································································································· 11分 图中折线A-B-C-D、线段AC、线段BE分别表示王老师、甲、乙离车站的路程y（km）与王老师所用时间x（h）之间的函数图象．
······································································································· 12分
（说明：没有图象或图象错误不得分．）
13