20 ÷ 2 = 10, we can know that the coordinates of AB are A (- 10, m) and B A(- 10, m), where m < 0.

10 ÷ 2 = 5, we can know that the coordinates of two points on the CD are C (-5, m+3) and D C(-5, m+3), and it can be seen from the figure that m+3 is less than 0, that is, m.

According to the water level width, the water level width must be 0 when it reaches the arch bridge top O (0,0). At this time, the water level width is shorter than the warning water level width10m, 20m shorter than the normal water level width and 20m shorter than the warning water level width.

20: 10=2: 1

So m: (m+3) = 2: 1.

The equation m = 2 (m+3) is obtained.

M = 2m+6。

Then m =-6

Coordinates are a (- 10, -6), b A(- 10, -6), c (-5, -3) and d C(-5, -3).

Substitute b into 10a =-6.

Get a =-0.6.

So y =-0.6x?

According to 1, the distance from the warning line to the top of the arch bridge is 10 m, and now the speed of water level rising is 0.2 m/h.

So it takes 10 ÷ 0.2 = 50 (hours) for the water level to reach the top of the arch bridge.